Chapter 7: TECHNIQUES OF INTEGRATION. Work and Answers

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7.1 Integration by Parts 1. Let  = ,  = 2  ⇒  = ,  = 1 22. Then by Equation 2,  2  = 1 22 −  1 22  = 1 2... 2 − 1 42 + . 2. Let  = ln,  = √  ⇒  = 1  ,  = 2 332. Then by Equation 2,  √ln  = 2 332 ln −  2 332 · 1  = 2 332 ln −  2 312  = 2 332 ln − 4 932 + . Note: A mnemonic device which is helpful for selecting  when using integration by parts is the LIATE principle of precedence for : Logarithmic Inverse trigonometric Algebraic Trigonometric Exponential If the integrand has several factors, then we try to choose among them a  which appears as high as possible on the list. For example, in  2  the integrand is 2, which is the product of an algebraic function () and an exponential function (2). Since Algebraic appears before Exponential, we choose  = . Sometimes the integration turns out to be similar regardless of the selection of  and , but it is advisable to refer to LIATE when in doubt. 3. Let  = ,  = cos 5  ⇒  = ,  = 1 5 sin5. Then by Equation 2,  cos5  = 1 5sin5 −  1 5 sin5  = 1 5sin5 + 25 1 cos5 + . 4. Let  = ,  = 02 ⇒  = ,  = 01202. Then by Equation 2,  02 = 502 −  502 = 502 − 2502 + . 5. Let  = ,  = −3 ⇒  = ,  = − 1 3−3. Then by Equation 2,  −3 = − 1 3−3 −  − 1 3−3 = − 1 3−3 + 1 3  −3 = − 1 3−3 − 1 9−3 + . 6. Let  =  − 1,  = sin  ⇒  = ,  = − 1  cos . Then by Equation 2,  ( − 1) sin  = −1 ( − 1) cos −  −1 cos  = −1 ( − 1) cos + 1  cos  = − 1  ( − 1) cos + 1 2 sin +  7. First let  = 2 + 2,  = cos   ⇒  = (2 + 2),  = sin. Then by Equation 2,  =  (2 + 2)cos  = (2 + 2)sin −  (2 + 2) sin . Next let  = 2 + 2,  = sin  ⇒  = 2,  = −cos, so  (2 + 2) sin  = −(2 + 2) cos  −  −2cos   = −(2 + 2) cos + 2 sin. Thus,  = (2 + 2)sin + (2 + 2) cos − 2sin + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 12 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 8. First let  = 2,  = sin  ⇒  = 2 ,  = − 1 cos. Then by Equation 2,  =  2 sin  = − 1 2 cos −  − 2 cos  . Next let  = ,  = cos  ⇒  = ,  = 1  sin, so  cos  = 1 sin −  1 sin  = 1 sin + 12 cos . Thus,  = − 1  2 cos + 2 1 sin + 12 cos +  = −1 2 cos  + 22 sin + 23 cos  + . 9. Let  = cos−1 ,  =  ⇒  = √1−−12 ,  = . Then by Equation 2,  cos−1   = cos−1  −  √1−−2  = cos−1  −  √1 1 2    = 1 = −−2  2,  = cos−1  − 1 2 · 212 +  = cos−1  − √1 − 2 +  10. Let  = ln√,  =  ⇒  = √1 · 2√1  = 21 ,  = . Then by Equation 2,  ln√  = ln√ −   · 21  = ln√ −  1 2  = ln√ − 12 + . Note: We could start by using ln√ = 1 2 ln. 11. Let  = ln,  = 4  ⇒  = 1  ,  = 1 5 5. Then by Equation 2,  4 ln  = 155 ln −  155 · 1  = 155 ln −  1 54  = 155 ln − 25 1 5 + . 12. Let  = tan−1 2,  =  ⇒  = 2 1 + 42 ,  = . Then by Equation 2,  tan−1 2  =  tan−1 2 −  1 + 4 2 2  =  tan−1 2 −  1 14    = 1 + 4 = 8 2, =  tan−1 2 − 1 4 ln|| +  =  tan−1 2 − 1 4 ln(1 + 42) +  13. Let  = ,  = csc2   ⇒  = ,  = −cot. Then by Equation 2,  csc2   = −cot −  −cot  = −cot +  cos sin  = −cot +  1    = sin = cos  ,  = −cot + ln|| +  = −cot + ln|sin| +  14. Let  = ,  = cosh  ⇒  = ,  = 1  sinh. Then by Equation 2,  cosh  = 1sinh −  1 sinh  = 1sinh − 12 cosh + . 15. First let  = (ln)2,  =  ⇒  = 2ln · 1 ,  = . Then by Equation 2,  =  (ln)2  = (ln)2 − 2 ln · 1  = (ln)2 − 2 ln . Next let  = ln,  =  ⇒  = 1 ,  =  to get  ln  = ln −   · (1) = ln −   = ln −  + 1. Thus,  = (ln)2 − 2(ln −  + 1) = (ln)2 − 2ln + 2 + , where  = −21. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.1 INTEGRATION BY PARTS ¤ 3 16.  10   =   10− . Let  = ,  = 10−  ⇒  = ,  = −ln 10 10− . Then by Equation 2,   10−  = −ln 10  10− −  −ln 10 10−  = 10−ln 10  − (ln 10)(ln 10) 10− +  = −10 ln 10 − 10(ln 10) 1 2 + . 17. First let  = sin3,  = 2  ⇒  = 3cos3 ,  = 1 2 2. Then  =  2 sin3  = 1 2 2 sin3 − 3 2  2 cos3 . Next let  = cos3,  = 2  ⇒  = −3sin3 ,  = 1 2 2 to get  2 cos3  = 1 2 2 cos 3 + 3 2  2 sin3 . Substituting in the previous formula gives  = 1 2 2 sin3 − 3 4 2 cos3 − 9 4  2 sin3  = 1 2 2 sin3 − 3 4 2 cos 3 − 9 4  ⇒ 13 4  = 1 2 2 sin3 − 3 4 2 cos3 + 1. Hence,  = 13 1 2(2sin3 − 3cos3) + , where  = 13 4 1. 18. First let  = −,  = cos2  ⇒  = −− ,  = 1 2 sin2. Then  =  − cos2  = 1 2 − sin2 −  1 2 sin2 −−  = 1 2 − sin2 + 1 2  − sin2 . Next let  = −,  = sin2  ⇒  = −− ,  = − 1 2 cos2, so  − sin2  = − 1 2 − cos2 −  − 1 2cos2−−  = − 1 2 − cos2 − 1 2  − cos2 . So  = 1 2 − sin2 + 1 2 − 1 2 − cos2 − 1 2  = 1 2 − sin2 − 1 4 − cos 2 − 1 4  ⇒ 5 4  = 1 2 − sin2 − 1 4 − cos2 +1 ⇒  = 4 5 1 2 − sin2 − 1 4 − cos 2 + 1 = 2 5 − sin2 − 1 5 − cos2 +. 19. First let  = 3,  =  ⇒  = 32,  = . Then 1 =  3 = 3 − 3 2. Next let 1 = 2, 1 =  ⇒ 1 = 2 , 1 = . Then 2 = 2 − 2 . Finally, let 2 = , 2 =  ⇒ 2 = , 2 = . Then   =  −   =  −  + 1. Substituting in the expression for 2, we get 2 = 2 − 2( −  + 1) = 2 − 2 + 2 − 21. Substituting the last expression for 2 into 1 gives 1 = 3 − 3(2 − 2 + 2 − 21) = 3 − 32 + 6 − 6 + , where  = 61. 20.  tan2   =  (sec2  − 1) =  sec2   −   . Let  = ,  = sec2   ⇒  = ,  = tan. Then by Equation 2,  sec2   = tan −  tan  = tan − ln|sec|, and thus,  tan2   = tan − ln|sec| − 1 2 2 + . 21. Let  = 2,  = 1 (1 + 2)2  ⇒  = ( · 22 + 2 · 1) = 2(2 + 1),  = −2(1 + 2 1 ). Then by Equation 2,  (1 + 2 2)2  = −2(1 + 2 2 ) + 1 2  21 + 2 (2 + 1)   = −2(1 + 2 2 ) + 1 2  2 = −2(1 + 2 2 ) + 1 42 + . The answer could be written as 2 4(2 + 1) + . 22. First let  = (arcsin)2,  =  ⇒  = 2 arcsin · √1 1− 2 ,  = . Then  =  (arcsin)2 = (arcsin)2 − 2 √arcsin 1 − 2 . To simplify the last integral, let  = arcsin [ = sin], so °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.4 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION  = √1 1− 2 , and  √arcsin 1 − 2  =  sin . To evaluate just the last integral, now let  = ,  = sin  ⇒  = ,  = −cos . Thus,  sin  = −cos +  cos  = −cos + sin +  = −arcsin · √1 − 2 1 +  + 1 [refer to the figure] Returning to , we get  = (arcsin)2 + 2√1 − 2 arcsin − 2 + , where  = −21. 23. Let  = ,  = cos  ⇒  = ,  = 1  sin. By (6), 012 cos  = 1 sin1 02 − 012 1 sin  = 21 − 0 − 1 −1 cos1 02 = 1 2 + 1 2 (0 − 1) = 1 2 − 1 2 or  − 2 22 24. First let  = 2 + 1,  = −  ⇒  = 2 ,  = −−. By (6), 01(2 + 1)−  = −(2 + 1)−1 0 + 01 2−  = −2−1 + 1 + 201 − . Next let  = ,  = −  ⇒  = ,  = −−. By (6) again, 01 −  = −−1 0 + 01 −  = −−1 + −−1 0 = −−1 − −1 + 1 = −2−1 + 1. So 01(2 + 1)−  = −2−1 + 1 + 2(−2−1 + 1) = −2−1 + 1 − 4−1 + 2 = −6−1 + 3. 25. Let  = ,  = sinh  ⇒  = ,  = cosh. By (6), 02  sinh  =  cosh2 0 − 02 cosh  = 2 cosh 2 − 0 − sinh2 0 = 2 cosh 2 − sinh 2. 26. Let  = ln,  = 2  ⇒  = 1  ,  = 1 33. By (6), 12 2 ln  =  1 33 ln2 1 − 12 1 32  = 8 3 ln 2 − 0 −  1 932 1 = 8 3 ln 2 −  8 9 − 1 9 = 8 3 ln 2 − 7 9 . 27. Let  = ln,  = 1 2  ⇒  = 1 ,  = −1 . By (6), 15 ln2  = −1 ln5 1 − 15 −12  = − 1 5 ln 5 − 0 − 1 5 1 = − 1 5 ln 5 −  1 5 − 1 = 4 5 − 1 5 ln 5. 28. First let  = 2,  = sin 2  ⇒  = 2 ,  = − 1 2 cos 2. By (6), 02 2 sin 2  = − 1 22 cos 22 0 + 02 cos 2  = −22 + 02 cos 2 . Next let  = ,  = cos 2  ⇒  = ,  = 1 2 sin 2. By (6) again, 02 cos 2  =  1 2sin 22 0 − 02 1 2 sin 2  = 0 − − 1 4 cos 22 0 = 1 4 − 1 4 = 0. Thus, 02 2 sin 2  = −22. 29. sin 2 = 2 sin cos, so 0  sin cos  = 1 2 0  sin 2 . Let  = ,  = sin 2  ⇒  = ,  = − 1 2 cos 2. By (6), 1 2 0  sin 2  = 1 2− 1 2cos 2 0 − 1 2 0 − 1 2 cos 2  = − 1 4 − 0 + 1 4 1 2 sin 2 0 = −4 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.1 INTEGRATION BY PARTS ¤ 5 30. Let  = arctan(1),  =  ⇒  = 1 1 + (1)2 · −1 2  = −2  + 1,  = . By (6), 1√3 arctan1  = arctan1√ 1 3 + 1√3   2 + 1 = √3 6 − 1 · 4 + 12ln(2 + 1)√ 1 3 =  √3 6 −  4 + 1 2 (ln 4 − ln 2) =  √3 6 −  4 + 1 2 ln 4 2 =  √3 6 −  4 + 1 2 ln 2 31. Let  = ,  = −  ⇒  = ,  = −−. By (6), 15   = 15 −  = −−5 1 − 15 −−  = −5−5 + −1 − −5 1 = −5−5 + −1 − (−5 − −1) = 2−1 − 6−5 32. Let  = (ln)2,  = −3  ⇒  = 2ln    = − 1 2−2. By (6),  = 12 (ln3)2  = −(ln 22)2 2 1 + 12 ln 3 . Now let  = ln,  = −3  ⇒  = 1 ,  = − 1 2−2. Then 12 ln3  = −ln 22 2 1 + 1 2 12 −3 = − 1 8 ln 2 + 0 + 1 2 − 212 2 1 = − 1 8 ln 2 + 1 2 − 1 8 + 1 2 = 16 3 − 1 8 ln 2. Thus  = − 1 8 (ln 2)2 + 0 +  16 3 − 1 8 ln 2 = − 1 8 (ln 2)2 − 1 8 ln 2 + 16 3 . 33. Let  = ln(cos),  = sin  ⇒  = 1 cos (−sin),  = −cos. By (6), 03 sin ln(cos) =  − cos ln(cos) 0 3 − 03 sin  = − 1 2 ln 1 2 − 0 −  − cos  0 3 = − 1 2 ln 1 2 +  1 2 − 1 = 1 2 ln 2 − 1 2 34. Let  = 2,  = √4 +  2  ⇒  = 2 ,  = √4 + 2. By (6), 01 √4 + 3 2  = 2√4 + 2 1 0 − 201 4 + 2  = √5 − 2 3(4 + 2)321 0 = √5 − 2 3(5)32 + 2 3(8) = √51 − 10 3  + 16 3 = 16 3 − 7 3√5 35. Let  = (ln)2,  = 4  ⇒  = 2 ln  ,  = 5 5 . By (6), 12 4(ln)2  = 55 (ln)22 1 − 212 54 ln  = 32 5 (ln 2)2 − 0 − 212 54 ln . Let  = ln,  = 4 5  ⇒  = 1  ,  = 5 25 . Then 12 54 ln  = 255 ln2 1 − 12 254  = 32 25 ln 2 − 0 − 125 5 2 1 = 32 25 ln 2 −  125 32 − 125 1 . So 12 4(ln)2  = 32 5 (ln 2)2 − 2 32 25 ln 2 − 125 31  = 32 5 (ln 2)2 − 64 25 ln 2 + 125 62 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.6 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 36. Let  = sin( − ),  =   ⇒  = −cos( − ),  = . Then  = 0  sin( − ) =  sin( − ) 0 + 0  cos( − ) =  sin0 − 0 sin + 1. For 1, let  = cos( − ),  =   ⇒  = sin( − ),  = . So 1 =  cos( − ) 0 − 0  sin( − ) =  cos0 − 0 cos − . Thus,  = −sin +  − cos −  ⇒ 2 =  − cos − sin ⇒  = 1 2( − cos − sin). 37. Let  = √, so that 2 =  and 2  = . Thus,  √  =  (2). Now use parts with  = ,  =  ,  = , and  =  to get 2   = 2  − 2   = 2  − 2 +  = 2√ √ − 2√ + . 38. Let  = ln, so that  =  and   = . Thus,  cos(ln) =  cos  ·   = . Now use parts with  = cos,  =  ,  = −sin , and  =  to get   cos  =  cos  −  − sin  =  cos +   sin . Now use parts with  = sin,  =  ,  = cos  , and  =  to get   sin  =  sin −   cos . Thus,  =  cos +  sin −  ⇒ 2 =  cos +  sin ⇒  = 1 2 cos + 1 2 sin +  = 1 2cos(ln) + 1 2sin(ln) + . 39. Let  = 2, so that  = 2 . Thus, √√/2 3 cos2  = √√/2 2 cos2 · 1 2(2 ) = 1 2 /2 cos . Now use parts with  = ,  = cos  ,  = ,  = sin to get 1 2 /2 cos  = 1 2 sin /2 − /2 sin  = 1 2 sin + cos /2 = 1 2 ( sin + cos) − 1 2  2 sin 2 + cos 2  = 1 2( · 0 − 1) − 1 2  2 · 1 + 0 = − 1 2 − 4 40. Let  = cos, so that  = −sin . Thus, 0 cos  sin 2  = 0 cos (2 sin cos ) = 1−1  · 2(−) = 2−11  . Now use parts with  = ,  =  ,  = ,  =  to get 2−11   = 21 −1 − −11   = 21 + −1 −   1 −1 = 2( + −1 − [1 − −1]) = 2(2−1) = 4. 41. Let  = 1 +  so that  = . Thus,  ln(1 + ) =  ( − 1) ln . Now use parts with  = ln  = ( − 1),  = 1  ,  = 1 22 −  to get  ( − 1) ln  =  1 22 − ln −   1 2 − 1  = 1 2( − 2) ln − 1 42 +  +  = 1 2 (1 + )( − 1) ln(1 + ) − 1 4(1 + )2 + 1 +  + , which can be written as 1 2(2 − 1) ln(1 + ) − 1 42 + 1 2 + 3 4 + . 42. Let  = ln, so that  = 1  . Thus,  arcsin(ln  )  =  arcsin . Now use parts with  = arcsin,  = ,  = 11− 2 , and  =  to get  arcsin  =  arcsin −  1− 2  =  arcsin + 1 − 2 +  = (ln)arcsin(ln) + 1 − (ln)2 + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.1 INTEGRATION BY PARTS ¤ 7 43. Let  = ,  = −2  ⇒  = ,  = − 1 2−2. Then  −2  = − 1 2−2 +  1 2−2  = − 1 2−2 − 1 4−2 + . We see from the graph that this is reasonable, since  has a minimum where  changes from negative to positive. Also,  increases where  is positive and  decreases where  is negative. 44. Let  = ln,  = 32  ⇒  = 1  ,  = 2 552. Then  32 ln  = 2 552 ln − 2 5  32  = 2 552 ln −  2 5 2 52 +  = 2 5 52 ln − 4 2552 +  We see from the graph that this is reasonable, since  has a minimum where  changes from negative to positive. 45. Let  = 1 22,  = 2 √1 + 2  ⇒  =  ,  = 2 3(1 + 2)32. Then  3 √1 + 2  = 1 22 2 3(1 + 2)32 − 2 3  (1 + 2)32 = 1 3 2(1 + 2)32 − 2 3 · 2 5 · 1 2(1 + 2)52 +  = 1 3 2(1 + 2)32 − 15 2 (1 + 2)52 +  We see from the graph that this is reasonable, since  increases where  is positive and  decreases where  is negative. Note also that  is an odd function and  is an even function. Another method: Use substitution with  = 1 + 2 to get 1 5(1 + 2)52 − 1 3(1 + 2)32 + . 46. First let  = 2,  = sin2  ⇒  = 2 ,  = − 1 2 cos2. Then  =  2 sin2  = − 1 22 cos2 +  cos2 . Next let  = ,  = cos2  ⇒  = ,  = 1 2 sin2, so  cos2  = 1 2sin2 −  1 2 sin2  = 1 2sin2 + 1 4 cos2 + . Thus,  = − 1 22 cos 2 + 1 2sin2 + 1 4 cos2 + . We see from the graph that this is reasonable, since  increases where  is positive and  decreases where  is negative. Note also that  is an odd function and  is an even function. 47. (a) Take  = 2 in Example 6 to get  sin2   = −12 cossin + 12  1 = 2 − sin2 4  + . (b)  sin4   = − 1 4 cos sin3  + 3 4  sin2   = − 1 4 cossin3  + 3 8 − 16 3 sin2 + . 48. (a) Let  = cos−1 ,  = cos  ⇒  = −( − 1)cos−2  sin ,  = sin in (2):  cos   = cos−1  sin + ( − 1) cos−2  sin2   = cos−1  sin + ( − 1) cos−2 (1 − cos2 ) = cos−1  sin + ( − 1) cos−2   − ( − 1) cos   [continued] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.8 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION Rearranging terms gives   cos   = cos−1  sin + ( − 1) cos−2   or  cos   = 1 cos−1  sin +  − 1  cos−2   (b) Take  = 2 in part (a) to get  cos2   = 1 2 cos sin + 1 2  1 =  2 + sin 2 4 + . (c)  cos4   = 1 4 cos3  sin + 3 4  cos2   = 1 4 cos3  sin + 3 8 + 16 3 sin 2 +  49. (a) From Example 6,  sin   = −1 cos sin−1  +  − 1  sin−2  . Using (6), 02 sin   = −cos  sin  −1  0 2 +  − 1 02 sin−2   = (0 − 0) +  − 1  02 sin−2   =  − 1 02 sin−2   (b) Using  = 3 in part (a), we have 02 sin3   = 2 3 02 sin  = − 2 3 cos  0 2 = 2 3. Using  = 5 in part (a), we have 02 sin5   = 4 5 02 sin3   = 4 5 · 2 3 = 15 8 . (c) The formula holds for  = 1 (that is, 2 + 1 = 3) by (b). Assume it holds for some  ≥ 1. Then 02 sin2+1   = 3 ·25··47·· · · · · 6 · · · · · (2(2  + 1) ) . By Example 6, 02 sin2+3   = 22 + 2 + 3 02 sin2+1   = 2 2  + 2 + 3 · 3 ·25··47·· · · · · 6 · · · · · (2(2  + 1) ) = 2 · 4 · 6 · · · · · (2)[2 ( + 1)] 3 · 5 · 7 · · · · · (2 + 1)[2 ( + 1) + 1], so the formula holds for  =  + 1. By induction, the formula holds for all  ≥ 1. 50. Using Exercise 49(a), we see that the formula holds for  = 1, because 02 sin2   = 1 2 02 1 = 1 2    0 2 = 1 2 · 2 . Now assume it holds for some  ≥ 1. Then 02 sin2   = 1 ·23··45·· · · · · 6 · · · · · (2(2  −)1) 2 . By Exercise 49(a), 02 sin2(+1)   = 22 + 1 + 2 02 sin2   = 2 2  + 1 + 2 · 1 ·23··45·· · · · · 6 · · · · · (2(2  −)1) 2 = 1 · 3 · 5 · · · · · (2 − 1)(2 + 1) 2 · 4 · 6 · · · · · (2)(2 + 2) ·  2 , so the formula holds for  =  + 1. By induction, the formula holds for all  ≥ 1. 51. Let  = (ln),  =  ⇒  = (ln)−1(),  = . By Equation 2,  (ln)  = (ln) −  (ln)−1() = (ln) −   (ln)−1 . 52. Let  = ,  =   ⇒  = −1 ,  = . By Equation 2,    =  −   −1 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.1 INTEGRATION BY PARTS ¤ 9 53.  tan   =  tan−2  tan2   =  tan−2 (sec2  − 1) =  tan−2  sec2   −  tan−2   =  −  tan−2  . Let  = tan−2 ,  = sec2   ⇒  = ( − 2) tan−3  sec2  ,  = tan. Then, by Equation 2,  = tan−1  − ( − 2) tan−2  sec2   1 = tan−1  − ( − 2) ( − 1) = tan−1   = tan−1   − 1 Returning to the original integral,  tan   = tan−1   − 1 −  tan−2  . 54. Let  = sec−2 ,  = sec2   ⇒  = ( − 2) sec−3  sec tan ,  = tan. Then, by Equation 2,  sec   = tan sec−2  − ( − 2) sec−2  tan2   = tan sec−2  − ( − 2) sec−2 (sec2  − 1) = tan sec−2  − ( − 2) sec   + ( − 2) sec−2   so ( − 1) sec   = tan sec−2  + ( − 2) sec−2  . If  − 1 6= 0, then  sec   = tansec − 1−2  +  −− 2 1  sec−2  . 55. By repeated applications of the reduction formula in Exercise 51,  (ln)3  = (ln)3 − 3 (ln)2  = (ln)3 − 3(ln)2 − 2 (ln)1  = (ln)3 − 3(ln)2 + 6(ln)1 − 1 (ln)0  = (ln)3 − 3(ln)2 + 6ln − 6 1 = (ln)3 − 3(ln)2 + 6ln − 6 +  56. By repeated applications of the reduction formula in Exercise 52,  4  = 4 − 4 3  = 4 − 43 − 3 2  = 4 − 43 + 122 − 2 1  = 4 − 43 + 122 − 241 −  0  = 4 − 43 + 122 − 24+ 24 +  or (4 − 43 + 122 − 24 + 24) +  57. The curves  = 2 ln and  = 4 ln intersect when 2 ln = 4 ln ⇔ 2 ln − 4ln = 0 ⇔ (2 − 4) ln = 0 ⇔  = 1 or 2 [since   0]. For 1    2, 4ln  2 ln Thus, area = 12(4 ln − 2 ln) = 12[(4 − 2)ln]. Let  = ln,  = (4 − 2) ⇒  = 1 ,  = 4 − 1 33. Then area = (ln)4 − 1 332 1 − 12 4 − 1 331  = (ln 2) 16 3  − 0 − 12 4 − 1 32  = 16 3 ln 2 − 4 − 1 932 1 = 16 3 ln 2 −  64 9 − 35 9  = 16 3 ln 2 − 29 9 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.10 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 58. The curves  = 2− and  = − intersect when 2− = − ⇔ 2 −  = 0 ⇔ ( − 1) = 0 ⇔  = 0 or 1. For 0    1, −  2−. Thus, area = 01(− − 2−) = 01( − 2)− . Let  =  − 2,  = − ⇒  = (1 − 2),  = −−. Then area = ( − 2)(−−)1 0 − 01[−−(1 − 2)] = 0 + 01(1 − 2)− . Now let  = 1 − 2,  = − ⇒  = −2,  = −−. Now area = (1 − 2)(−−)1 0 − 01 2−  = −1 + 1 − −2−1 0 = −1 + 1 + 2(−1 − 1) = 3−1 − 1. 59. The curves  = arcsin 1 2  and  = 2 − 2 intersect at  =  ≈ −175119 and  =  ≈ 117210. From the figure, the area bounded by the curves is given by  = [(2 − 2) − arcsin 1 2 ] = 2 − 1 3 3  −  arcsin 1 2 . Let  = arcsin 1 2 ,  =  ⇒  = 1 −1 1 2 2 · 1 2 ,  = . Then  = 2 − 133  −   arcsin1 2  −  21− 1 4 2    = 2 − 1 3 3 − arcsin 1 2  − 21 − 1 42   ≈ 399926 60. The curves  = ln( + 1) and  = 3 − 2 intersect at  = 0 and  =  ≈ 192627. From the figure, the area bounded by the curves is given by  = 0[(3−2)−ln(+1)]  =  3 2 2 − 1 33 0 −0 ln(+1). Let  = ln( + 1),  =   ⇒  = 1  + 1 ,  = 1 2 2. Then  = 322 − 133 0 − 1 22 ln( + 1) 0 − 12 0 + 1 2  = 322 − 133 0 − 1 22 ln( + 1) 0 + 12 0  − 1 +  + 1 1   =  3 2 2 − 1 33 − 1 2 2 ln( + 1) + 1 42 − 1 2  + 1 2 ln| + 1| 0 ≈ 169260 61. Volume = 01 2cos(2). Let  = ,  = cos(2) ⇒  = ,  = 2 sin(2).  = 2 2 sin  2 1 0 − 2 · 2 01 sin  2   = 22 − 0 − 4−2 cos  2 1 0 = 4 + 8 (0 − 1) = 4 − 8 . 62. Volume = 01 2( − −) = 2 01( − −) = 201   − 01 −  [both integrals by parts] = 2( − ) − −− − − 1 0 = 2[2 − 0] = 4 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.1 INTEGRATION BY PARTS ¤ 11 63. Volume = −01 2(1 − )− . Let  = 1 − ,  = −  ⇒  = − ,  = −−.  = 2(1 − )(−−)0 −1 − 2 −01 −  = 2( − 1)(−) + −0 −1 = 2−0 −1 = 2(0 + ) = 2. 64.  =  ⇔  = ln. Volume = 13 2 ln . Let  = ln,  =   ⇒  = 1  ,  = 1 22.  = 2  1 22 ln3 1 − 2 13 1 2  = 2  1 22 ln − 1 423 1 = 2  9 2 ln 3 − 9 4  − 0 − 1 4  = 2 9 2 ln 3 − 2 = (9 ln 3 − 4) 65. (a) Use shells about the -axis:  = 12 2ln    = ln = 1 , ,   ==   1 2 2  = 2 1 22 ln2 1 − 12 1 2  = 2(2 ln 2 − 0) −  1 422 1 = 22ln2 − 3 4  (b) Use disks about the -axis:  = 12 (ln)2    = (ln = 2 ln)2·, 1 ,   ==   = (ln)22 1 − 12 2ln    = ln = 1 , ,   ==   = 2(ln 2)2 − 2ln2 1 − 12  = 2(ln 2)2 − 4ln2 + 22 1 = [2(ln 2)2 − 4ln2 + 2] = 2[(ln 2)2 − 2ln2 + 1] 66. ave = 1  −   () = 41− 0 04 sec2     ==  , ,   = tan = sec2   = 4  tan 0 4 − 04 tan  = 4 4 − ln|sec|  0 4 = 4 4 − ln√2 = 1 − 4  ln√2 or 1 − 2  ln 2 67. () = 0 sin 1 22  ⇒  () =  0 sin 1 22  . Let  = 0 sin 1 22  = (),  =  ⇒  = sin 1 22 ,  = . Thus,  () = () −  sin 1 22  = () −  sin  1    ==   1 2 2  = () + 1 cos  +  = () + 1 cos  1 22 +  68. The rocket will have height  = 060 () after 60 seconds.  = 060 − −  ln −   = − 1 2260 0 −  060 ln( − ) − 060 ln  = −(1800) + (ln)(60) −  060 ln( − ) Let  = ln( − ),  =  ⇒  = 1  −  (−),  = . Then °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.12 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 060 ln( − ) = ln( − )60 0 + 060   −   = 60 ln( − 60) + 060−1 + −  = 60 ln( − 60) + − −  ln( − )60 0 = 60 ln( − 60) − 60 −  ln( − 60) +  ln So  = −1800 + 60 ln − 60 ln( − 60) + 60 +    ln( − 60) −    ln. Substituting  = 98,  = 30,000,  = 160, and  = 3000 gives us  ≈ 14,844 m. 69. Since ()  0 for all , the desired distance is () = 0 () = 0 2− . First let  = 2,  = −  ⇒  = 2 ,  = −−. Then () = −2− 0 + 20 − . Next let  = ,  = −  ⇒  = ,  = −−. Then () = −2− + 2−− 0 + 0 −  = −2− + 2−− + 0 + −− 0 = −2− + 2(−− − − + 1) = −2− − 2− − 2− + 2 = 2 − −(2 + 2 + 2) meters 70. Suppose (0) = (0) = 0 and let  = (),  = 00() ⇒  =  0(),  = 0(). Then 0 ()00() = ()0() 0 − 0  0()0() = ()0() − 0  0()0(). Now let  =  0(),  = 0() ⇒  =  00() and  = (), so 0 0()0() = 0()() 0 − 0  00()() =  0()() − 0 00()(). Combining the two results, we get 0 ()00() = ()0() −  0()() + 0  00()(). 71. For  = 14  00(), let  = ,  =  00() ⇒  = ,  = 0(). Then  =  0()4 1 − 14  0() = 4 0(4) − 1 ·  0(1) − [(4) − (1)] = 4 · 3 − 1 · 5 − (7 − 2) = 12 − 5 − 5 = 2. We used the fact that 00 is continuous to guarantee that  exists. 72. (a) Take () =  and 0() = 1 in Equation 1. (b) By part (a),  () = () −  () −   0(). Now let  = (), so that  = () and  =  0(). Then    0() = (()) (). The result follows. (c) Part (b) says that the area of region  is = () − () − (()) () = (area of rectangle ) − (area of rectangle ) − (area of region ) °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.1 INTEGRATION BY PARTS ¤ 13 (d) We have () = ln, so −1() = , and since  = −1, we have () = . By part (b), 1 ln  = ln − 1ln1 − ln 1 ln    =  − 01   =  −   1 0 =  − ( − 1) = 1. 73. Using the formula for volumes of rotation and the figure, we see that Volume = 0 2  − 0 2  −  [()]2  = 2 − 2 −  [()]2 . Let  = (), which gives  =  0() and () = , so that  = 2 − 2 −   2 0(). Now integrate by parts with  = 2, and  =  0() ⇒  = 2 ,  = (), and  2  0() = 2 ()  −  2 () = 2 () − 2 () −  2 (), but () =  and () =  ⇒  = 2 − 2 − 2 − 2 −  2 () =  2 (). 74. (a) We note that for 0 ≤  ≤ 2 , 0 ≤ sin ≤ 1, so sin2+2  ≤ sin2+1  ≤ sin2 . So by the second Comparison Property of the Integral, 2+2 ≤ 2+1 ≤ 2. (b) Substituting directly into the result from Exercise 50, we get 2+2 2 = 1 · 3 · 5 · · · · · [2( + 1) − 1] 2 · 4 · 6 · · · · · [2( + 1)]  2 1 · 3 · 5 · · · · · (2 − 1) 2 · 4 · 6 · · · · · (2)  2 = 2( + 1) − 1 2( + 1) = 2 + 1 2 + 2 (c) We divide the result from part (a) by 2. The inequalities are preserved since 2 is positive: 2+2 2 ≤ 22+1  ≤ 2 2  . Now from part (b), the left term is equal to 2 + 1 2 + 2 , so the expression becomes 2 + 1 2 + 2 ≤ 2+1 2 ≤ 1. Now lim →∞ 2 + 1 2 + 2 = lim →∞ 1 = 1, so by the Squeeze Theorem, lim →∞ 2+1 2 = 1. (d) We substitute the results from Exercises 49 and 50 into the result from part (c): 1 = lim →∞ 2+1 2 = lim →∞ 2 · 4 · 6 · · · · · (2) 3 · 5 · 7 · · · · · (2 + 1) 1 · 3 · 5 · · · · · (2 − 1) 2 · 4 · 6 · · · · · (2)  2 = lim →∞ 3 ·25··47·· · · · · 6 · · · · · (2(2  + 1) ) 1 ·23··45·· · · · · 6 · · · · · (2(2  −)1)2  = lim →∞ 2 1 · 2 3 · 4 3 · 4 5 · 6 5 · 6 7 · · · · · 2 2 − 1 · 2 2 + 1 · 2  [rearrange terms] Multiplying both sides by 2 gives us the Wallis product:  2 = 2 1 · 2 3 · 4 3 · 4 5 · 6 5 · 6 7 · · · · (e) The area of the th rectangle is . At the 2th step, the area is increased from 2 − 1 to 2 by multiplying the width by 2 2 − 1 , and at the (2 + 1)th step, the area is increased from 2 to 2 + 1 by multiplying the height by 2 + 1 2 . These two steps multiply the ratio of width to height by 2 2 − 1 and 1 (2 + 1)(2) = 2 2 + 1 respectively. So, by part (d), the limiting ratio is 2 1 · 2 3 · 4 3 · 4 5 · 6 5 · 6 7 · · · · =  2 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.14 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 7.2 Trigonometric Integrals The symbols =s and =c indicate the use of the substitutions { = sin  = cos } and { = cos  = −sin }, respectively. 1.  sin2  cos3   =  sin2  cos2  cos  =  sin2 (1 − sin2 ) cos  s=  2(1 − 2) =  (2 − 4) = 1 33 − 1 55 +  = 1 3 sin3  − 1 5 sin5  +  2.  sin3  cos4   =  sin2  cos4  sin  =  (1 − cos2 ) cos4  sin  c=  (1 − 2)4(−) =  (6 − 4) = 1 77 − 1 55 +  = 1 7 cos7  − 1 5 cos5  +  3. 02 sin7  cos5   = 02 sin7  cos4  cos  = 02 sin7  (1 − sin2 )2 cos  s= 01 7(1 − 2)2  = 01 7(1 − 22 + 4) = 01(7 − 29 + 11) = 1 88 − 1 510 + 12 1 121 0 = 18 − 1 5 + 12 1  − 0 = 15 −120 24 + 10 = 120 1 4. 02 sin5   = 02 sin4  sin  = 02(1 − cos2 )2 sin  =c 10(1 − 2)2(−) = 01(1 − 22 + 4) =  − 233 + 1 551 0 = 1 − 23 + 15 − 0 = 15 −15 10 + 3 = 15 8 5.  sin5(2)cos2(2)  =  sin4(2) cos2(2) sin(2) =  [1 − cos2(2)]2 cos2(2) sin(2) =  (1 − 2)2 2 − 1 2  [ = cos(2),  = −2sin(2)] = − 1 2  (4 − 22 + 1)2  = − 1 2  (6 − 24 + 2) = − 1 2  1 77 − 2 55 + 1 33 +  = − 14 1 cos7(2) + 1 5 cos5(2) − 1 6 cos3(2) +  6.  cos5(2) =  cos4(2) cos(2) =  [1 − sin2(2)]2 cos(2) =  1 2(1 − 2)2   = sin(2),  = 2cos(2) = 1 2  (4 − 22 + 1) = 1 2( 1 55 − 2 33 + ) +  = 10 1 sin5(2) − 1 3 sin3(2) + 1 2 sin(2) +  7. 02 cos2   = 02 1 2(1 + cos 2) [half-angle identity] = 1 2  + 1 2 sin 2 0 2 = 1 2  2 + 0 − (0 + 0) = 4 8. 02 sin2 1 3  = 02 1 2 1 − cos2 · 1 3  [half-angle identity] = 1 2  − 3 2 sin2 32 0 = 1 22 − 3 2−√23  − 0 =  + 3 8 √3 9. 0 cos4(2) = 0[cos2(2)]2  = 0  1 2(1 + cos(2 · 2))2  [half-angle identity] = 1 4 0[1 + 2 cos 4 + cos2(4)] = 1 4 0[1 + 2 cos 4 + 1 2(1 + cos 8)] = 1 4 0  3 2 + 2 cos 4 + 1 2 cos 8  = 1 4  3 2 + 1 2 sin 4 + 16 1 sin 8 0 = 1 4  3 2 + 0 + 0 − 0 = 3 8 10. 0 sin2  cos4   = 1 4 0(4 sin2  cos2 )cos2   = 1 4 0(2 sin cos )2 1 2(1 + cos 2) = 1 8 0(sin 2)2(1 + cos 2) = 1 8 0(sin2 2 + sin2 2 cos 2) = 1 8 0 sin2 2  + 1 8 0 sin2 2 cos 2  = 1 8 0 12(1 − cos 4) + 1 8  1 3 · 1 2 sin3 2 0 = 1 16  − 1 4 sin 4 0 + 1 8(0 − 0) = 16 1 [( − 0) − 0] = 16  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 15 11. 02 sin2  cos2   = 02 1 4(4 sin2  cos2 ) = 02 14(2 sin cos)2 = 1 4 02 sin2 2  = 1 4 02 1 2(1 − cos 4) = 1 8 02(1 − cos 4) = 1 8  − 1 4 sin 4 0 2 = 1 8  2  = 16  12. 02(2 − sin)2  = 02(4 − 4sin + sin2 ) = 02 4 − 4sin + 1 2(1 − cos 2)  = 02  9 2 − 4sin − 1 2 cos 2  =  9 2 + 4 cos − 1 4 sin 2 0 2 =  94 + 0 − 0 − (0 + 4 − 0) = 9 4 − 4 13.  √cos sin3   =  √cos sin2  sin  =  (cos)12(1 − cos2 )sin  c=  12(1 − 2)(−) =  (52 − 12) = 2 7 72 − 2 332 +  = 2 7(cos)72 − 2 3(cos)32 +  14.  sin2(1 2 )  =  sin2 (−)  = 1 ,  = −12  = −  1 2(1 − cos 2) = −12 − 12 sin 2 +  = −21 + 1 4 sin2  +  15.  cot cos2   =  cos sin (1 − sin2 ) s=  1 −2  =  1 −   = ln|| − 1 22 +  = ln|sin| − 1 2 sin2  +  16.  tan2  cos3   =  cos sin22  cos3   =  sin2  cos  =s  2  = 1 33 +  = 1 3 sin3  +  17.  sin2  sin 2  =  sin2 (2 sin cos ) =s  23  = 1 24 +  = 1 2 sin4  +  18.  sin cos 1 2  =  sin2 · 1 2cos 1 2  =  2sin 1 2cos2 1 2  =  22 (−2) [ = cos 1 2 ,  = − 1 2 sin 1 2  ] = − 4 3 3 +  = − 4 3 cos3 1 2 +  19.  sin2   =   1 2(1 − cos 2)  = 1 2  ( − cos 2) = 1 2    − 1 2  cos 2  = 1 2  1 22 − 1 2  1 2sin 2 −  1 2 sin 2    ==  , ,   == cos 2 1 2 sin 2     = 1 4 2 − 1 4sin 2 + 1 2 − 1 4 cos 2 +  = 1 42 − 1 4sin 2 − 1 8 cos 2 +  20.  =  sin3  . First, evaluate  sin3   =  (1 − cos2 ) sin  =c  (1 − 2)(−) =  (2 − 1) = 1 33 −  + 1 = 1 3 cos3  − cos + 1. Now for , let  = ,  = sin3  ⇒  = ,  = 1 3 cos3  − cos, so  = 1 3cos3  − cos  −   1 3 cos3  − cos  = 1 3cos3  − cos − 1 3  cos3   + sin = 1 3 cos3  − cos  − 1 3(sin − 1 3 sin3 ) + sin +  [by Example 1] = 1 3 cos3  − cos  + 2 3 sin + 1 9 sin3  +  21.  tan sec3   =  tan sec sec2   =  2  [ = sec ,  = sec  tan  ] = 1 3 3 +  = 1 3 sec3  +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.16 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 22.  tan2  sec4   =  tan2  sec2  sec2   =  tan2  (tan2  + 1) sec2   =  2(2 + 1) [ = tan ,  = sec2  ] =  (4 + 2) = 1 5 5 + 1 3 3 +  = 1 5 tan5  + 1 3 tan3  +  23.  tan2   =  (sec2  − 1) = tan −  +  24.  (tan2  + tan4 ) =  tan2 (1 + tan2 ) =  tan2  sec2   =  2  [ = tan ,  = sec2  ] = 1 3 3 +  = 1 3 tan3  +  25. Let  = tan. Then  = sec2 , so  tan4sec6  =  tan4sec4(sec2 ) =  tan4(1 + tan2)2 (sec2 ) =  4(1 + 2)2  =  (8 + 26 + 4) = 1 9 9 + 2 7 7 + 1 5 5 +  = 1 9 tan9 + 2 7 tan7 + 1 5 tan5 +  26. 04 sec6  tan6   = 04 tan6  sec4  sec2   = 04 tan6 (1 + tan2 )2 sec2   = 01 6(1 + 2)2    = tan = sec2  ,  = 01 6(4 + 22 + 1) = 01(10 + 28 + 6) =  11 1 11 + 2 9 9 + 1 7 71 0 = 11 1 + 2 9 + 1 7 = 63 + 154 + 99 693 = 316 693 27.  tan3  sec  =  tan2  sec tan  =  (sec2  − 1) sec tan  =  (2 − 1) [ = sec,  = sec tan ] = 1 3 3 −  +  = 1 3 sec3  − sec +  28. Let  = sec, so  = sectan . Thus,  tan5sec3  =  tan4sec2(sectan) =  (sec2 − 1)2 sec2(sectan ) =  (2 − 1)22  =  (6 − 24 + 2) = 1 7 7 − 2 5 5 + 1 3 3 +  = 1 7 sec7  − 2 5 sec5  + 1 3 sec3  +  29.  tan3  sec6   =  tan3  sec4  sec2   =  tan3 (1 + tan2 )2 sec2   =  3(1 + 2)2    = tan = sec2  ,  =  3(4 + 22 + 1) =  (7 + 25 + 3) = 1 8 8 + 1 3 6 + 1 4 4 +  = 1 8 tan8  + 1 3 tan6  + 1 4 tan4  +  30. 04 tan4   = 04 tan2 (sec2  − 1) = 04 tan2  sec2   − 04 tan2   = 01 2  [ = tan ] − 04(sec2  − 1) =  1 3 31 0 − tan −  0 4 = 1 3 − 1 − 4  − 0 = 4 − 2 3 31.  tan5   =  (sec2  − 1)2 tan  =  sec4  tan  − 2 sec2  tan  +  tan  =  sec3  sec tan  − 2 tan sec2   +  tan  = 1 4 sec4  − tan2  + ln|sec| +  [or 1 4 sec4  − sec2  + ln|sec| +  ] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 17 32.  tan2  sec  =  (sec2  − 1) sec  =  sec3   −  sec  = 1 2 (sec tan + ln|sec + tan|) − ln|sec + tan| +  [by Example 8 and (1)] = 1 2 (sec tan − ln|sec + tan|) +  33. Let  =   = sec tan  ⇒  =   = sec. Then   sec tan  = sec −  sec  = sec − ln|sec + tan| + . 34.  cos sin3  =  cos sin · cos12   =  tan sec2   =     = tan,  = sec2   = 1 2 2 +  = 1 2 tan2  +  Alternate solution: Let  = cos to get 1 2 sec2  + . 35.  62 cot2   =  62(csc2  − 1) = −cot −  2 6 = 0 − 2  − −√3 − 6  = √3 − 3 36.   42 cot3   =   42 cot(csc2  − 1) =   42 cot csc2   −   42 cos sin  = − 1 2 cot2  − ln|sin|  2 4 = (0 − ln 1) − − 1 2 − ln √12  = 1 2 + ln √12 = 1 2(1 − ln 2) 37.  42 cot5  csc3   =  42 cot4  csc2  csc cot  =  42(csc2  − 1)2 csc2  csc cot  = √12(2 − 1)22 (−) [ = csc ,  = − csc  cot  ] = 1√2(6 − 24 + 2) =  1 77 − 2 55 + 1 33√ 1 2 =  8 7√2 − 8 5√2 + 2 3√2 −  1 7 − 2 5 + 1 3  = 120 − 168 + 70 105 √2 − 15 − 42 + 35 105 = 22 105 √2 − 8 105 38.  42 csc4  cot4   =  42 cot4  csc2  csc2   =  42 cot4  (cot2  + 1) csc2   = 10 4(2 + 1) (−)   = cot = − csc ,2   = 01(6 + 4) =  1 77 + 1 551 0 = 1 7 + 1 5 = 12 35 39.  =  csc  =  csccsc (csc  −cot − cot  )  =  −csccsc  cot  −cot + csc  2  . Let  = csc − cot ⇒  = (−csc cot + csc2 ). Then  =   = ln|| = ln|csc − cot| + . 40. Let  = csc,  = csc2  . Then  = −csc cot ,  = −cot ⇒  csc3   = −csc cot −  csc cot2   = −csc cot −  csc(csc2  − 1) = −csc cot +  csc  −  csc3   Solving for  csc3   and using Exercise 39, we get °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.18 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION  csc3   = − 1 2 csc cot + 1 2  csc  = − 1 2 csc cot + 1 2 ln|csc − cot| + . Thus,  63 csc3   = − 1 2 csc cot + 1 2 ln|csc − cot|  3 6 = − 1 2 · 2√ 3 · 1√ 3 + 1 2 ln   √23 − √13    + 1 2 · 2 · √3 − 1 2 ln 2 − √3  = − 1 3 + √3 + 1 2 ln √13 − 1 2 ln2 − √3 ≈ 17825 41.  sin 8 cos 5  =2a  1 2[sin(8 − 5) + sin(8 + 5)] = 1 2  (sin 3 + sin 13) = 1 2 (− 1 3 cos 3 − 13 1 cos 13) +  = − 1 6 cos 3 − 26 1 cos 13 +  42.  sin 2 sin 6  =2b  1 2[cos(2 − 6) − cos(2 + 6)]  = 1 2  [cos(−4) − cos 8] = 1 2  (cos 4 − cos 8) = 1 2  1 4 sin 4 − 1 8 sin 8 +  = 1 8 sin 4 − 16 1 sin 8 +  43. 02 cos 5 cos 10  =2c 02 1 2[cos(5 − 10) + cos(5 + 10)] = 1 2 02[cos(−5) + cos 15]  = 1 2 02(cos 5 + cos 15) = 1 2  1 5 sin 5 + 15 1 sin 15 0 2 = 1 2  1 5 − 15 1  = 15 1 44.  sin sec5   =  cos sin5  =c  15 (−) = 414 +  = 4cos 1 4  +  = 1 4 sec4  +  45. 06 √1 + cos 2  = 06 1 + (2 cos2  − 1) = 06 √2cos2   = √206 √cos2   = √206 |cos|  = √206 cos  [since cos   0 for 0 ≤  ≤ 6] = √2sin 0 6 = √2 1 2 − 0 = 1 2√2 46. 04 √1 − cos 4  = 04 1 − (1 − 2sin2(2)) = 04 2sin2(2) = √204 sin2(2) = √204 |sin 2|  = √204 sin 2  [since sin 2 ≥ 0 for 0 ≤  ≤ 4] = √2− 1 2 cos 2 0 4 = − 1 2√2(0 − 1) = 1 2√2 47.  1 −sec tan 2 2   =  cos2  − sin2   =  cos 2  = 1 2 sin 2 +  48.  cos  − 1 =  cos1− 1 · cos cos + 1 + 1  =  cos cos2+ 1 − 1  =  cos −sin  + 1 2   =  −cot csc − csc2   = csc + cot +  49.  tan2   =  (sec2  − 1) =  sec2   −    = tan −  tan  − 1 22   ==  , ,   = tan = sec2   = tan − ln|sec| − 1 22 +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 19 50. Let  = tan7 ,  = sec tan  ⇒  = 7 tan6  sec2  ,  = sec. Then  tan8  sec  =  tan7  · sec tan  = tan7  sec −  7tan6 sec2 sec  = tan7  sec − 7 tan6 (tan2  + 1) sec  = tan7  sec − 7 tan8 sec  − 7 tan6 sec . Thus, 8 tan8 sec  = tan7  sec − 7 tan6 sec  and 04 tan8 sec  = 1 8tan7  sec 0 4 − 78 04 tan6  sec  = √82 − 7 8. In Exercises 51–54, let () denote the integrand and () its antiderivative (with  = 0). 51. Let  = 2, so that  = 2 . Then  sin2(2) =  sin2   1 2  = 1 2  1 2(1 − cos 2) = 1 4  − 1 2 sin 2 +  = 1 4 − 1 4  1 2 · 2sin cos +  = 1 4 2 − 1 4 sin(2) cos(2) +  We see from the graph that this is reasonable, since  increases where  is positive and  decreases where  is negative. Note also that  is an odd function and  is an even function. 52.  sin5  cos3   =  sin5  cos2  cos  =  sin5 (1 − sin2 ) cos  s=  5(1 − 2) =  (5 − 7) = 1 6 sin6  − 1 8 sin8  +  We see from the graph that this is reasonable, since  increases where  is positive and  decreases where  is negative. Note also that  is an odd function and  is an even function. 53.  sin 3 sin 6  =  1 2[cos(3 − 6) − cos(3 + 6)] = 1 2  (cos 3 − cos 9) = 1 6 sin 3 − 1 18 sin 9 +  Notice that () = 0 whenever  has a horizontal tangent. 54.  sec4 1 2  =  tan2 2 + 1sec2 2  =  (2 + 1) 2  = tan 2 ,  = 1 2 sec2 2  = 2 3 3 + 2 +  = 2 3 tan3 2 + 2 tan 2 +  Notice that  is increasing and  is positive on the intervals on which they are defined. Also,  has no horizontal tangent and  is never zero. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.20 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 55. ave = 21 − sin2  cos3   = 21 − sin2 (1 − sin2 ) cos  = 1 2 00 2(1 − 2) [where  = sin ] = 0 56. (a) Let  = cos. Then  = −sin  ⇒  sin cos  =  (−) = − 1 2 2 +  = − 1 2 cos2  + 1. (b) Let  = sin. Then  = cos  ⇒  sin cos  =    = 1 2 2 +  = 1 2 sin2  + 2. (c)  sin cos  =  1 2 sin 2  = − 1 4 cos 2 + 3 (d) Let  = sin,  = cos  . Then  = cos ,  = sin, so  sin cos   = sin2  −  sin cos , by Equation 7.1.2, so  sin cos  = 1 2 sin2  + 4. Using cos2  = 1 − sin2  and cos 2 = 1 − 2sin2 , we see that the answers differ only by a constant. 57.  = 0(sin2  − sin3 ) = 0  1 2(1 − cos 2) − sin(1 − cos2 )  = 0  1 2 − 1 2 cos 2  + 1−1(1 − 2)   = cos = − sin ,  =  1 2  − 1 4 sin 2 0 + 201(2 − 1) =  1 2  − 0 − (0 − 0) + 2 1 3 3 − 1 0 = 1 2  + 2 1 3 − 1 = 1 2  − 4 3 58.  = 04(tan − tan2 ) = 04(tan − sec2  + 1) = ln|sec| − tan +  0 4 = ln√2 − 1 + 4  − (ln 1 − 0 + 0) = ln√2 − 1 + 4 59. It seems from the graph that 02 cos3   = 0, since the area below the -axis and above the graph looks about equal to the area above the axis and below the graph. By Example 1, the integral is sin − 1 3 sin3 2 0 = 0. Note that due to symmetry, the integral of any odd power of sin or cos between limits which differ by 2 ( any integer) is 0. 60. It seems from the graph that 02 sin 2 cos 5  = 0, since each bulge above the -axis seems to have a corresponding depression below the -axis. To evaluate the integral, we use a trigonometric identity: 01 sin 2 cos 5  = 1 2 02[sin(2 − 5) + sin(2 + 5)] = 1 2 02[sin(−3) + sin 7] = 1 2  31 cos(−3) − 71 cos 72 0 = 1 2  31(1 − 1) − 71(1 − 1) = 0 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 21 61. Using disks,  =   2  sin2   =    2 1 2(1 − cos 2) =  1 2 − 1 4 sin 2 2 =  2 − 0 − 4 + 0 = 42 62. Using disks,  = 0 (sin2 )2  = 2 02  1 2(1 − cos 2)2  = 2 02(1 − 2cos 2 + cos2 2) = 2 02 1 − 2cos 2 + 1 2(1 − cos 4)  = 2 02  3 2 − 2cos 2 − 1 2 cos 4  = 2  3 2 − sin 2 + 1 8 sin 4 0 2 = 2  34 − 0 + 0 − (0 − 0 + 0) = 3 82 63. Using washers,  = 04 (1 − sin)2 − (1 − cos)2  =  04 (1 − 2sin + sin2 ) − (1 − 2cos + cos2 )  =  04(2 cos − 2sin + sin2  − cos2 ) =  04(2 cos − 2sin − cos 2) = 2sin + 2 cos − 1 2 sin 2 0 4 = √2 + √2 − 1 2  − (0 + 2 − 0) = 2√2 − 5 2  64. Using washers,  = 03 [sec − (−1)]2 − [cos − (−1)]2  =  03[(sec2  + 2 sec + 1) − (cos2  + 2 cos  + 1)] =  03 sec2  + 2 sec − 1 2(1 + cos 2) − 2cos  = tan + 2 ln|sec + tan| − 1 2 − 1 4 sin 2 − 2sin 0 3 = √3 + 2 ln2 + √3 − 6 − 1 8 √3 − √3 − 0 = 2 ln2 + √3 − 1 62 − 1 8 √3 65.  = () = 0 sin cos2  . Let  = cos ⇒  = − sin . Then  = − 1  1cos  2 = − 1  1 33cos 1  = 31(1 − cos3 ). 66. (a) We want to calculate the square root of the average value of [()]2 = [155 sin(120)]2 = 1552 sin2(120). First, we calculate the average value itself, by integrating [()]2 over one cycle (between  = 0 and  = 60 1 , since there are 60 cycles per second) and dividing by  60 1 − 0: [()]2 ave = 1160 0160[1552 sin2(120)]  = 60 · 1552 0160 1 2[1 − cos(240)] = 60 · 1552 1 2  − 240 1  sin(240)1 060 = 60 · 1552 1 2  60 1 − 0 − (0 − 0) = 155 2 2 The RMS value is just the square root of this quantity, which is 155 √2 ≈ 110 V. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.22 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (b) 220 = [()]2 ave ⇒ 2202 = [()]2 ave = 1160 0160 2 sin2(120) = 602 0160 1 2[1 − cos(240)]  = 302 − 240 1  sin(240)1 060 = 302 60 1 − 0 − (0 − 0) = 1 22 Thus, 2202 = 1 22 ⇒  = 220√2 ≈ 311 V. 67. Just note that the integrand is odd [(−) = −()]. Or: If  6= , calculate − sin cos  = − 1 2[sin( − ) + sin( + )]  = 1 2−cos( −−) − cos(++) − = 0 If  = , then the first term in each set of brackets is zero. 68. − sin sin  = − 1 2[cos( − ) − cos( + )]. If  6= , this is equal to 1 2sin(−−) − sin(++) − = 0. If  = , we get − 1 2[1 − cos( + )] =  1 2 − − sin( 2(++)) − =  − 0 = . 69. − cos  cos  = − 1 2[cos( − ) + cos( + )]. If  6= , this is equal to 1 2sin(−−) + sin(++) − = 0. If  = , we get − 1 2[1 + cos( + )] =  1 2 − + sin( 2(++)) − =  + 0 = . 70. 1  − () sin  = 1 − =1  sinsin  = =1  − sin sin . By Exercise 68, every term is zero except the th one, and that term is   ·  = . 7.3 Trigonometric Substitution 1. Let  = 2 sin, where −2 ≤  ≤ 2. Then  = 2 cos  and √4 − 2 = 4 − 4sin2 = √4cos2 = 2|cos| = 2 cos . Thus,  2√ 4 − 2 =  4sin2cos 2(2 cos  )  = 14  csc2  = − 1 4 cot +  = − √4 − 2 4 +  [see figure] 2. Let  = 2 tan, where − 2    2 . Then  = 2 sec2   and √2 + 4 = √4tan2  + 4 = 4(tan2  + 1) = √4sec2  = 2|sec| = 2 sec for the relevant values of . [continued] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 23  √23+ 4  =  8tan 2sec3 2sec2   = 8 tan2  sec tan  = 8 (sec2  − 1) sec tan  = 8 (2 − 1) [ = sec ] = 81 33 −  +  = 8 3 sec3  − 8sec +  = 83√22+ 43 − 8√22+ 4 +  = 1 3 (2 + 4)32 − 4√2 + 4 +  3. Let  = 2 sec, where 0 ≤   2 or  ≤   32 . Then  = 2 sec tan  and √2 − 4 = √4sec2  − 4 = 4(sec2  − 1) = √4tan2  = 2|tan| = 2 tan for the relevant values of   √2− 4  =  2tan 2sec 2sec tan  = 2 tan2   = 2 (sec2  − 1) = 2 (tan − ) +  = 2√22− 4 − sec−1 2 +  = √2 − 4 − 2sec−1 2 +  4. Let  = 3 sin, where −2 ≤  ≤ 2. Then  = 3 cos  and √9 − 2 = 9 − 9sin2  = √9cos2  = 3|cos| = 3 cos.  √9−2 2  =  9sin 3cos2 3cos  = 9 sin2   = 9 1 2(1 − cos 2) = 9 2 − 1 2 sin 2 +  = 9 2 − 9 4(2 sin cos) +  = 9 2 sin−13 − 92 · 3 · √9 3− 2 +  = 9 2 sin−13 − 12√9 − 2 +  5. Let  = sec, where 0 ≤  ≤ 2 or  ≤   32 . Then  = sec tan  and √2 − 1 = √sec2  − 1 = √tan2  = |tan| = tan for the relevant values of , so  √24− 1  =  sec tan4 sec tan  =  tan2  cos3   =  sin2  cos  =s  2  = 1 33 +  = 1 3 sin3  +  = 1 3 √2− 13 +  = 1 3 (2 −31)32 +  6. Let  = 36 − 2, so  = −2 . When  = 0,  = 36; when  = 3,  = 27. Thus, 03 √36− 2  = 3627 √1−12  = −122√ 27 36 = − √27 − √36 = 6 − 3√3 [continued] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.24 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION Another method: Let  = 6sin, so  = 6cos ,  = 0 ⇒  = 0, and  = 3 ⇒  = 6 . Then 03 √36− 2  = 06 36(1 6sin − sin  2 ) 6cos  = 06 6cos 6sin 6cos  = 606 sin  = 6−cos  0 6 = 6− √23 + 1 = 6 − 3√3 7. Let  = tan, where   0 and − 2    2 . Then  =  sec2  ,  = 0 ⇒  = 0, and  =  ⇒  = 4 . Thus, 0 (2 + 2)32 = 04 [2(1 + tan sec2   2 )]32 = 04 sec 3 sec 2   3  = 12 04 cos   = 1 2 √22 − 0 = √212 . = 1 2 sin 0 4 8. Let  = 4 sec, where 0 ≤   2 or  ≤   32 . Then  = 4 sec tan  and √2 − 16 = √16 sec2  − 16 = √16 tan2  = 4 tan for the relevant values of , so  2√ 2 − 16 =  16 sec 4sec2tan · 4tan   = 16 1  sec 1   = 16 1  cos  = 1 16 sin +  = 1 16 √2 − 16  +  = √2 − 16 16 +  9. Let  = sec, so  = sec tan ,  = 2 ⇒  = 3 , and  = 3 ⇒  = sec−1 3. Then 23 (2 −1)32 =  sec 3 −1 3 sectan tan 3   =  sec 3 −1 3 sin cos2  s= √√3823 12  = −1√ √8 3 3 2 = √−38 + √23 = −34√2 + 23√3 10. Let  = 2 3 sin, so  = 2 3 cos ,  = 0 ⇒  = 0, and  = 2 3 ⇒  =  2 . Thus, 023 4 − 92  = 02 4 − 9 · 49 sin2  23 cos  = 02 2cos · 2 3 cos  = 43 02 cos2   = 4 3 02 1 2(1 + cos 2) = 23 + 12 sin 2 0 2 = 2 32 + 0 − (0 + 0) = 3 11. 012 √1 − 42  = 10 12 − 1 8    = 1 = −−8  42, = 1 8  2 3321 0 = 12 1 (1 − 0) = 12 1 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 25 12. Let  = 2tan, so  = 2sec2  ,  = 0 ⇒  = 0, and  = 2 ⇒  = 4 . Thus, 02 √4 +  2 = 04 √2sec 4 + 4 tan 2  2  = 04 2sec 2sec 2    = 04 sec  = ln|sec + tan| 0 4 = ln √2 + 1  − ln|1 + 0| = ln(√2 + 1) 13. Let  = 3 sec, where 0 ≤   2 or  ≤   32 . Then  = 3 sec tan  and √2 − 9 = 3 tan, so  √23− 9  =  27 sec 3tan3 3sec tan  = 13  tan sec22   = 1 3  sin2   = 1 3  1 2(1 − cos 2) = 1 6  − 12 1 sin 2 +  = 1 6  − 1 6 sin cos +  = 1 6 sec−13 − 16 √2− 9 3 +  = 1 6 sec−13 − √22−2 9 +  14. Let  = tan, so  = sec2  ,  = 0 ⇒  = 0, and  = 1 ⇒  = 4 . Then 01 (2 + 1)2 = 04 (tan sec22  + 1)2 = 04 (sec sec22  )2 = 04 cos2   = 04 1 2 (1 + cos 2)  = 1 2  + 1 2 sin 2 0 4 = 1 2( 4 + 1 2) − 0 = 8 + 1 4 15. Let  = sin,  = cos ,  = 0 ⇒  = 0 and  =  ⇒  = 2 . Then 0 2 √2 − 2  = 02 2 sin2  (cos)cos   = 4 02 sin2  cos2   = 4 02  1 2(2 sin cos)2  = 44 02 sin2 2  = 44 02 12(1 − cos 4) = 4 8  − 1 4 sin 4 0 2 = 84 2 − 0 − 0 = 16  4 16. Let  = 1 3 sec, so  = 1 3 sec tan ,  = √23 ⇒  = 4 ,  = 2 3 ⇒  = 3 . Then √2233 5 √9 2 − 1 =   43 131 3sec 5 sec  5tan  tan   = 34   43 cos4   = 81  43  1 2(1 + cos 2)2  = 81 4  43(1 + 2 cos 2 + cos2 2) = 81 4  43 1 + 2 cos 2 + 1 2(1 + cos 4)  = 81 4  43  3 2 + 2 cos 2 + 1 2 cos 4  = 81 4  3 2  + sin 2 + 1 8 sin 4 3 4 = 81 4  2 + √23 − √163 −  38 + 1 + 0 = 81 4  8 + 16 7 √3 − 1 17. Let  = 2 − 7, so  = 2 . Then  √2− 7  = 12  √1  = 1 2 · 2√ +  = 2 − 7 + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.26 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 18. Let  = sec, so ()2 = 2 sec2  ⇒ ()2 − 2 = 2 sec2  − 2 = 2(sec2  − 1) = 2 tan2 . So ()2 − 2 = tan,  =   sec tan , and  [()2 − 2]32 =   sec 3 tan  tan 3    =  12  tan sec2  = 1 2  cos sin2   =  12  csc cot  = − 1 2 csc +  = − 12 ( )2 − 2 +  = −  2()2 − 2 +  19. Let  = tan, where − 2    2 . Then  = sec2   and √1 + 2 = sec, so  √1 +  2  =  tan sec sec2   =  tan sec (1 + tan2 ) =  (csc + sec tan) = ln|csc − cot| + sec +  [by Exercise 7.2.39] = ln  √1 + 2  − 1   + √1 + 2 1 +  = ln  √1 + 2 − 1   + √1 + 2 +  20. Let  = 1 + 2, so  = 2 . Then  √1 +  2  =  √1 12  = 12  −12  = 1 2 · 212 +  = 1 + 2 +  21. Let  = 3 5 sin, so  = 3 5 cos ,  = 0 ⇒  = 0, and  = 06 ⇒  = 2 . Then 006 √9 −2252  = 02  3 53cos 2 sin2   3 5 cos  = 125 9 02 sin2   = 9 125 02 1 2(1 − cos 2) = 250 9  − 1 2 sin 2 0 2 = 9 250 2 − 0 − 0 = 500 9  22. Let  = tan, where − 2    2 . Then  = sec2  , √2 + 1 = sec and  = 0 ⇒  = 0,  = 1 ⇒  = 4 , so 01 √2 + 1 = 04 sec sec2   = 04 sec3   = 1 2 sec tan + ln|sec + tan|  0 4 [by Example 7.2.8] = 1 2 √2 · 1 + ln1 + √2 − 0 − ln(1 + 0) = 1 2√2 + ln1 + √2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 27 23.  √2 + 2  + 5 =  ( + 1)  2 + 4 =  √2sec 4tan22   + 4  + 1 = 2 tan  = 2 sec2  ,  =  2sec 2sec 2    =  sec  = ln|sec + tan| + 1 = ln  √2 + 2 + 5 2 +  + 1 2  + 1, or ln √2 + 2 + 5 +  + 1  + , where  = 1 − ln 2. 24. 01  − 2  = 01  1 4 − 2 −  + 1 4  = 01  1 4 −  − 1 22  = − 22  1 4 − 1 4 sin2  1 2 cos   − 1 2 == 1 21 2 sin cos  ,  = 202 1 2 cos 1 2 cos   = 1 2 02 cos2   = 1 2 02 1 2(1 + cos 2) = 1 4  + 1 2 sin 2 0 2 = 1 4 2  = 8 25.  2√3 + 2 − 2  =  24 − (2 + 2 + 1) =  222 − ( − 1)2  =  (1 + 2 sin)2√4cos2  2cos   −1 = 2 sin = 2 cos  ,  =  (1 + 4 sin + 4 sin2 )4cos2   = 4 (cos2  + 4 sin cos2  + 4 sin2  cos2 ) = 4 1 2(1 + cos 2) + 4 4sin cos2   + 4 (2 sin cos)2  = 2 (1 + cos 2) + 16 sin cos2   + 4 sin2 2  = 2 + 1 2 sin 2 + 16− 1 3 cos3  + 4 1 2(1 − cos 4) = 2 + sin 2 − 16 3 cos3  + 2 − 1 4 sin 4 +  = 4 − 1 2 sin 4 + sin 2 − 16 3 cos3  +  = 4 − 1 2(2 sin 2 cos 2) + sin 2 − 16 3 cos3  +  = 4 + sin 2(1 − cos 2) − 16 3 cos3  +  = 4 + (2 sin cos)(2 sin2 ) − 16 3 cos3  +  = 4 + 4 sin3  cos − 16 3 cos3  +  = 4 sin−1 −2 1 + 4 −2 13 √3 + 22 − 2 − 16 3 (3 + 22−3 2)32 +  = 4 sin−1 −2 1 + 1 4( − 1)3√3 + 2 − 2 − 2 3(3 + 2 − 2)32 +  26. 3 + 4 − 42 = −(42 − 4 + 1) + 4 = 22 − (2 − 1)2. Let 2 − 1 = 2 sin, so 2 = 2 cos  and √3 + 4 − 42 = 2 cos. Then °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.28 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION  (3 + 4−242)32  =   1 2(1 + 2 sin (2 cos)3)2 cos  = 1 32  1 + 4 sin cos  2+ 4 sin  2   = 32 1  (sec2  + 4 tan sec + 4 tan2 ) = 1 32  [sec2  + 4 tan sec + 4(sec2  − 1)]  = 1 32  (5 sec2  + 4 tan sec − 4) = 32 1 (5 tan + 4 sec − 4) +  = 1 325 · √3 + 4 2 −−1 42 + 4 · √3 + 42 − 42 − 4 · sin−122− 1 +  = 10 + 3 32√3 + 4 − 42 − 1 8 sin−122− 1 +  27. 2 + 2 = (2 + 2 + 1) − 1 = ( + 1)2 − 1. Let  + 1 = 1 sec, so  = sec tan  and √2 + 2 = tan. Then  √2 + 2  =  tan (sec tan ) =  tan2  sec  =  (sec2  − 1) sec  =  sec3   −  sec  = 1 2 sec tan + 1 2 ln|sec + tan| − ln|sec + tan| +  = 1 2 sec tan − 1 2 ln|sec + tan| +  = 1 2( + 1)√2 + 2 − 1 2 ln  + 1 + √2 + 2   +  28. 2 − 2 + 2 = (2 − 2 + 1) + 1 = ( − 1)2 + 1. Let  − 1 = 1 tan, so  = sec2   and √2 − 2 + 2 = sec. Then  (2 −22+ 1  + 2)2  =  (tansec + 1) 4 2 + 1 sec2   =  tan2  + 2 tan sec2   + 2  =  (sin2  + 2 sin cos + 2 cos2 ) =  (1 + 2 sin cos + cos2 ) =  1 + 2 sin cos + 1 2(1 + cos 2)  =   3 2 + 2 sin cos + 1 2 cos 2  = 3 2  + sin2  + 1 4 sin 2 +  = 3 2 + sin2  + 1 2 sin cos +  = 3 2 tan−1  −1 1 + 2(−−21)+ 2 2 + 12 √2−−21  + 2 √2 −12 + 2 +  = 3 2 tan−1( − 1) + 2(2 − 2 + 1) +  − 1 2(2 − 2 + 2) +  = 32 tan−1( − 1) + 2( 22 2 − − 3 2  + 1 + 2) +  We can write the answer as 3 2 tan−1( − 1) + (22 − 4 + 4) +  − 3 2(2 − 2 + 2) +  = 32 tan−1( − 1) + 1 + 2(2 −−23+ 2) +  = 3 2 tan−1( − 1) +  − 3 2(2 − 2 + 2) + 1, where 1 = 1 +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 29 29. Let  = 2,  = 2 . Then   √1 − 4  =  √1 − 2  1 2  = 1 2  cos · cos  whereand  = sin √1 −,  2 = cos = cos   ,  = 1 2  1 2(1 + cos 2) = 1 4 + 1 8 sin 2 +  = 1 4 + 1 4 sin cos  +  = 1 4 sin−1  + 1 4 √1 − 2 +  = 1 4 sin−1(2) + 1 42 √1 − 4 +  30. Let  = sin,  = cos . Then 02 1 + sin cos 2   = 01 √1 + 1 2  = 04 sec 1  sec2   where and = tan √1 +,  2 = sec = sec2  ,  = 04 sec  = ln|sec + tan|  0 4 [by (1) in Section 7.2] = ln√2 + 1 − ln(1 + 0) = ln√2 + 1 31. (a) Let  = tan, where − 2    2 . Then √2 + 2 = sec and  √2+ 2 =  sec sec 2    =  sec  = ln|sec + tan| + 1 = ln   √2+ 2 +      + 1 = ln + √2 + 2  +  where  = 1 − ln|| (b) Let  = sinh, so that  = cosh  and √2 + 2 = cosh. Then  √2+ 2 =  cosh cosh   =  +  = sinh−1  + . 32. (a) Let  = tan, − 2    2 . Then  =  (2 +2 2)32  =  23tan sec32  sec2   =  tan sec2  =  secsec 2  − 1  =  (sec − cos ) = ln|sec + tan| − sin +  = ln  √2 + 2  +    −  √2 + 2 +  = ln + √2 + 2  − √2+ 2 + 1 (b) Let  = sinh. Then  =  32cosh sinh23  cosh  =  tanh2   =  (1 − sech2 ) =  − tanh +  = sinh−1   −  √2 + 2 +  33. The average value of () = √2 − 1 on the interval [17] is 1 7 − 1 17 √2− 1  = 1 6 0 tan sec · sec tan  where √2 −= sec 1 = tan ,  , and = sec  = sec  tan−  1 7 ,  = 1 6 0 tan2   = 1 6 0(sec2  − 1) = 1 6 tan −  0 = 1 6 (tan − ) = 1 6 √48 − sec−1 7 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.30 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 34. 92 − 42 = 36 ⇒  = ± 3 2√2 − 4 ⇒ area = 223 3 2√2 − 4 = 323 √2 − 4 = 30 2tan 2sec tan   where  = 2 sec ,  = 2 sec  tan  ,  = sec−1 3 2   = 120 sec2  − 1sec  = 120 sec3  − sec  = 12 1 2(sec tan + ln|sec + tan|) − ln|sec + tan| 0 = 6sec tan − ln|sec + tan|  0 = 6 3√4 5 − ln 3 2 + √25  = 9√2 5 − 6ln 3 +2√5  35. Area of 4  = 1 2( cos)( sin) = 1 22 sin cos. Area of region   = cos  √2 − 2 . Let  =  cos ⇒  = − sin  for  ≤  ≤ 2 . Then we obtain  √2 − 2  =   sin(− sin) = −2  sin2   = − 1 22( − sin cos ) +  = − 1 2 2 cos−1() + 1 2 √2 − 2 +  so area of region  = 1 2 −2 cos−1() + √2 − 2   cos  = 1 2 0 − (−2 +  cos  sin) = 1 22 − 1 22 sin cos and thus, (area of sector ) = (area of 4) + (area of region  ) = 1 22. 36. Let  = √2sec, where 0 ≤   2 or  ≤   32 , so  = √2 sec tan . Then  4 √ 2 − 2 =  √4sec 2sec 4 √tan 2 tan    = 1 4  cos3   = 1 4  1 − sin2 cos   = 1 4 sin − 1 3 sin3  +  [substitute  =sin] = 1 4 √2− 2 − 2 − 32332  +  From the graph, it appears that our answer is reasonable. [Notice that () is large when  increases rapidly and small when  levels out.] 37. Use disks about the -axis:  = 03 29+ 92  = 81 03 (2 + 9) 1 2  Let  = 3 tan, so  = 3 sec2  ,  = 0 ⇒  = 0 and  = 3 ⇒  =  4 . Thus,  = 81 04 (9 sec 12 )2 3sec2   = 3 04 cos2   = 3 04 1 2(1 + cos 2) = 3 2  + 1 2 sin 2 0 4 = 32  4 + 1 2  − 0 = 3 82 + 3 4 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 31 38. Use shells about  = 1:  = 01 2(1 − )√1 − 2  = 2 01 √1 − 2  − 2 01 2√1 − 2  = 21 − 22 For 1, let  = 1 − 2, so  = −2 , and 1 = 10 √ − 1 2  = 1 2 01 12  = 1 2  2 3321 0 = 1 2  2 3  = 1 3. For 2, let  = sin, so  = cos , and 2 = 02 sin2  √cos2  cos  = 02 sin2  cos2   = 02 1 4(2 sin cos)2  = 1 4 02 sin2 2  = 1 4 02 1 2(1 − cos 2) = 1 8  − 1 2 sin 2 0 2 = 1 8  2  = 16  Thus,  = 2 1 3  − 2 16   = 2 3 − 1 82. 39. (a) Let  = sin,  = cos ,  = 0 ⇒  = 0 and  =  ⇒  = sin−1(). Then 0 2 − 2  = 0sin−1() cos (cos ) = 2 0sin−1() cos2   = 2 2 0sin−1()(1 + cos 2) = 22  + 1 2 sin 2sin 0 −1() = 22  + sin cossin 0 −1() = 2 2 sin−1  +  · √2− 2  − 0 = 1 22 sin−1() + 1 2 √2 − 2 (b) The integral 0 √2 − 2  represents the area under the curve  = √2 − 2 between the vertical lines  = 0 and  = . The figure shows that this area consists of a triangular region and a sector of the circle 2 + 2 = 2. The triangular region has base  and height √2 − 2, so its area is 1 2 √2 − 2. The sector has area 1 22 = 1 22 sin−1(). 40. The curves intersect when 2 +  1 222 = 8 ⇔ 2 + 1 44 = 8 ⇔ 4 + 42 − 32 = 0 ⇔ (2 + 8)(2 − 4) = 0 ⇔  = ±2. The area inside the circle and above the parabola is given by 1 = −22 √8 − 2 − 1 22  = 202 √8 − 2  − 202 122  = 2 1 2(8) sin−1 √28  + 1 2(2)√8 − 22 − 1 2  1 332 0  [by Exercise 39] = 8 sin−1 √12  + 2√4 − 8 3 = 8 4  + 4 − 8 3 = 2 + 4 3 Since the area of the disk is √82 = 8, the area inside the circle and below the parabola ia 2 = 8 − 2 + 4 3  = 6 − 4 3. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.32 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 41. We use cylindrical shells and assume that   . 2 = 2 − ( − )2 ⇒  = ±2 − ( − )2, so () = 22 − ( − )2 and  = −+ 2 · 22 − ( − )2  = − 4( + )√2 − 2  [where  =  − ] = 4 −  √2 − 2  + 4 − √2 − 2  where  =in the second integral  sin  ,  =  cos    = 4− 1 3(2 − 2)32 − + 4 −  22 2 cos2   = − 43(0 − 0) + 42 −  22 cos2   = 22 −  22(1 + cos 2) = 22 + 1 2 sin 2 − 2 2 = 222 Another method: Use washers instead of shells, so  = 8 0 2 − 2  as in Exercise 6.2.63(a), but evaluate the integral using  =  sin. 42. Let  = tan, so that  = sec2   and √2 + 2 = sec. () = −− 40(2 + 2)32  = 4 0 12 (sec 1 )3 sec2   =  40 12 sec 1   = 4 0 12 cos  = 4 0 sin 2 1 =  40√2+ 2  −−  = 4 0(−−)2 + 2 + √2+ 2  43. Let the equation of the large circle be 2 + 2 = 2. Then the equation of the small circle is 2 + ( − )2 = 2, where  = √2 − 2 is the distance between the centers of the circles. The desired area is  = −  + √2 − 2  − √2 − 2   = 20 + √2 − 2 − √2 − 2  = 20   + 20 √2 − 2  − 20 √2 − 2  The first integral is just 2 = 2 √2 − 2. The second integral represents the area of a quarter-circle of radius , so its value is 1 42. To evaluate the other integral, note that  √2 − 2  =  2 cos2   [ = sin,  = cos ] =  1 22 (1 + cos 2) = 1 2 2 + 1 2 sin 2 +  = 1 22( + sin cos) +  = 2 2 arcsin + 22 √2− 2 +  = 22 arcsin + 2 √2 − 2 +  Thus, the desired area is  = 2 √2 − 2 + 2 1 42 − 2 arcsin() + √2 − 2  0 = 2 √2 − 2 + 1 22 − 2 arcsin() +  √2 − 2  =  √2 − 2 + 2 2 − 2 arcsin() °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 33 44. Note that the circular cross-sections of the tank are the same everywhere, so the percentage of the total capacity that is being used is equal to the percentage of any cross-section that is under water. The underwater area is  = 2−25 25 − 2  = 25 arcsin(5) +  25 − 2 2 −5 [substitute  = 5 sin] = 25 arcsin 2 5 + 2√21 + 25 2  ≈ 5872 ft2 so the fraction of the total capacity in use is  (5)2 ≈ 5872 25 ≈ 0748 or 748%. 7.4 Integration of Rational Functions by Partial Fractions 1. (a) 4 +  (1 + 2)(3 − ) =  1 + 2 +  3 −  (b) 1 −  3 + 4 = 1 −  3(1 + ) =   +  2 +  3 +  1 +  2. (a)  − 6 2 +  − 6 =  − 6 ( + 3)( − 2) =   + 3 +   − 2 (b) 2 2 +  + 6 = (2 +  + 6) − ( + 6) 2 +  + 6 = 1 − 2 ++ 6  + 6 Notice that 2 +  + 6 can’t be factored because its discriminant is 2 − 4 = −23  0. 3. (a) 1 2 + 4 = 1 2(1 + 2) =   +  2 +  +  1 + 2 (b) 3 + 1 3 − 32 + 2 = (3 − 32 + 2) + 32 − 2 + 1 3 − 32 + 2 = 1 + 3(22 −− 23 + 1 + 2) [or use long division] = 1 + 32 − 2 + 1 ( − 1)( − 2) = 1 +  +  − 1 +  − 2 4. (a) 4 − 23 + 2 + 2 − 1 2 − 2 + 1 = 2(2 − 2 + 1) + 2 − 1 2 − 2 + 1 = 2 + (2−−1) 12 [or use long division] = 2 +   − 1 +  ( − 1)2 (b) 2 − 1 3 + 2 +  = 2 − 1 (2 +  + 1) =   +  +  2 +  + 1 5. (a) 6 2 − 4 = 4 + 42 + 16 + ( + 2)( 64 − 2) [by long division] = 4 + 42 + 16 +   + 2 +   − 2 (b) 4 (2 −  + 1)(2 + 2)2 =  +  2 −  + 1 +  +  2 + 2 +  +  (2 + 2)2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.34 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 6. (a) 6 + 1 6 + 3 = (6 + 3) − 3 + 1 6 + 3 = 1+ 3−(33 + 1 + 1) = 1+ 3( + 1)( −32+ 1 −  + 1) = 1+  + 2 + 3 +  + 1  + 2 −+ + 1  (b) 5 + 1 (2 − )(4 + 22 + 1) = 5 + 1 ( − 1)(2 + 1)2 =   +   − 1 +  +  2 + 1 +  +  (2 + 1)2 7.  −4 1  =  3 + 2 +  + 1 +  −1 1  [by division] = 144 + 1 33 + 122 +  + ln| − 1| +  8.  3+ 1 − 2  =  3 −  + 1 5   = 3 − 5ln| + 1| +  9. 5 + 1 (2 + 1)( − 1) =  2 + 1 +   − 1 . Multiply both sides by (2 + 1)( − 1) to get 5 + 1 = ( − 1) + (2 + 1) ⇒ 5 + 1 =  −  + 2 +  ⇒ 5 + 1 = ( + 2) + (− + ). The coefficients of  must be equal and the constant terms are also equal, so  + 2 = 5 and − +  = 1. Adding these equations gives us 3 = 6 ⇔  = 2, and hence,  = 1. Thus,  (2 + 1)( 5 + 1  − 1)  =  21+ 1 +  −2 1  = 1 2 ln|2 + 1| + 2 ln| − 1| + . Another method: Substituting 1 for  in the equation 5 + 1 = ( − 1) + (2 + 1) gives 6 = 3 ⇔  = 2. Substituting − 1 2 for  gives − 3 2 = − 3 2  ⇔  = 1. 10.  ( + 4)(2 − 1) =   + 4 +  2 − 1 . Multiply both sides by ( + 4)(2 − 1) to get  = (2 − 1) + ( + 4) ⇒  = 2 −  +  + 4 ⇒  = (2 + ) + (− + 4). The coefficients of  must be equal and the constant terms are also equal, so 2 +  = 1 and − + 4 = 0. Adding 2 times the second equation and the first equation gives us 9 = 1 ⇔  = 1 9 and hence,  = 4 9. Thus,  ( + 4)(2   − 1) =   + 4 4 9 + 2 1 9− 1  = 49 ln| + 4| + 1 9 · 12 ln|2 − 1| +  = 4 9 ln| + 4| + 18 1 ln|2 − 1| +  Another method: Substituting 1 2 for  in the equation  = (2 − 1) + ( + 4) gives 1 2 = 9 2  ⇔  = 1 9. Substituting −4 for  gives −4 = −9 ⇔  = 4 9. 11. 2 22 + 3 + 1 = 2 (2 + 1)( + 1) =  2 + 1 +   + 1 . Multiply both sides by (2 + 1)( + 1) to get 2 = ( + 1) + (2 + 1). The coefficients of  must be equal and the constant terms are also equal, so  + 2 = 0 and  +  = 2. Subtracting the second equation from the first gives  = −2, and hence,  = 4. Thus, 01 22 + 3 2  + 1  = 01 24+ 1 −  + 1 2   = 42 ln|2 + 1| − 2ln| + 1|1 0 = (2ln3 − 2ln2) − 0 = 2ln 3 2. Another method: Substituting −1 for  in the equation 2 = ( + 1) + (2 + 1) gives 2 = − ⇔  = −2. Substituting − 1 2 for  gives 2 = 1 2  ⇔  = 4. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 35 12.  − 4 2 − 5 + 6 =   − 2 +   − 3 . Multiply both sides by ( − 2)( − 3) to get  − 4 = ( − 3) + ( − 2) ⇒  − 4 =  − 3 +  − 2 ⇒  − 4 = ( + ) + (−3 − 2). The coefficients of  must be equal and the constant terms are also equal, so  +  = 1 and −3 − 2 = −4. Adding twice the first equation to the second gives us − = −2 ⇔  = 2, and hence,  = −1.Thus, 01 2 − −54+ 6  = 01  −2 2 −  −1 3  = [2 ln| − 2| − ln| − 3|]1 0 = (0 − ln 2) − (2 ln 2 − ln 3) = −3ln 2 + ln3 [or ln 3 8 ] Another method: Substituting 3 for  in the equation  − 4 = ( − 3) + ( − 2) gives −1 = . Substituting 2 for  gives −2 = − ⇔  = 2. 13.  2 −   =  ( − )  =   −   = ln| − | +  14. If  6= , 1 ( + )( + ) = 1  −  +1  −  +1 , so if  6= , then  ( +  )( + ) =  −1  (ln| + | − ln| + |) +  =  −1  ln    ++     +  If  = , then  ( +)2 = − +1  + . 15. 3 − 4 + 1 2 − 3 + 2 =  + 3 + ( −31)( −5− 2). Write ( −31)( −5− 2) =  − 1 +  − 2. Multiplying both sides by ( − 1)( − 2) gives 3 − 5 = ( − 2) + ( − 1). Substituting 2 for  gives 1 = . Substituting 1 for  gives −2 = − ⇔  = 2. Thus, −01  3 2 − − 4 3  + 1 + 2  = −01  + 3 +  −2 1 +  −1 2  =  1 22 + 3 + 2 ln| − 1| + ln| − 2|0 −1 = (0 + 0 + 0 + ln 2) −  1 2 − 3 + 2ln2 + ln3 = 5 2 − ln 2 − ln 3, or 5 2 − ln 6 16. 3 + 42 +  − 1 3 + 2 = 1 + 322(+ + 1)  − 1. Write 322(+ + 1)  − 1 =  + 2 +  + 1. Multiplying both sides by 2( + 1) gives 32 +  − 1 = ( + 1) + ( + 1) + 2. Substituting 0 for  gives −1 = . Substituting −1 for  gives 1 = . Equating coefficients of 2 gives 3 =  +  =  + 1, so  = 2. Thus, 12 3 + 4 3+2 +2 − 1  = 12 1 + 2 − 12 +  + 1 1   =  + 2 ln|| + 1 + ln| + 1|2 1 = 2 + 2 ln 2 + 1 2 + ln 3 − (1 + 0 + 1 + ln 2) = 1 2 + ln 2 + ln 3, or 1 2 + ln 6. 17. 42 − 7 − 12 ( + 2)( − 3) =   +   + 2 +   − 3 ⇒ 42 − 7 − 12 = ( + 2)( − 3) + ( − 3) + ( + 2). Setting  = 0 gives −12 = −6, so  = 2. Setting  = −2 gives 18 = 10, so  = 9 5 . Setting  = 3 gives 3 = 15, so  = 1 5 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.36 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION Now 12 4(2+ 2)( − 7−−12 3)  = 122 + 9+ 2 5 + 1−53  = 2ln|| + 9 5 ln| + 2| + 1 5 ln| − 3| 2 1 = 2 ln 2 + 9 5 ln 4 + 1 5 ln 1 − 2ln1 − 9 5 ln 3 − 1 5 ln 2 = 2 ln 2 + 18 5 ln 2 − 1 5 ln 2 − 9 5 ln 3 = 27 5 ln 2 − 9 5 ln 3 = 9 5(3 ln 2 − ln 3) = 9 5 ln 8 3 18. 32 + 6 + 2 2 + 3 + 2 = 3 + ( + 1)( −3 − + 2) 4 . Write ( + 1)( −3 − + 2) 4 =  + 1 +  + 2. Multiplying both sides by ( + 1)( + 2) gives −3 − 4 = ( + 2) + ( + 1). Substituting −2 for  gives 2 = − ⇔  = −2. Substituting −1 for  gives −1 = . Thus, 12 322+ 3 + 6+ 2 + 2  = 12 3 −  + 1 1 −  + 2 2   = 3 − ln| + 1| − 2ln| + 2|2 1 = (6 − ln 3 − 2ln4) − (3 − ln 2 − 2 ln 3) = 3 + ln 2 + ln 3 − 2ln4, or 3 + ln 3 8 19. 2 +  + 1 ( + 1)2( + 2) =   + 1 +  ( + 1)2 +   + 2 . Multiplying both sides by ( + 1)2( + 2) gives 2 +  + 1 = ( + 1)( + 2) + ( + 2) + ( + 1)2. Substituting −1 for  gives 1 =  Substituting −2 for  gives 3 = . Equating coefficients of 2 gives 1 =  +  =  + 3, so  = −2. Thus, 01 (+ 1) 2 +2(+ 1 + 2)  = 01 −+ 1 2 + ( + 1) 1 2 +  + 2 3   = −2ln| + 1| −  + 1 1 + 3 ln| + 2|1 0 = −2ln2 − 1 2 + 3 ln 3 − (0 − 1 + 3 ln 2) = 1 2 − 5ln2 + 3 ln 3, or 1 2 + ln 27 32 20. (3 − 5) (3 − 1)( − 1)2 =  3 − 1 +   − 1 +  ( − 1)2 . Multiplying both sides by (3 − 1)( − 1)2 gives (3 − 5) = ( − 1)2 + ( − 1)(3 − 1) + (3 − 1). Substituting 1 for  gives −2 = 2 ⇔  = −1. Substituting 1 3 for  gives 4 9 = 4 9 ⇔  = 1. Substituting 0 for  gives 0 =  +  −  = 1 +  + 1, so  = −2. Thus, 23 (3−(31)( −5−)1)2  = 23 31− 1 −  −2 1 − ( −11)2   = 13 ln|3 − 1| − 2ln| − 1| +  −1 13 2 =  1 3 ln 8 − 2ln2 + 1 2 −  1 3 ln 5 − 0 + 1 = −ln 2 − 1 3 ln 5 − 1 2 21. 1 (2 − 1)2 = 1 ( + 1)2( − 1)2 =   + 1 +  ( + 1)2 +   − 1 +  ( − 1)2 . Multiplying both sides by ( + 1)2( − 1)2 gives 1 = ( + 1)( − 1)2 + ( − 1)2 + ( − 1)( + 1)2 + ( + 1)2. Substituting 1 for  gives 1 = 4 ⇔  = 1 4. Substituting −1 for  gives 1 = 4 ⇔  = 1 4. Substituting 0 for  gives 1 =  +  −  +  =  + 1 4 −  + 1 4, so 1 2 =  − . Equating coefficients of 3 gives 0 =  + . Adding the last two equations gives 2 = 1 2 ⇔  = 1 4, and so  = − 1 4. Thus,  (2 −1)2 =  1+ 1 4 + (1+ 1) 4 2 − 1−41 + (1−41)2   = 1 4 ln| + 1| −  + 1 1 − ln| − 1| −  −1 1 + , or 1 4ln     + 1 − 1    + 1 −22  +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 37 22.  4 + 92 2+ 9 +  + 2  =  2 + 2+ 2 + 9  =  2 + 2+ 9 + 22+ 9  = 1 3 3 + 1 2 ln(2 + 9) + 2 3 tan−1  3 +  23. 10 ( − 1)(2 + 9) =   − 1 +  +  2 + 9 . Multiply both sides by ( − 1)2 + 9 to get 10 = 2 + 9 + ( + )( − 1) (). Substituting 1 for  gives 10 = 10 ⇔  = 1. Substituting 0 for  gives 10 = 9 −  ⇒  = 9(1) − 10 = −1. The coefficients of the 2-terms in () must be equal, so 0 =  +  ⇒  = −1 Thus,  ( − 1)( 102 + 9)  =   −1 1 + −2+ 9 − 1  =   −1 1 − 2+ 9 − 21+ 9  = ln| − 1| − 1 2 ln(2 + 9) − 1 3 tan−1 3  +  In the second term we used the substitution  = 2 + 9 and in the last term we used Formula 10. 24. 2 −  + 6 3 + 3 = 2 −  + 6 (2 + 3) =   +  +  2 + 3 . Multiply by 2 + 3 to get 2 −  + 6 = 2 + 3 + ( + ). Substituting 0 for  gives 6 = 3 ⇔  = 2. The coefficients of the 2-terms must be equal, so 1 =  +  ⇒  = 1 − 2 = −1. The coefficients of the -terms must be equal, so −1 = . Thus,  23−+ 3  + 6   =  2 + −2+ 3 − 1  =  2 − 2+ 3 − 21+ 3  = 2 ln|| − 1 2 ln2 + 3 − √13 tan−1 √3 +  25. 4 3 + 2 +  + 1 = 4 2( + 1) + 1( + 1) = 4 ( + 1)(2 + 1) =   + 1 +  +  2 + 1 . Multiply both sides by ( + 1)(2 + 1) to get 4 = (2 + 1) + ( + )( + 1) ⇔ 4 = 2 +  + 2 +  +  +  ⇔ 4 = ( + )2 + ( + ) + ( + ). Comparing coefficients gives us the following system of equations:  +  = 0 (1)  +  = 4 (2)  +  = 0 (3) Subtracting equation (1) from equation (2) gives us − +  = 4, and adding that equation to equation (3) gives us 2 = 4 ⇔  = 2, and hence  = −2 and  = 2. Thus,  3 + 24+  + 1  =  −+ 1 2 + 22 + 1 + 2  =  −+ 1 2 + 22+ 1  + 22+ 1  = −2ln| + 1| + ln(2 + 1) + 2 tan−1  +  26.  (2+ 2 + 1)  + 1 2  =  (22+ 1) + 1 2  +  (2 + 1)  2  =  21+ 1  + 12  12   = 2 + 1,  = 2  = tan−1  + 1 2 −1 +  = tan−1  − 2(21+ 1) +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.38 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 27. 3 + 4 + 3 4 + 52 + 4 = 3 + 4 + 3 (2 + 1)(2 + 4) =  +  2 + 1 +  +  2 + 4 . Multiply both sides by (2 + 1)(2 + 4) to get 3 + 4 + 3 = ( + )(2 + 4) + ( + )(2 + 1) ⇔ 3 + 4 + 3 = 3 + 2 + 4 + 4 + 3 + 2 +  +  ⇔ 3 + 4 + 3 = ( + )3 + ( + )2 + (4 + ) + (4 + ). Comparing coefficients gives us the following system of equations:  +  = 1 (1)  +  = 0 (2) 4 +  = 4 (3) 4 +  = 3 (4) Subtracting equation (1) from equation (3) gives us  = 1 and hence,  = 0. Subtracting equation (2) from equation (4) gives us  = 1 and hence,  = −1. Thus,  43+ 5 + 42+ 3 + 4  =  2+ 1 + 1 + 2−+ 4 1   =  2+ 1 + 21+ 1 − 21+ 4  = 1 2 ln(2 + 1) + tan−1  − 1 2 tan−12 +  28. 3 + 6 − 2 4 + 62 = 3 + 6 − 2 2(2 + 6) =   +  2 +  +  2 + 6 . Multiply both sides by 2(2 + 6) to get 3 + 6 − 2 = (2 + 6) + (2 + 6) + ( + )2 ⇔ 3 + 6 − 2 = 3 + 6 + 2 + 6 + 3 + 2 ⇔ 3 + 6 − 2 = ( + )3 + ( + )2 + 6 + 6. Substituting 0 for  gives −2 = 6 ⇔  = − 1 3 . Equating coefficients of 2 gives 0 =  + , so  = 1 3 . Equating coefficients of  gives 6 = 6 ⇔  = 1. Equating coefficients of 3 gives 1 =  + , so  = 0. Thus,  34+ 6 + 6 −2 2  =  1 + −123 + 21+ 6 3   = ln|| + 31 + 3√16 tan−1√6 + . 29.  2 + 2  + 4  + 5  =  2 + 2  + 1  + 5  +  2 + 23 + 5  = 12  (22+ 2 + 2)  + 5  +  ( + 1) 32 + 4 = 1 2 ln 2 + 2 + 5  + 3 4(22 + 1) where and   + 1 = 2 = 2 ,  = 1 2 ln(2 + 2 + 5) + 3 2 tan−1  +  = 1 2 ln(2 + 2 + 5) + 3 2 tan−1 + 1 2  +  30. 3 − 22 + 2 − 5 4 + 42 + 3 = 3 − 22 + 2 − 5 (2 + 1)(2 + 3) =  +  2 + 1 +  +  2 + 3 . Multiply both sides by (2 + 1)(2 + 3) to get 3 − 22 + 2 − 5 = ( + )(2 + 3) + ( + )(2 + 1) ⇔ 3 − 22 + 2 − 5 = 3 + 2 + 3 + 3 + 3 + 2 +  +  ⇔ 3 − 22 + 2 − 5 = ( + )3 + ( + )2 + (3 + ) + (3 + ). Comparing coefficients gives us the following system of equations:  +  = 1 (1)  +  = −2 (2) 3 +  = 2 (3) 3 +  = −5 (4) Subtracting equation (1) from equation (3) gives us 2 = 1 ⇔  = 1 2 , and hence,  = 1 2 . Subtracting equation (2) from equation (4) gives us 2 = −3 ⇔  = − 3 2 , and hence,  = − 1 2 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 39 Thus,  3 −42+ 4 2 + 2 2 + 3  − 5  =   1 22 + 1 − 3 2 + 1 22 + 3 − 1 2   =  21 2+ 1  − 2 3 2+ 1 + 21 2+ 3  − 2 1 2+ 3  = 1 4 ln(2 + 1) − 3 2 tan−1  + 1 4 ln(2 + 3) − 1 2√3 tan−1 √3 +  31. 1 3 − 1 = 1 ( − 1)(2 +  + 1) =   − 1 +  +  2 +  + 1 ⇒ 1 = 2 +  + 1 + ( + )( − 1). Take  = 1 to get  = 1 3. Equating coefficients of 2 and then comparing the constant terms, we get 0 = 1 3 + , 1 = 1 3 − , so  = − 1 3,  = − 2 3 ⇒  31− 1 =   −1 3 1  +  −2 1 3+−+ 1 2 3  = 1 3 ln| − 1| − 13  2 ++ 2  + 1 = 1 3 ln| − 1| − 1 3  2++ 1 + 1 2  − 13  ( + 1 (32) 2)2 + 34 = 1 3 ln| − 1| − 1 6 ln2 +  + 1 − 1 2√23tan−1 √+31 22 +  = 1 3 ln| − 1| − 1 6 ln(2 +  + 1) − √13 tan−1 √13(2 + 1) +  32. 01 2 + 4 + 13  = 01 21 2+ 4 (2+ 4) + 13  − 201 ( + 2) 2 + 9 = 1 2 1318   − 2213 932 + 9 where  = + 2 = 3 2 + 4+ 13 , and,  = 3 = (2  + 4) ,  = 1 2 ln18 13 − 2 3 tan−1 1 23 = 1 2 ln 18 13 − 2 3 4 − tan−1 2 3 = 1 2 ln 18 13 − 6 + 2 3 tan−1 2 3 33. Let  = 4 + 42 + 3 so that  = (43 + 8) = 4(3 + 2),  = 0 ⇒  = 3, and  = 1 ⇒  = 8. Then 01 4+ 4 3 + 2 2+ 3  = 38 114  = 14ln|| 8 3 = 14(ln 8 − ln 3) = 1 4 ln 83. 34. 5 +  − 1 3 + 1 = 2 + −2 3++ 1  − 1 = 2 + ( + 1)( −2 +2−−1+ 1) = 2 + −+ 1 1 , so  5+3 + 1  − 1  =  2 −  + 1 1   = 133 − ln| + 1| +  35. 54 + 72 +  + 2 (2 + 1)2 =   +  +  2 + 1 +  +  (2 + 1)2 . Multiply by (2 + 1)2 to get 54 + 72 +  + 2 = (2 + 1)2 + ( + )(2 + 1) + ( + ) ⇔ 54 + 72 +  + 2 = (4 + 22 + 1) + (2 + )(2 + 1) + 2 +  ⇔ 54 + 72 +  + 2 = 4 + 22 +  + 4 + 3 + 2 +  + 2 +  ⇔ 54 + 72 +  + 2 = ( + )4 + 3 + (2 +  + )2 + ( + ) + . Equating coefficients gives us  = 0, °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.40 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION  = 2,  +  = 5 ⇒  = 3,  +  = 1 ⇒  = 1, and 2 +  +  = 7 ⇒  = 0. Thus,  54 + 7 (22+ 1) + 2 + 2  =  2 + 23+ 1  + (2 + 1) 1 2   =  Now  (2 + 1)2 =  (tan sec22  + 1)2   = tan = sec2  ,  =  sec sec2 4    =  cos2   =  1 2(1 + cos 2) = 1 2  + 1 4 sin 2 +  = 1 2 + 1 2 sin cos +  = 1 2 tan−1  + 1 2  √2 + 1 1 √2 + 1 +  Therefore,  = 2 ln|| + 3 2 ln(2 + 1) + 1 2 tan−1  +  2(2 + 1) + . 36. Let  = 5 + 53 + 5, so that  = (54 + 152 + 5) = 5(4 + 32 + 1). Then  54+ 5 + 332+ 5 + 1  =  115  = 1 5 ln|| +  = 1 5 ln 5 + 53 + 5  +  37. 2 − 3 + 7 (2 − 4 + 6)2 =  +  2 − 4 + 6 +  +  (2 − 4 + 6)2 ⇒ 2 − 3 + 7 = ( + )(2 − 4 + 6) +  +  ⇒ 2 − 3 + 7 = 3 + (−4 + )2 + (6 − 4 + ) + (6 + ). So  = 0, −4 +  = 1 ⇒  = 1, 6 − 4 +  = −3 ⇒  = 1, 6 +  = 7 ⇒  = 1. Thus,  =  (22−−43+ 6) + 7 2  =  2 − 41 + 6 + (2 −4+ 1  + 6)2   =  ( − 2) 1 2 + 2  +  (2 −4−2+ 6)2  +  (2 − 43 + 6)2  = 1 + 2 + 3. 1 =  ( − 2)21+ √22  = √12 tan−1√−22 + 1 2 = 1 2  (2 −24− + 6) 4 2  = 1 2  12  = 1 2−1 + 2 = −2(2 −14 + 6) + 2 3 = 3 ( − 2)2 1+ √222  = 3 [2(tan21 + 1)]2 √2sec2   −=2 = √2 sec √2 tan 2   ,  = 3√2 4  sec sec2 4    = 3√4 2  cos2   = 3√4 2  1 2(1 + cos 2) = 3√2 8  + 1 2 sin 2 + 3 = 3√8 2 tan−1√−22 + 3√8 2  1 2 · 2sin cos + 3 = 3√2 8 tan−1√−22 + 3√8 2 · √2−−42 + 6 · √2 −√42 + 6 + 3 = 3√2 8 tan−1√−22 + 4(3( 2 − −42) + 6) + 3 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 41 So  = 1 + 2 + 3 [ = 1 + 2 + 3] = 1 √ 2 tan−1√−22 + 2(2 −−41 + 6) + 3√8 2 tan−1√−22 + 4(3( 2 − −42) + 6) +  = 4√8 2 + 3√8 2tan−1√−22 + 4(3( 2 −−42) −+ 6) 2 +  = 7√8 2 tan−1√−22 + 4(23− −48+ 6) +  38. 3 + 22 + 3 − 2 (2 + 2 + 2)2 =  +  2 + 2 + 2 +  +  (2 + 2 + 2)2 ⇒ 3 + 22 + 3 − 2 = ( + )(2 + 2 + 2) +  +  ⇒ 3 + 22 + 3 − 2 = 3 + (2 + )2 + (2 + 2 + ) + 2 + . So  = 1, 2 +  = 2 ⇒  = 0, 2 + 2 +  = 3 ⇒  = 1, and 2 +  = −2 ⇒  = −2. Thus,  =  3(+ 2 2 + 2 2 + 3 + 2)  −2 2  =  2 + 2  + 2 + (2 + 2  −2+ 2)2   =  2 + 2  + 1  + 2  +  2 + 2 −1 + 2  +  (2 + 2  + 1  + 2)2  +  (2 + 2 −3+ 2)2  = 1 + 2 + 3 + 4. 1 =  2 + 2  + 1  + 2  =  112   = = 2( 2+ 2 + 1)  + 2  , = 12 ln 2 + 2 + 2  + 1 2 = −  ( + 1) 1 2 + 1  = −11 tan−1 + 1 1  + 2 = −tan−1( + 1) + 2 3 =  (2 + 2  + 1  + 2)2  =  12 1 2  = −21 + 3 = −2(2 + 2 1  + 2) + 3 4 = −3 [( + 1) 12 + 1]2  = −3 (tan2 1+ 1)2 sec2    + 1 = 1 tan = sec2  , = −3 sec12   = −3 cos2   = −3 2  (1 + cos 2) = − 3 2  + 1 2 sin 2 + 4 = − 3 2  − 3 2 1 2 · 2sin cos + 4 = − 3 2 tan−1 + 1 1  − 3 2 · √2+ 2 + 1  + 2 · √2 + 2 1  + 2 + 4 = − 3 2 tan−1( + 1) − 3( + 1) 2(2 + 2 + 2) + 4 So  = 1 + 2 + 3 + 4 [ = 1 + 2 + 3 + 4] = 1 2 ln(2 + 2 + 2) − tan−1( + 1) − 1 2(2 + 2 + 2) − 3 2 tan−1( + 1) − 3( + 1) 2(2 + 2 + 2) +  = 1 2 ln(2 + 2 + 2) − 5 2 tan−1( + 1) − 3 + 4 2(2 + 2 + 2) +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.42 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 39.  √  − 1 =  (22+ 1)  2 == √  −−1,1,  == 2   2 + 1  = 2 21+ 1  = 2 tan−1  +  = 2 tan−1 √ − 1 +  40. Let  = √ + 3, so 2 =  + 3 and 2  = . Then  2√  + 3 +  =  2 + ( 2  2 − 3) =  2 + 2 2 − 3  =  ( + 3)( 2 − 1) . Now 2 ( + 3)( − 1) =   + 3 +   − 1 ⇒ 2 = ( − 1) + ( + 3). Setting  = 1 gives 2 = 4, so  = 1 2 . Setting  = −3 gives −6 = −4, so  = 3 2 . Thus,  ( + 3)( 2 − 1)  =   + 3 3 2 +  −1 2 1  = 3 2 ln| + 3| + 1 2 ln| − 1| +  = 3 2 ln√ + 3 + 3 + 1 2 ln √ + 3 − 1  +  41. Let  = √, so 2 =  and 2  = . Then  2 +√ =  24  + 3 =  32+2 =  2(2 + 1). 2 2( + 1) =   +  2 +   + 1 ⇒ 2 = ( + 1) + ( + 1) + 2. Setting  = 0 gives  = 2. Setting  = −1 gives  = 2. Equating coefficients of 2, we get 0 =  + , so  = −2. Thus,  2(2 + 1) =  −2 + 22 +  + 1 2   = −2ln||− 2 +2 ln| + 1|+ = −2ln√− √2 +2 ln√ + 1+. 42. Let  = √3 . Then  = 3,  = 32  ⇒ 01 1 +1√3   = 01 31 + 2   = 01 3 − 3 + 1 +3   =  3 22 − 3 + 3 ln(1 + )1 0 = 3ln 2 − 1 2. 43. Let  = √3 2 + 1. Then 2 = 3 − 1, 2  = 32  ⇒  √332 + 1 =  (3 − 1)  3 22  = 3 2  (4 − ) = 3 105 − 3 42 +  = 10 3 (2 + 1)53 − 3 4(2 + 1)23 +  44.  (1 + √)2 =  2(−2 1)   = ( = 1 +  −√1) 2, ,  = 2( − 1)   = 2 1 − 12   = 2 ln|| + 2 +  = 2 ln(1 + √) + 1 +2√ +  45. If we were to substitute  = √, then the square root would disappear but a cube root would remain. On the other hand, the substitution  = √3  would eliminate the cube root but leave a square root. We can eliminate both roots by means of the substitution  = √6 . (Note that 6 is the least common multiple of 2 and 3.) Let  = √6 . Then  = 6, so  = 65  and √ = 3, √3  = 2. Thus, °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 43  √ − √3  =  63−5  2 = 6 2(5− 1)  = 6 −3 1  = 6 2 +  + 1 +  −1 1  [by long division] = 6 1 33 + 1 22 +  + ln| − 1| +  = 2√ + 3 √3  + 6 √6  + 6 ln  √6  − 1   +  46. Let  = 1 + √, so that 2 = 1 + √,  = (2 − 1)2, and  = 2(2 − 1) · 2  = 4(2 − 1). Then  1 + √  =  (2 − 1)2 · 4(2 − 1) =  24−2 1  =  4 + 24− 1 . Now 4 2 − 1 =   + 1 +   − 1 ⇒ 4 = ( − 1) + ( + 1). Setting  = 1 gives 4 = 2, so  = 2. Setting  = −1 gives 4 = −2, so  = −2. Thus,  4 + 24− 1  =  4 −  + 1 2 +  −2 1  = 4 − 2ln| + 1| + 2 ln| − 1| +  = 41 + √ − 2ln1 + √ + 1 + 2 ln1 + √ − 1 +  47. Let  = . Then  = ln,  =   ⇒  2 + 3 2   + 2 =  22+ 3 (  + 2 ) =  ( + 1)(   + 2) =  −+ 1 1 +  + 2 2   = 2 ln| + 2| − ln| + 1| +  = ln ( + 2)2  + 1 +  48. Let  = cos, so that  = −sin . Then  cos2 sin −3cos   =  2 −1 3 (−) =  (−−1 3) . −1 ( − 3) =   +   − 3 ⇒ −1 = ( − 3) + . Setting  = 3 gives  = − 1 3. Setting  = 0 gives  = 1 3. Thus,  (−−1 3)  =   1 3 −  −1 3 3  = 13 ln|| − 13 ln| − 3| +  = 1 3 ln|cos| − 1 3 ln|cos  − 3| + . 49. Let  = tan, so that  = sec2  . Then  tan2  + 3 tan sec2   + 2  =  2 + 31 + 2  =  ( + 1)( 1 + 2) . Now 1 ( + 1)( + 2) =   + 1 +   + 2 ⇒ 1 = ( + 2) + ( + 1). Setting  = −2 gives 1 = −, so  = −1. Setting  = −1 gives 1 = . Thus, ( + 1)( 1 + 2)  =   + 1 1 −  + 2 1   = ln| + 1|−ln| + 2|+ = ln|tan + 1|−ln|tan + 2|+. 50. Let  = , so that  =  . Then  ( − 2)( 2 + 1)  =  ( − 2)( 12 + 1) . Now 1 ( − 2)(2 + 1) =   − 2 +  +  2 + 1 ⇒ 1 = (2 + 1) + ( + )( − 2). Setting  = 2 gives 1 = 5, so  = 1 5. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.44 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION Setting  = 0 gives 1 = 1 5 − 2, so  = − 2 5. Comparing coefficients of 2 gives 0 = 1 5 + , so  = − 1 5. Thus,  ( − 2)( 12 + 1)  =   −1 5 2 + −1 52+ 1 − 2 5   = 1 5   −1 2  − 1 5  2+ 1  − 2 5  21+ 1  = 1 5 ln| − 2| − 1 5 · 1 2 ln 2 + 1  − 2 5 tan−1  +  = 1 5 ln| − 2| − 10 1 ln(2 + 1) − 2 5 tan−1  +  51. Let  = , so that  =   and  =   . Then  1 + =  (1 +). (1+ 1) =  +  + 1 ⇒ 1 = ( + 1) + . Setting  = −1 gives  = −1. Setting  = 0 gives  = 1. Thus,  ( + 1) =  1 −  + 1 1   = ln|| − ln| + 1| +  = ln − ln( + 1) +  =  − ln( + 1) + . 52. Let  = sinh, so that  = cosh . Then  sinh2cosh  + sinh  4   =  2 +1 4  =  2(21+ 1) . 1 2(2 + 1) =   +  2 +  +  2 + 1 ⇒ 1 = (2 + 1) + (2 + 1) + ( + )2. Setting  = 0 gives  = 1. Comparing coefficients of 2, we get 0 =  + , so  = −1. Comparing coefficients of , we get 0 = . Comparing coefficients of 3, we get 0 =  + , so  = 0. Thus,  2(21+ 1)  =  12 − 21+ 1  = −1 − tan−1  +  = −sinh 1  − tan−1(sinh) +  = −csch − tan−1(sinh) +  53. Let  = ln(2 −  + 2),  = . Then  = 2 − 1 2 −  + 2 ,  = , and (by integration by parts)  ln(2 −  + 2) = ln(2 −  + 2) −  22−2 −+ 2   = ln(2 −  + 2) −  2 + 2 −−4+ 2  = ln(2 −  + 2) − 2 −  1 22(2−−+ 2 1)  + 72  ( −  1 2)2 + 7 4 = ln(2 −  + 2) − 2 − 1 2 ln(2 −  + 2) + 7 2  √7 2  7 4 (2 + 1)  where  − 1 2 = √7 2 ,  = √7 2 , ( − 1 2 )2 + 7 4 = 7 4 (2 + 1)  = ( − 1 2)ln(2 −  + 2) − 2 + √7tan−1  +  = ( − 1 2)ln(2 −  + 2) − 2 + √7tan−1 2√−7 1 +  54. Let  = tan−1 ,  =   ⇒  = (1 + 2),  = 1 22. Then  tan−1   = 1 22 tan−1  − 12  1 +22 . To evaluate the last integral, use long division or observe that  1 +22  =  (1 +1 +2)2− 1  =  1 −  1 +12  =  − tan−1  + 1. So  tan−1   = 1 22 tan−1  − 1 2( − tan−1  + 1) = 1 2(2 tan−1  + tan−1  − ) + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 45 55. From the graph, we see that the integral will be negative, and we guess that the area is about the same as that of a rectangle with width 2 and height 03, so we estimate the integral to be −(2 · 03) = −06. Now 1 2 − 2 − 3 = 1 ( − 3)( + 1) =   − 3 +   + 1 ⇔ 1 = ( + ) +  − 3, so  = − and  − 3 = 1 ⇔  = 1 4 and  = − 1 4, so the integral becomes 02 2 − 2 − 3 = 14 02  − 3 − 14 02  + 1 = 14ln| − 3| − ln| + 1|2 0 = 1 4 ln    −+ 1 3    2 0 = 1 4 ln 1 3 − ln 3 = − 1 2 ln 3 ≈ −055 56.  = 0:  2 +  =   2 = −1 +    0:  2 +  =  2 + ( √ )2 = √1 tan−1√  +    0:  2 +  =  2 −(−) =  2 −  √− 2 = 2√1− ln     − + √ √− −      +  [by Example 3] 57.  2  − 2 =  ( − 1)2 − 1 =  2 − 1 [put  =  − 1] = 1 2 ln   − 1  + 1  +  [by Equation 6] = 1 2 ln   − 2   +  58.  4(22+ 12 + 1)  − 7 = 14  4(8 2+ 12 + 12)  −7 −  (2 + 3) 22 − 16 = 1 4 ln 42 + 12 − 7  −  2 − 16 [put  = 2 + 3] = 1 4 ln 42 + 12 − 7  − 1 8 ln|( − 4)( + 4)| +  [by Equation 6] = 1 4 ln 42 + 12 − 7  − 1 8 ln|(2 − 1)(2 + 7)| +  59. (a) If  = tan2, then 2 = tan−1 . The figure gives cos2 = √1 + 1 2 and sin2 = √1 +  2 . (b) cos = cos2 · 2 = 2 cos22 − 1 = 2√1 + 1 2 2 − 1 = 1 +2 2 − 1 = 1 1 +−  2 2 (c)  2 = arctan ⇒  = 2 arctan ⇒  = 2 1 + 2  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.46 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 60. Let  = tan(2). Then, by using the expressions in Exercise 59 we have  1 − cos =  1 − (1 2 − (1 + 2)(1 + 2) 2) =  (1 + 2)2−(1 − 2) =  22 2 =  12  = − 1  +  = − 1 tan(2) +  = −cot(2) +  Another method:  1 − cos =  1 −1cos · 1 + cos 1 + cos    =  11 + cos − cos2  =  1 + cos sin2    =  sin12  + sin cos2  =  (csc2  + csc cot) = −cot − csc +  61. Let  = tan(2). Then, using the expressions in Exercise 59, we have  3sin −1 4cos  =  1 31 +22  − 411 +− 2 2  2 1 + 2 = 2 3(2) − 4(1 − 2) =  22 + 3  − 2 =  (2 − 1)(   + 2) =  25 2 1− 1 − 15  + 2 1   [using partial fractions] = 1 5 ln|2 − 1| − ln| + 2|  +  = 15 ln   2+ 2 − 1    +  = 1 5 ln   2tan ( tan ( 2) + 2 2) − 1    +  62. Let  = tan(2). Then, by Exercise 59,   32 1 + sin  − cos = 11√3 1 + 2(1 +2 2) −(1 + (1 −2)2)(1 + 2) = 11√3 1 + 2 + 2 2  − 1 + 2 = 11√3 1 −  + 1 1   = ln − ln( + 1)1 1√3 = ln 1 2 − ln √3 + 1 1 = ln √3 + 1 2 63. Let  = tan (2). Then, by Exercise 59, 02 2 + cos sin 2  = 02 2sin 2 + cos  cos   = 01 2 · 1 +22 · 1 1 +−  2 2 2 + 1 − 2 1 + 2 2 1 + 2  = 01 8(1 − 2) (1 + 2)2 2(1 + 2) + (1 − 2)  = 01 8 · (2 + 3)( 1 −22+ 1)2  =  If we now let  = 2, then 1 − 2 (2 + 3)(2 + 1)2 = 1 −  ( + 3)( + 1)2 =   + 3 +   + 1 +  ( + 1)2 ⇒ 1 −  = ( + 1)2 + ( + 3)( + 1) + ( + 3). Set  = −1 to get 2 = 2, so  = 1. Set  = −3 to get 4 = 4, so  = 1. Set  = 0 to get 1 = 1 + 3 + 3, so  = −1. So  = 01 28+ 3  − 28+ 1  + (2 + 1) 8 2   = 4ln(2 + 3) − 4ln(2 + 1) − 2 4+ 11 0 = (4 ln 4 − 4ln2 − 2) − (4 ln 3 − 0 − 4) = 8 ln 2 − 4ln2 − 4ln3 + 2 = 4 ln 2 3 + 2 64. 1 3 +  = 1 (2 + 1) =   +  +  2 + 1 ⇒ 1 = (2 + 1) + ( + ). Set  = 0 to get 1 = . So 1 = (1 + )2 +  + 1 ⇒  + 1 = 0 [ = −1] and  = 0. Thus, the area is °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 47 12 3 1+   = 121 − 2+ 1  = ln|| − 1 2 ln 2 + 1 2 1 = ln 2 − 1 2 ln 5 − 0 − 1 2 ln 2 = 3 2 ln 2 − 1 2 ln 5 or 1 2 ln 8 5 65. By long division, 2 + 1 3 − 2 = −1 + 33−+ 1 2 . Now 3 + 1 3 − 2 = 3 + 1 (3 − ) =   +  3 −  ⇒ 3 + 1 = (3 − ) + . Set  = 3 to get 10 = 3, so  = 10 3 . Set  = 0 to get 1 = 3, so  = 1 3 . Thus, the area is 12 32−+ 1 2  = 12 −1 + 1 3 + 3 −10 3  = − + 1 3 ln|| − 10 3 ln|3 − |2 1 = −2 + 1 3 ln 2 − 0 − −1 + 0 − 10 3 ln 2 = −1 + 11 3 ln 2 66. (a) We use disks, so the volume is  =  01 2 + 31 + 22  =  01 ( + 1) 2( + 2)2 . To evaluate the integral, we use partial fractions: 1 ( + 1)2( + 2)2 =   + 1 +  ( + 1)2 +   + 2 +  ( + 2)2 ⇒ 1 = ( + 1)( + 2)2 + ( + 2)2 + ( + 1)2( + 2) + ( + 1)2. We set  = −1, giving  = 1, then set  = −2, giving  = 1. Now equating coefficients of 3 gives  = −, and then equating constants gives 1 = 4 + 4 + 2(−) + 1 ⇒  = −2 ⇒  = 2. So the expression becomes  =  01 −+ 1 2 + ( + 1) 1 2 + ( + 2) 2 + ( + 2) 1 2   = 2ln    + 2 + 1    −  + 1 1 −  + 2 1 1 0 = 2ln 3 2 − 1 2 − 1 3 − 2ln2 − 1 − 1 2 = 2ln 322 + 2 3 =  2 3 + ln 16 9  (b) In this case, we use cylindrical shells, so the volume is  = 2 01 2 + 3    + 2 = 2 01 ( + 1)(   + 2). We use partial fractions to simplify the integrand:  ( + 1)( + 2) =   + 1 +   + 2 ⇒  = ( + ) + 2 + . So  +  = 1 and 2 +  = 0 ⇒  = −1 and  = 2. So the volume is 2 01 −+ 1 1 +  + 2 2   = 2−ln| + 1| + 2 ln| + 2| 1 0 = 2(−ln 2 + 2 ln 3 + ln 1 − 2ln2) = 2(2 ln 3 − 3ln2) = 2 ln 9 8 67.  =  [( − 1) + − ]  =  (01+ − )  [ = 11]. Now (01+ − ) =  + 01−  ⇒  +  = (01 − ) + . Substituting 0 for  gives  = − ⇒  = −1. Substituting 10 for  gives 11 = 10 ⇒  = 11 10 . Thus,  =  −1 + 011 110 −    ⇒  = −ln + 11 ln(01 − ) + . When  = 0,  = 10,000 and  = 900, so 0 = −ln 10,000 + 11 ln(1000 − 900) +  ⇒  = ln 10,000 − 11 ln 100 [= ln 10−18 ≈ −4145]. Therefore,  = −ln + 11 ln 10 1  − 900 + ln 10,000 − 11 ln 100 ⇒  = ln 10,000 + 11 ln  − 1000 9000. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.48 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 68. If we subtract and add 22, we get 4 + 1 = 4 + 22 + 1 − 22 = 2 + 12 − 22 = 2 + 12 − √22 = (2 + 1) − √22 + 1 + √2 = 2 − √2 + 12 + √2 + 1 So we can decompose 1 4 + 1 =  +  2 + √2 + 1 +  +  2 − √2 + 1 ⇒ 1 = ( + )2 − √2 + 1 + ( + )2 + √2 + 1. Setting the constant terms equal gives  +  = 1, then from the coefficients of 3 we get  +  = 0. Now from the coefficients of  we get  +  + ( − )√2 = 0 ⇔ [(1 − ) − ]√2 = 0 ⇒  = 1 2 ⇒  = 1 2, and finally, from the coefficients of 2 we get √2( − ) +  +  = 0 ⇒  −  = − √12 ⇒  = − √42 and  = √42. So we rewrite the integrand, splitting the terms into forms which we know how to integrate: 1 4 + 1 = √2 4  + 1 2 2 + √2 + 1 + − √2 4  + 1 2 2 − √2 + 1 = 1 4√2 22+ + 2 √2√+ 1 2 − 22− −√22√+ 1 2  = √2 8 2 2+√+2√2+ 1 − 2 2−√−2√2+ 1 + 1 4    + √112 2 + 1 2 +  − √112 2 + 1 2    Now we integrate:  4 + 1 = √82 ln 2 2 + − √ √2 2  + 1 + 1 + √42 tan−1√2 + 1 + tan−1√2 − 1 + . 69. (a) In Maple, we define (), and then use convert(f,parfrac,x); to obtain () = 24,1104879 5 + 2 − 668323 2 + 1 − 943880,155 3 − 7 + (22,098 + 48,935)260,015 2 +  + 5 In Mathematica, we use the command Apart, and in Derive, we use Expand. (b)  () = 24 4879 ,110 · 1 5 ln|5 + 2| − 668 323 · 1 2 ln|2 + 1| − 80 9438 ,155 · 1 3 ln|3 − 7| + 1 260,015  22,098 ++1 2 1 22++ 37 19 4 ,886  +  = 24,110 4879 · 1 5 ln|5 + 2| − 668 323 · 1 2 ln|2 + 1| − 80 9438 ,155 · 1 3 ln|3 − 7| + 1 260,015 22,098 · 1 2 ln2 +  + 5 + 37,886 ·  19 4 tan−1√19 1 4  + 1 2  +  = 4822 4879 ln|5 + 2| − 334 323 ln|2 + 1| − 80 3146 ,155 ln|3 − 7| + 260 11,049 ,015 ln2 +  + 5 + 75,772 260,015√19 tan−1 √119 (2 + 1) +  Using a CAS, we get 4822 ln(5 + 2) 4879 − 334 ln(2 + 1) 323 − 3146 ln(3 − 7) 80,155 + 11,049 ln(2 +  + 5) 260,015 + 3988√19 260,015 tan−1√19 19 (2 + 1) The main difference in this answer is that the absolute value signs and the constant of integration have been omitted. Also, the fractions have been reduced and the denominators rationalized. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 49 70. (a) In Maple, we define (), and then use convert(f,parfrac,x); to get () = 58281815 (5 − 2)2 − 59,09619,965 5 − 2 + 2(2843 + 816)3993 22 + 1 + (313 − 251)363 (22 + 1)2 . In Mathematica, we use the command Apart, and in Derive, we use Expand. (b) As we saw in Exercise 69, computer algebra systems omit the absolute value signs in  (1) = ln||. So we use the CAS to integrate the expression in part (a) and add the necessary absolute value signs and constant of integration to get  () = −9075(5 5828  − 2) − 59,096 ln 99,825 |5 − 2| + 2843 ln7986 22 + 1 + 503 15,972 √2tan−1√2 − 2904 1 1004 22 + 1 + 626 +  (c) From the graph, we see that  goes from negative to positive at  ≈ −078, then back to negative at  ≈ 08, and finally back to positive at  = 1. Also, lim→04 () = ∞. So we see (by the First Derivative Test) that  () has minima at  ≈ −078 and  = 1, and a maximum at  ≈ 080, and that  () is unbounded as  → 04. Note also that just to the right of  = 04,  has large values, so  () increases rapidly, but slows down as  drops toward 0.  () decreases from about 08 to 1, then increases slowly since  stays small and positive. 71. 4(1 − )4 1 + 2 = 4(1 − 4 + 62 − 43 + 4) 1 + 2 = 8 − 47 + 66 − 45 + 4 1 + 2 = 6 − 45 + 54 − 42 + 4 − 1 +42 , so 01 41 + (1 −2)4  = 177 − 2 36 + 5 − 433 + 4 − 4tan−1 1 0 = 17 − 23 + 1 − 4 3 + 4 − 4 · 4  − 0 = 22 7 − . 72. (a) Let  = (2 + 2)−,  =  ⇒  = −(2 + 2)−−1 2 ,  = .  =  (2 +2) = (2 +2) −  (2−+2 2)2+1  [by parts] =  (2 + 2) + 2  ((22++22))−+1 2  =  (2 + 2) + 2  (2 +2) − 22  (2 + 2)+1 Recognizing the last two integrals as  and +1, we can solve for +1 in terms of . 22+1 =  (2 + 2) + 2 −  ⇒ +1 = 22(2 + 2) + 22−21 ⇒  =  22( − 1)(2 + 2)−1 + 2 − 3 22( − 1)−1 [decrease -values by 1], which is the desired result. (b) Using part (a) with  = 1 and  = 2, we get  (2 + 1)2 = 2(2+ 1) + 12  2 + 1 = 2(2+ 1) + 12 tan−1  +  Using part (a) with  = 1 and  = 3, we get °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.50 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION  (2 + 1)3 = 2(2)(2 + 1)2 + 2(2) 3  (2 + 1)2 = 4(2+ 1)2 + 342(2+ 1) + 12 tan−1  +  =  4(2 + 1)2 + 3 8(2 + 1) + 3 8 tan−1  +  73. There are only finitely many values of  where () = 0 (assuming that  is not the zero polynomial). At all other values of , ()() = ()(), so () = (). In other words, the values of  and  agree at all except perhaps finitely many values of . By continuity of  and , the polynomials  and  must agree at those values of  too. More explicitly: if  is a value of  such that () = 0, then () 6= 0 for all  sufficiently close to . Thus, () = lim → () [by continuity of ] = lim → () [whenever () 6= 0] = () [by continuity of ] 74. Let () = 2 +  + . We calculate the partial fraction decomposition of () 2( + 1)3 . Since (0) = 1, we must have  = 1, so () 2( + 1)3 = 2 +  + 1 2( + 1)3 =   +  2 +   + 1 +  ( + 1)2 +  ( + 1)3 . Now in order for the integral not to contain any logarithms (that is, in order for it to be a rational function), we must have  =  = 0, so 2 +  + 1 = ( + 1)3 + 2( + 1) + 2. Equating constant terms gives  = 1, then equating coefficients of  gives 3 =  ⇒  = 3. This is the quantity we are looking for, since  0(0) =  75. If  6= 0 and  is a positive integer, then () = 1 ( − ) = 1  + 2 2 + ··· +   +   −  . Multiply both sides by ( − ) to get 1 = 1−1( − ) + 2−2( − ) + ·· · + ( − ) + . Let  =  in the last equation to get 1 =  ⇒  = 1. So () −   −  = 1 ( − ) − 1 ( − ) =  −  ( − ) = −  −  ( − ) = − ( − )(−1 + −2 + −32 + ·· · + −2 + −1) ( − ) = − −1 + −2 + −32 + ··· +  −2 + −1  = − 1  − 1 −12 − 1 −23 − ·· · − 1 2−1 − 1  Thus, () = 1 ( − ) = − 1  − 1 −12 − ·· · − 1  + 1 ( − ). 7.5 Strategy for Integration 1. Let  = 1 − sin. Then  = −cos  ⇒  1 −cos sin    =  1 (−) = −ln|| +  = −ln|1 − sin| +  = −ln(1 − sin) +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 51 2. Let  = 3 + 1. Then  = 3 ⇒ 01(3 + 1)√2  = 14 √2 1 3  = 13√2 + 1 1 √2+14 1 = 3√2 + 1 1 4√2+1 − 1 3. Let  = ln,  = √  ⇒  = 1  ,  = 2 3 32. Then 14 √ ln  = 2 332 ln4 1 −14 2 312  = 2 3 ·8ln4−0−4 9324 1 = 16 3 (2 ln 2)−49 · 8 − 49 = 32 3 ln 2− 28 9 4.  sin cos3  =  sin2cos  sin    =  (1 − cos cos 2 )sin  =  1 −2 (−)  = cos = − sin    =   − 1  = 1 22 − ln|| +  = 1 2 cos2  − ln|cos| +  5. Let  = 2. Then  = 2  ⇒  4 + 2   =  21+ 2 1 2  = 1 2 √12 tan−1√2 +  [by Formula 17] = 2√1 2 tan−1√22 +  6. Let  = 2 + 1. Then  = 2 ⇒ 01 (2 + 1)  3  = 13 ( −31)2 1 2  = 14 13 12 − 13   = 14−1 + 212 3 1 = 1 4 − 1 3 + 18 1  − −1 + 1 2 = 1 4 2 9 = 18 1 7. Let  = arctan. Then  =  1 + 2 ⇒ −11 1 + arctan 2  = − 44   =  − 4 4 = 4 − −4. 8.  sin cos  =   · 1 2(2 sin cos ) = 1 2   sin 2  = 1 2 − 1 2cos 2 −  − 1 2 cos 2    ==  , ,   == sin 2 − 1 2 cos 2     = − 1 4 cos 2 + 1 4  cos 2  = − 1 4cos 2 + 1 8 sin 2 +  9.  + 2 2 + 3 − 4 =  + 2 ( + 4)( − 1) =   + 4 +   − 1 . Multiply by ( + 4)( − 1) to get  + 2 = ( − 1) + ( + 4). Substituting 1 for  gives 3 = 5 ⇔  = 3 5. Substituting −4 for  gives −2 = −5 ⇔  = 2 5. Thus, 24 2 + 3  + 2  − 4  = 24 2+ 4 5 + 3−51  = 2 5 ln| + 4| + 35 ln| − 1|4 2 =  2 5 ln 8 + 3 5 ln 3 −  2 5 ln 6 + 0 = 2 5(3 ln 2) + 3 5 ln 3 − 2 5(ln2 + ln3) = 4 5 ln 2 + 1 5 ln 3, or 1 5 ln 48 10. Let  = 1  ,  = cos(1) 2 ⇒  = −12 ,  = −sin1. Then  cos(1 3)  = −1 sin1 −  12 sin1  = −1 sin1 − cos1 + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.52 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 11. Let  = sec, where 0 ≤  ≤ 2 or  ≤   32 . Then  = sec tan  and √2 − 1 = √sec2  − 1 = √tan2  = |tan| = tan for the relevant values of , so  3√12 − 1  =  sec sec3tan tan  =  cos2   =  1 2(1 + cos 2) = 1 2  + 1 4 sin 2 +  = 1 2 + 1 2 sin cos +  = 1 2 sec−1  + 1 2 √2 − 1  1  +  = 1 2 sec−1  + √2 − 1 22 +  12. 2 − 3 3 + 3 = 2 − 3 (2 + 3) =   +  +  2 + 3 . Multiply by (2 + 3) to get 2 − 3 = (2 + 3) + ( + ) ⇔ 2 − 3 = ( + )2 +  + 3. Equating coefficients gives us  = 2, 3 = −3 ⇔  = −1, and  +  = 0, so  = 1. Thus,  23+ 3 − 3  =  −1 + 2+ 2 + 3  =  −1 + 2+ 3 + 22+ 3  = −ln|| + 1 2 ln(2 + 3) + √23 tan−1√3 +  13.  sin5  cos4   =  sin4  cos4  sin  =  (sin2 )2 cos4  sin  =  (1 − cos2 )2 cos4  sin  =  (1 − 2)24 (−) [ = cos ,  = − sin  ] =  (−4 + 26 − 8) = − 1 55 + 2 77 − 1 99 +  = − 1 5 cos5  + 2 7 cos7  − 1 9 cos9  +  14. Let  = ln(1 + 2),  =  ⇒  = 2 1 + 2 ,  = . Then  ln(1 + 2) = ln(1 + 2) −  1 + 222  = ln(1 + 2) − 2 (21 + + 1) 2− 1  = ln(1 + 2) − 2 1 − 1 +12   = ln(1 + 2) − 2 + 2 tan−1  +  15. Let  = ,  = sec tan  ⇒  = ,  = sec. Then  sec tan  = sec −  sec  = sec − ln|sec + tan| + . 16. 0√22 √1−2 2  = 04 sin cos2 cos    = sin = cos  ,  = 04 1 2(1 − cos 2) = 1 2 − 1 2 sin 2 0 4 = 1 2 4 − 1 2 − (0 − 0) = 8 − 1 4 17. 0  cos2   = 0  1 2(1 + cos 2)  = 1 2 0   + 1 2 0 cos 2  = 1 2  1 22 0 + 1 2 1 2sin 2 0 − 1 2 0 1 2 sin 2    ==  , ,   == cos 2 1 2 sin 2     = 1 4 2 + 0 − 1 4− 1 2 cos 2 0 = 1 42 + 1 8(1 − 1) = 1 42 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 53 18. Let  = √. Then  = 1 2√  ⇒ 14 √√  = 12  (2) = 22 1 = 2(2 − ). 19. Let  = . Then  +  =    =    =  +  =  + . 20. Since 2 is a constant,  2  = 2 + . 21. Let  = √, so that 2 =  and 2  = . Then  arctan√  =  arctan (2 ) = . Now use parts with  = arctan,  = 2  ⇒  = 1 1 + 2 ,  = 2. Thus,  = 2 arctan −  1 +22  = 2 arctan −  1 − 1 +1 2   = 2 arctan −  + arctan +  = arctan√ − √ + arctan√ +  or ( + 1) arctan√ − √ +  22. Let  = 1 + (ln)2, so that  = 2ln  . Then   1 + (ln ln )2  = 12  √1  = 122√  +  = 1 + (ln)2 + . 23. Let  = 1 + √. Then  = ( − 1)2,  = 2( − 1) ⇒ 01 1 + √ 8  = 12 8 · 2( − 1) = 212(9 − 8) =  1 510 − 2 · 1 992 1 = 1024 5 − 1024 9 − 1 5 + 2 9 = 4097 45 . 24.  (1 + tan)2 sec  =  (1 + 2 tan + tan2 )sec  =  [sec + 2 sec tan + (sec2  − 1) sec]  =  (2 sec tan + sec3 ) = 2 sec + 1 2(sec tan + ln|sec + tan| + ) [by Example 7.2.8] 25. 01 1 + 12 1 + 3  = 01 (123+ 4)  + 1− 3  = 014 − 3 3+ 1  = 4 − ln|3 + 1|1 0 = (4 − ln 4) − (0 − 0) = 4 − ln 4 26. 32 + 1 3 + 2 +  + 1 = 32 + 1 (2 + 1)( + 1) =   + 1 +  +  2 + 1 . Multiply by ( + 1)(2 + 1) to get 32 + 1 = (2 + 1) + ( + )( + 1) ⇔ 32 + 1 = ( + )2 + ( + ) + ( + ). Substituting −1 for  gives 4 = 2 ⇔  = 2. Equating coefficients of 2 gives 3 =  +  = 2 +  ⇔  = 1. Equating coefficients of  gives 0 =  +  = 1 +  ⇔  = −1. Thus, 01 3 +322 ++ 1  + 1  = 01  + 1 2 + 2−+ 1 1   = 01  + 1 2 + 2+ 1 − 21+ 1  = 2ln| + 1| + 1 2 ln(2 + 1) − tan−1 1 0 = (2 ln 2 + 1 2 ln 2 − 4 ) − (0 + 0 − 0) = 5 2 ln 2 −  4 27. Let  = 1 + , so that  =   = ( − 1). Then  1 +1  =  1 ·  − 1 =  (1− 1)  = . Now 1 ( − 1) =   +   − 1 ⇒ 1 = ( − 1) + . Set  = 1 to get 1 = . Set  = 0 to get 1 = −, so  = −1. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.54 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION Thus,  =  −1 +  −1 1  = −ln|| + ln| − 1| +  = −ln(1 + ) + ln +  =  − ln(1 + ) + . Another method: Multiply numerator and denominator by − and let  = − + 1. This gives the answer in the form −ln(− + 1) + . 28.  sin√  =  sin · 2   [ = √, 2 = , 2  =  ] = 2  sin  = 2 [−cos + sin] +  [integration by parts] = − 2√ cos √ + 2 sin√ +  = −2  cos √ + 2 sin√ +  29. Use integration by parts with  = ln + √2 − 1,  =  ⇒  = 1  + √2 − 11 + √2− 1  =  + √12 − 1√√2−2 −1 +1   = √21− 1 ,  = . Then  ln + 2 − 1  = ln + 2 − 1 −  √2− 1  = ln + 2 − 1 − 2 − 1 + . 30. | − 1| =  −(−1− 1) if if   − − 1 1 ≥  0 0 =  1−−1 if if   ≥ 00 Thus, −21 | − 1|  = −01(1 − ) + 02( − 1) =  − 0 −1 +  − 2 0 = (0 − 1) − (−1 − −1) + (2 − 2) − (1 − 0) = 2 + −1 − 3 31. As in Example 5,  1 + 1 −    =  √√1 + 1 −   · √ √1 + 1 +    =  √1 + 1 −2  =  √1 − 2 +  √1  − 2 = sin−1  − 1 − 2 + . Another method: Substitute  = (1 + )(1 − ). 32. 13 3 2  = 31  − 1 3    = 3 = − 3 , 2  = − 1 3 1 3 = − 1 3( − 3) = 1 3(3 − ) 33. 3 − 2 − 2 = −(2 + 2 + 1) + 4 = 4 − ( + 1)2. Let  + 1 = 2 sin, where −  2 ≤  ≤ 2 . Then  = 2 cos   and  √3 − 2 − 2  =  4 − ( + 1)2  =  4 − 4sin2  2cos  = 4 cos2   = 2 (1 + cos 2) = 2 + sin 2 +  = 2 + 2 sin cos +  = 2 sin−1 + 1 2  + 2 ·  + 1 2 · √3 − 22 − 2 +  = 2 sin−1 + 1 2  +  + 1 2 √3 − 2 − 2 +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 55 34.   42 1 + 4cot 4 − cot  =   42 (1 + 4 cos (4 − cos sin sin)) · sin sin   =   42 sin 4sin+ 4 cos − cos   = 34√2 1   = 4 sin = (4 cos−+ sin cos , )  = ln|| 4 3√2 = ln 4 − ln √32 = ln 34√2 = ln43√2 35. The integrand is an odd function, so − 22 1 + cos  2   = 0 [by 5.5.7(b)]. 36.  1 + cos 1 + sin  =  (1 + cos (1 + sin)(1 )(1−−cos cos))  =  1 − cos  + sin sin2− sin cos  =  csc2  − sin cos2 + csc − cos sin  s= −cot + 1 sin + ln|csc − cot| − ln|sin| +  [by Exercise 7.2.39] The answer can be written as 1 − cos sin − ln(1 + cos) + . 37. Let  = tan. Then  = sec2   ⇒ 04 tan3  sec2   = 01 3  =  1 441 0 = 1 4 . 38.   63 sinsec  cot    =   63 cos2   = 1 2   63(1 + cos 2) = 12 + 1 2 sin 2 3 6 = 1 2 3 + √43  − 6 + √43  = 126  = 12  39. Let  = sec, so that  = sec tan . Then  sec sec 2 −tan sec  =  2 1−   =  (1− 1)  = . Now 1 ( − 1) =   +   − 1 ⇒ 1 = ( − 1) + . Set  = 1 to get 1 = . Set  = 0 to get 1 = −, so  = −1. Thus,  =  −1 +  −1 1  = −ln|| + ln| − 1| +  = ln|sec − 1| − ln|sec| +  [or ln|1 − cos| + ]. 40. Using product formula 2(a) in Section 7.2, sin 6 cos 3 = 1 2[sin(6 − 3) + sin(6 + 3)] = 1 2(sin 3 + sin 9). Thus, 0 sin 6 cos 3  = 0 1 2(sin 3 + sin 9) = 1 2− 1 3 cos 3 − 1 9 cos 9 0 = 1 2  1 3 + 1 9 − − 1 3 − 1 9 = 1 2 4 9 + 4 9 = 4 9 41. Let  = ,  = tan2   = sec2  − 1  ⇒  =  and  = tan − . So   tan2   = (tan − ) −  (tan − ) =  tan − 2 − ln|sec| + 1 22 +  =  tan − 1 22 − ln|sec| +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.56 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 42. Let  = tan−1 ,  = 1 2  ⇒  = 1 +12 ,  = −1. Then  =  tan−21   = −1 tan−1  −  −(1 +1 2)  = −1 tan−1  +   +  1 ++2   1 (1 + 2) =   +  +  1 + 2 ⇒ 1 = (1 + 2) + ( + ) ⇒ 1 = ( + )2 +  + , so  = 0,  = 1, and  +  = 0 ⇒  = −1. Thus,  = −1  tan−1  +  1 − 1 +2  = −1 tan−1  + ln|| − 1 2 ln 1 + 2  +  = − tan−1   + ln   √2 + 1  +  Or: Let  = tan, so that  = sec2  . Then  tan−21   =  tan2  sec2   =   csc2   = . Now use parts with  = ,  = csc2   ⇒  = ,  = −cot. Thus,  = − cot −  (−cot) = − cot + ln|sin| +  = −tan−1  · 1  + ln   √2 + 1  +  = −tan−1   + ln   √2 + 1  +  43. Let  = √ so that  = 1 2√ . Then  1 + √3  =  1 +6 (2 ) = 2 1 + ( 23)2  = 2 1 +1 2 1 3    == 3 3 2   = 2 3 tan−1  +  = 2 3 tan−1 3 +  = 2 3 tan−1(32) +  Another method: Let  = 32 so that 2 = 3 and  = 3 212  ⇒ √  = 2 3 . Then  1 + √3  =  1 +2 3 2  = 23 tan−1  +  = 2 3 tan−1(32) + . 44. Let  = √1 + . Then 2 = 1 + , 2  =   = (2 − 1), and  = 2 2 − 1, so  √1 +   =   · 22− 1  =  22−21  =  2 + 22− 1  =  2 +  −1 1 −  + 1 1   = 2 + ln| − 1| − ln| + 1| +  = 2√1 +  + ln√1 +  − 1 − ln√1 +  + 1 +  45. Let  = 3. Then  = 32  ⇒  =  5−3  = 1 3  − . Now integrate by parts with  = ,  = − :  = − 1 3− + 1 3  −  = − 1 3− − 1 3− +  = − 1 3−3(3 + 1) + . 46. Use integration by parts with  = ( − 1),  = 1 2  ⇒  = [( − 1) + ] =  ,  = −1. Then  ( −21)  = ( − 1)−1 −  −  = − +  +  +  =  + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 57 47. Let  =  − 1, so that  = . Then  3( − 1)−4  =  ( + 1)3−4  =  (3 + 32 + 3 + 1)−4  =  (−1 + 3−2 + 3−3 + −4) = ln|| − 3−1 − 3 2−2 − 1 3−3 +  = ln| − 1| − 3( − 1)−1 − 3 2( − 1)−2 − 1 3( − 1)−3 +  48. Let  = √1 − 2, so 2 = 1 − 2, and 2  = −2 . Then 01 2 − √1 − 2  = 10 √2 − (− ). Now let  = √2 − , so 2 = 2 − , and 2  = −. Thus, 10 √2 − (− ) = 1√2 (2 − 2)(2 ) = 1√2(42 − 24) =  4 33 − 2 55√ 1 2 =  8 3√2 − 8 5√2 −  4 3 − 2 5  = 16 15√2 − 14 15 49. Let  = √4 + 1 ⇒ 2 = 4 + 1 ⇒ 2  = 4 ⇒  = 1 2 . So   √41 + 1  =  1 4(1 22  − 1) = 2 2 − 1 = 2 1 2 ln     − + 1 1    +  [by Formula 19] = ln  √4 + 1 − 1 √4 + 1 + 1  +  50. As in Exercise 49, let  = √4 + 1. Then  2 √ 4 + 1 =   1 4(1 22  − 1)2  = 8 (2 − 1)2 . Now 1 (2 − 1)2 = 1 ( + 1)2( − 1)2 =   + 1 +  ( + 1)2 +   − 1 +  ( − 1)2 ⇒ 1 = ( + 1)( − 1)2 + ( − 1)2 + ( − 1)( + 1)2 + ( + 1)2.  = 1 ⇒  = 1 4,  = −1 ⇒  = 1 4. Equating coefficients of 3 gives  +  = 0, and equating coefficients of 1 gives 1 =  +  −  +  ⇒ 1 =  + 1 4 −  + 1 4 ⇒ 1 2 =  − . So  = 1 4 and  = − 1 4. Therefore,  2√ 4 + 1 = 8 1+ 1 4 + (1+ 1) 4 2 + −1−41 + (1−41)2   =   + 1 2 + 2( + 1)−2 −  −2 1 + 2( − 1)−2  = 2 ln| + 1| − 2  + 1 − 2ln| − 1| − 2  − 1 +  = 2 ln√4 + 1 + 1 − √4 + 1 + 1 2 − 2ln √4 + 1 − 1  − √4 + 1 2 − 1 +  51. Let 2 = tan ⇒  = 1 2 tan,  = 1 2 sec2  , √42 + 1 = sec, so  √4 2 + 1 =  1 21 2tan sec2   sec =  tan sec  =  csc  = −ln|csc + cot| +  [or ln|csc − cot| + ] = −ln  √42 + 1 2 + 1 2  +  or ln   √422+ 1 − 21    +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.58 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 52. Let  = 2. Then  = 2  ⇒  ( 4 + 1) =  2(  4 + 1) = 12  ( 2 + 1) = 12  1 − 2+ 1  = 1 2 ln|| − 1 4 ln(2 + 1) +  = 1 2 ln(2) − 1 4 ln(4 + 1) +  = 1 4ln(4) − ln(4 + 1) +  = 1 4 ln4+ 1 4  +  Or: Write  =  4(34 + 1) and let  = 4. 53.  2 sinh() = 1 2 cosh() − 2  cosh()  = = 2  2,  = sinh( = 1 cosh(  ) ),  = 1  2 cosh() − 2  1 sinh() − 1  sinh()  = =  ,  = cosh( = 1 sinh(  ) ),  = 1  2 cosh() − 2 2 sinh() + 23 cosh() +  54.  ( + sin)2  =  2 + 2sin + sin2   = 1 33 + 2(sin − cos) + 1 2( − sincos) +  = 1 3 3 + 1 2 + 2 sin − 1 2 sincos − 2cos +  55. Let  = √, so that  = 2 and  = 2 . Then   + √ =  22+  2 ·  =  (1 + 2 )  = . Now 2 (1 + ) =   +  1 +  ⇒ 2 = (1 + ) + . Set  = −1 to get 2 = −, so  = −2. Set  = 0 to get 2 = . Thus,  =  2 − 1 +2   = 2 ln|| − 2ln|1 + | +  = 2 ln√ − 2ln1 + √  + . 56. Let  = √, so that  = 2 and  = 2 . Then  √ + √ =   +2  2 ·  =  1 +22  = 2 tan−1  +  = 2 tan−1 √ + . 57. Let  = √3  + . Then  = 3 −  ⇒   √3  +   =  (3 − ) · 32  = 3 (6 − 3) = 3 77 − 3 44 +  = 3 7( + )73 − 3 4( + )43 +  58. Let  = √2 − 1. Then  = √2 − 1 , 2 − 1 = 2,  = √2 + 1, so  =  √2ln−1  =  ln2 + 1 = 1 2  ln(2 + 1). Now use parts with  = ln(2 + 1),  = :  = 1 2ln2 + 1 −  2+ 1 2  = 1 2 ln2 + 1 −  1 − 2 1+ 1  = 1 2 ln2 + 1 −  + tan−1  +  = √2 − 1 ln − √2 − 1 + tan−1√2 − 1 +  Another method: First integrate by parts with  = ln,  = √2 − 1  and then use substitution  = sec or  = √2 − 1. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 59 59. 1 4 − 16 = 1 (2 − 4)(2 + 4) = 1 ( − 2)( + 2)(2 + 4) =   − 2 +   + 2 +  +  2 + 4 . Multiply by ( − 2)( + 2)(2 + 4) to get 1 = ( + 2)(2 + 4) + ( − 2)(2 + 4) + ( + )( − 2)( + 2). Substituting 2 for  gives 1 = 32 ⇔  = 32 1 . Substituting −2 for  gives 1 = −32 ⇔  = − 32 1 . Equating coefficients of 3 gives 0 =  +  +  = 1 32 − 1 32 + , so  = 0. Equating constant terms gives 1 = 8 − 8 − 4 = 1 4 + 1 4 − 4, so 1 2 = −4 ⇔  = − 1 8 . Thus,  4 − 16 =  1−322 − 1+ 2 32 − 21+ 4 8   = 32 1 ln| − 2| − 32 1 ln| + 2| − 18 · 12 tan−12 +  = 1 32 ln   − 2  + 2  − 1 16 tan−12 +  60. Let 2 = sec, so that 2 = sec tan . Then  2 √4 2 − 1 =  1 4 sec 1 2 sec 2 √tan sec2   − 1 =  sec 2tan  tan   = 2 cos   = 2 sin +  = 2 · √42 − 1 2 +  = √42 − 1  +  61.  1 + cos   =  1 + cos 1  · 1 1 − − cos cos     =  11−−cos cos2  =  1 sin − cos 2    =  sin12  − sin cos2   =  (csc2  − cot csc) = −cot + csc +  Another method: Use the substitutions in Exercise 7.4.59.  1 + cos   =  1 + (1 2(1 + − 2)2(1 + ) 2) =  (1 + 2) + (1 2 − 2) =   =  +  = tan2 +  62.  1 + cos  2  =  (1 + cos (1cos 2 2))cos  2  =  secsec 2 2+ 1   =  tansec 2 2+ 2   =  21+ 2    = tan = sec2    =  2 + 1√22  = √12 tan−1√2 +  = √12 tan−1tan √2  +  63. Let  = √ so that  = 1 2√  ⇒  = 2√  = 2 . Then  √ √  =  (2 ) =  22   = 2 = 4   2,  ==   ,  = 22 −  4   = 4 = 4 ,  = =    ,  = 22 − 4 −  4  = 22 − 4 + 4 +  = 2(2 − 2 + 2) +  = 2 − 2√ + 2 √ +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.60 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 64. Let  = √ + 1, so that  = ( − 1)2 and  = 2( − 1). Then  √1 + 1  =  2( −√1) =  (212 −2−12) = 4332 −412 + = 4 3√ + 132 −4√ + 1+. 65. Let  = cos2 , so that  = 2 cos(−sin). Then  1 + cos sin 24   =  1 + (cos 2sin cos 2 )2  =  1 +12 (−) = −tan−1  +  = −tan−1(cos2 ) + . 66. Let  = tan. Then   43 ln(tan sin cos )  =   43 ln(tan tan) sec2   = 1√3 ln  =  1 2(ln)2√ 1 3 = 1 2ln√32 = 1 8(ln 3)2. 67.  √ + 1 +  √ =  √ + 1 + 1 √ · √√ + 1  + 1 −−√√   =   + 1 − √   = 2 3 ( + 1)32 − 32 +  68.  6 + 3 2 3 + 2  =  (3 + 1)( 2  3 + 2) =  ( + 1)( 1 3  + 2)   == 3 32  . Now 1 ( + 1)( + 2) =   + 1 +   + 2 ⇒ 1 = ( + 2) + ( + 1). Setting  = −2 gives  = −1. Setting  = −1 gives  = 1. Thus, 1 3  ( + 1)(  + 2) = 1 3   + 1 1 −  + 2 1   = 1 3 ln| + 1| − 13 ln| + 2| +  = 1 3 ln 3 + 1  − 1 3 ln 3 + 2  +  69. Let  = tan, so that  = sec2  ,  = √3 ⇒  = 3 , and  = 1 ⇒  = 4 . Then 1√3 √1 + 2 2  =   43 tan sec2 sec2   =   43 sec (tan tan22 + 1)  =   43 sectan  tan 2 2  + tan sec2   =   43(sec + csc cot) = ln|sec + tan| − csc 3 4 = ln 2 + √3  − √23 − ln √2 + 1  − √2 = √2 − √23 + ln2 + √3 − ln1 + √2 70. Let  = . Then  = ln,  =  ⇒  1 + 2  − − =  1 + 2   − 1 =  22 + − 1 =  22−31 − 1+ 1 3   = 1 3 ln|2 − 1| − 1 3 ln| + 1| +  = 1 3 ln|(2 − 1)( + 1)| +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 61 71. Let  = . Then  = ln,  =  ⇒  1 + 2  =  1 + 2   =  1 +   =  1 − 1 +1   =  − ln|1 + | +  =  − ln(1 + ) + . 72. Use parts with  = ln( + 1),  = 2:  ln(+ 1) 2  = −1 ln( + 1) +  ( + 1) = −1 ln( + 1) +  1 −  + 1 1   = − 1  ln( + 1) + ln|| − ln( + 1) +  = − 1 + 1ln( + 1) + ln|| +  73. Let  = arcsin, so that  = √1 1− 2  and  = sin. Then   √+ arcsin 1 − 2   =  (sin + ) = −cos + 1 22 +  = −√1 − 2 + 1 2(arcsin)2 +  74.  4 + 10 2   =  42  + 10 2   =  (2 + 5) = ln 2 2 + ln 5 5 +  75.  ln  −  =  (ln  − 1) =      = ln = (1  −)  1, = ln|| +  = ln|ln − 1| +  76.  √22+ 1  =  tan sec2 sec2     = tan = sec2  ,  =  tan2  sec  =  (sec2  − 1) sec  =  (sec3  − sec) = 1 2 (sec tan + ln|sec + tan|) − ln|sec + tan| +  [by (1) and Example 7.2.8] = 1 2 (sec tan − ln|sec + tan|) +  = 1 2√2 + 1 − ln(√2 + 1 + ) +  77. Let  = √1 + , so that 2 = 1 + , 2  =  ,  = 2 − 1, and  = ln(2 − 1). Then  √1 +   =  ln(2− 1)(2 ) = 2 [ln( + 1) + ln( − 1)] = 2[( + 1) ln( + 1) − ( + 1) + ( − 1) ln( − 1) − ( − 1)] +  [by Example 7.1.2] = 2[ ln( + 1) + ln( + 1) −  − 1 +  ln( − 1) − ln( − 1) −  + 1] +  = 2[(ln( + 1) + ln( − 1)) + ln( + 1) − ln( − 1) − 2] +  = 2 ln(2 − 1) + ln  + 1 − 1 − 2 +  = 2√1 +  ln() + ln √ √1 + 1 +    + 1 − 1 − 2√1 +  +  = 2 √1 +  + 2 ln √1 +  + 1 √1 +  − 1 − 4√1 +  +  = 2( − 2)√1 +  + 2 ln √1 +  + 1 √1 +  − 1 +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.62 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 78. 1 + sin 1 − sin = 1 + sin 1 − sin · 1 + sin 1 + sin = 1 + 2 sin + sin2  1 − sin2  = 1 + 2sin + sin2  cos2  = 1 cos2  + 2sin cos2  + sin2  cos2  = sec2  + 2sec tan + tan2  = sec2  + 2sec tan + sec2  − 1 = 2sec2  + 2sec tan − 1 Thus,  1 + sin 1 − sin   =  (2 sec2  + 2 sec tan − 1) = 2 tan + 2 sec −  +  79. Let  = ,  = sin2 cos  ⇒  = ,  = 1 3 sin3 . Then  sin2  cos   = 1 3sin3  −  1 3 sin3   = 1 3sin3  − 1 3  (1 − cos2 )sin  = 1 3 sin3  + 1 3  (1 − 2)  = cos = − sin ,  = 1 3 sin3  + 1 3 − 1 93 +  = 1 3sin3  + 1 3 cos  − 1 9 cos3  +  80.  sin sec+ sec cos 2  =  sin sec+ sec cos 2 · 2cos 2cos   =  2sin2cos 2 cos + 2  =  sin 2 2cos 2  + 2   =  1   = sin 2 = 2 cos 2    + 2, = ln|| +  = ln|sin 2 + 2| +  = ln(sin 2 + 2) +  81.  √1 − sin  =  1 −1sin · 1 + sin 1 + sin   =  11 + sin − sin2  =  1 + sin cos2   =  √cos 1 + sin   [assume cos   0] =  √   = 1 + sin = cos    = 2√ +  = 2√1 + sin +  Another method: Let  = sin so that  = cos  = 1 − sin2   = √1 − 2 . Then  √1 − sin  =  √1 −  √1 − 2  =  √1 + 1   = 2√1 +  +  = 2√1 + sin + . 82.  sinsin 4 + cos cos4   =  (sin2sin )2+ (cos cos 2 )2  =  (sin2 )sin 2 + (1  cos −sin2 )2  =  2 + (1 1 − )2 1 2   = sin = 2 sin 2 , cos    =  42 −14 + 2  =  (42 − 41+ 1) + 1  =  (2 −11)2 + 1  = 1 2  2 1+ 1   = 2 = 2− 1, = 1 2 tan−1  +  = 1 2 tan−1(2 − 1) +  = 1 2 tan−1(2 sin2  − 1) +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ¤ 63 Another solution:  sinsin 4 + cos cos4   =  (sin (sin 4 + cos cos4))cos cos 4 4   =  tan tan4 sec + 1 2   =  21+ 1 12   = tan = 2 tan 2 , sec2   = 1 2 tan−1  +  = 1 2 tan−1(tan2 ) +  83. The function  = 22 does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.  (22 + 1)2  =  222  +  2  =  22  +  2  = 2 −  2  +  2   = =  , = 2 =  2 2 ,  = 2 +  84. (a) 12   = 0ln 2      ==   = 0ln 2   = (ln 2) (b) 23 ln1  = ln 2 ln 3 1 ( )   = ln = 1    = ln ln 2 ln ln 3      ==   = ln ln 2 0   + 0ln ln 3   [note that ln ln 2  0] = 0ln ln 3   − 0ln ln 2   = (lnln3) − (lnln2) Another method: Substitute  =  in the original integral. 7.6 Integration Using Tables and Computer Algebra Systems Keep in mind that there are several ways to approach many of these exercises, and different methods can lead to different forms of the answer. 1. 02 cos 5 cos 2  80 = sin(5 2(5 −−2) 2) + sin(5 + 2) 2(5 + 2) 0 2   = 2 = 5  = sin 3 6  + sin 7 14 0 2 = −16 − 14 1  − 0 = −742− 3 = −21 5 2. 01  − 2  = 01 2 1 2  − 2  113 =  −2 1 2 2 1 2  − 2 +  1 222 cos−1 1 2 −1 2 1 0 = 24− 1 √ − 2 + 18 cos−1(1 − 2)1 0 = 0 + 18 ·  − 0 + 18 · 0 = 18 3. 12 √42 − 3 = 1 2 24 2 − √32   = 2  = 2  39 = 1 2 22 − √32 − √232 ln    + 2 − √32    4 2 = 1 2 2√13 − 3 2 ln4 + √13 − 1 21 − 3 2 ln 3 = √13 − 3 4 ln4 + √13 − 1 2 + 3 4 ln 3 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.64 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 4. 01 tan36   = 6 06 tan3   [ = (6)  = (6) ] 69 = 6  12 tan2  + ln|cos| 0 6 = 6 12√132 + ln √23 − (0 + ln 1) = 1 + 6 ln √23 5. 08 arctan 2  = 1 2 04 arctan   = 2  = 2  89 = 1 2 arctan − 1 2 ln(1 + 2) 0 4 = 124 arctan 4 − 12 ln1 + 162  − 0 =  8 arctan  4 − 1 4 ln1 + 162  6. 02 24 − 2  31 = 8(22 − 4)4 − 2 + 16 8 sin−122 0 = 0 + 2 · 2  − 0 =  7.  sincos 2 − 9  =  21− 9    = sin = cos   20 = 2(3) 1 ln     − + 3 3    +  = 1 6 ln   sin sin −+ 3 3    +  8.  4 −2  =  4 −12    == , 19 = 2(2) 1 ln     + 2 − 2    +  = 1 4 ln     + 2 − 2    +  9.  √922+ 4  =  √22+ 4 9 13   = 3 = 3 ,  = 3 √4 + 2 2  24 = 3−√4 +  2 + ln( + 4 + 2) +  = − 3√4 + 92 3 + 3 ln(3 + √4 + 92) +  = − √92 + 4  + 3 ln(3 + √92 + 4) +  10. Let  = √2 and  = √3. Then  = √2 and  222 − 3  =  √1 22−2 2 √2 = √2 √2−2 2  =42 √2−√2− 2 + ln   + 2 − 2   +  = √2−2√22 − 3 + ln  √2 + 22 − 3   +  = − 22 − 3  + √2ln  √2 + 22 − 3   +  11. 0 cos6   74 =  1 6 cos5  sin 0 + 5 6 0 cos4   74 = 0 + 5 6 1 4 cos3  sin 0 + 3 4 0 cos2   64 = 5 6 0 + 3 4  1 2 + 1 4 sin 2 0  = 5 6 · 3 4 · 2 = 516  12.  √2 + 4  =  √2 + 2 ( 1 2 )   == 2 2  ,  21 = 1 2 2√2 + 2 + 22 ln + √2 + 2  +  = 42 √2 + 4 + 1 2 ln(2 + √2 + 4 ) +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ¤ 65 13.  arctan √√  =  arctan(2)   == 1 √(2 , √)  89 = 2arctan − 1 2 ln(1 + 2) +  = 2√arctan√ − ln(1 + ) +  14. 0 3 sin  =84 −3 cos  0 + 30 2 cos  =85 −3(−1) + 32 sin 0 − 20 sin  = 3 − 60 sin  =84 3 − 6−cos  0 + 0 cos   = 3 − 6[] − 6sin 0 = 3 − 6 15.  coth(1 2 )  =  coth(−)   = 1 = − 1 , 2  106 = −ln|sinh| +  = −ln|sinh(1)| +  16.  √23− 1  =  √22− 1 ( ) =  √22− 1    == , 44 =  2 √2 − 1 + 1 2 ln  + √2 − 1  +  = 1 2√2 − 1 + 1 2 ln + √2 − 1 +  17. Let  = 6 + 4 − 42 = 6 − (42 − 4 + 1) + 1 = 7 − (2 − 1)2,  = 2 − 1, and  = √7. Then  = 2 − 2,  = 2, and   6 + 4 − 42  =   √  =  1 2( + 1)√2 − 2 1 2  = 1 4   √2 − 2  + 1 4  √2 − 2  = 1 4  √2 − 2  − 1 8  (−2)√2 − 2  30 =  8 √2 − 2 + 2 8 sin−1  − 18  √   == −22−  2,  = 2 − 1 8 6 + 4 − 42 + 7 8 sin−1 2√−7 1 − 1 8 · 2332 +  = 2 − 1 8 6 + 4 − 42 + 7 8 sin−1 2√−7 1 − 12 1 (6 + 4 − 42)32 +  This can be rewritten as 6 + 4 − 42 1 8(2 − 1) − 12 1 (6 + 4 − 42) + 7 8 sin−1 2√−7 1 +  = 1 32 − 12 1  − 58 6 + 4 − 42 + 7 8 sin−12√−7 1 +  = 1 24(82 − 2 − 15)6 + 4 − 42 + 7 8 sin−12√−7 1 +  18.  23 − 32 =  2(− 3 + 2) 50 = −−13 + (−23)2 ln   −3 + 2      +  = 31 + 2 9 ln   2− 3    +  19. Let  = sin. Then  = cos , so  sin2  cos ln(sin) =  2 ln  101 = (2 + 1) 2+1 2 [(2 + 1) ln − 1] +  = 1 93(3 ln − 1) +  = 1 9 sin3 [3 ln(sin) − 1] +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.66 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 20. Let  = sin, so that  = cos  . Then  √5sin 2 − sin    =  2sin √5 − sin cos  = 2 √5−   55 = 2 · 3(−21)2 [−1 − 2(5)]√5 −  +  = 4 3 (− − 10)√5 −  +  = − 4 3(sin + 10)√5 − sin +  21. Let  =  and  = √3. Then  =   and  3 −2  =  2  − 2 =19 21 ln     + −      +  = 2√1 3 ln     + − √ √3 3    + . 22. Let  = 2 and  = 2. Then  = 2  and 02 3√42 − 4  = 1 2 02 2 2 · 2 · 2 − (2)2 · 2  = 1 2 04  √2 − 2  114 = 22 −  12 − 32 √2 − 2 + 43 cos−1 − 4 0 = 22 −12 2 − 12√4 − 2 + 8 4 cos−12 −2 4 0 = 2 −6 − 6√4 − 2 + 2 cos−12 −2 4 0 = [0 + 2 cos−1(−1)] − (0 + 2 cos−1 1) = 2 ·  − 2 · 0 = 2 23.  sec5   =77 1 4 tan sec3  + 3 4  sec3   =77 1 4 tan sec3  + 3 4 1 2 tan sec + 1 2  sec  14 = 1 4 tan sec3  + 3 8 tan sec + 3 8 ln|sec + tan| +  24.  3 arcsin(2) =  arcsin  1 2   == 2 2  ,  90 = 1 2 224− 1 arcsin + √14− 2  +  = 248− 1 arcsin(2) + 2√18− 4 +  25. Let  = ln and  = 2. Then  =  and  4 + (ln  )2  =  2 + 2  =21 22 + 2 + 22 ln + 2 + 2  +  = 1 2 (ln)4 + (ln)2 + 2 lnln + 4 + (ln)2  +  26.  4−  =97 −4− + 4 3−  =97 −4− + 4−3− + 3 2−  97 = −(4 + 43)− + 12−2− + 2 −  96 = −(4 + 43 + 122)− + 24[(− − 1)−] +  = −(4 + 43 + 122 + 24 + 24)− +  So 01 4−  = −(4 + 43 + 122 + 24 + 24)−1 0 = −(1 + 4 + 12 + 24 + 24)−1 + 240 = 24 − 65−1. 27.  cos−1(3−2)  = −12  cos−1     == −−22−3   88 = − 1 2 cos−1  − √1 − 2  +  = − 1 2−2 cos−1(−2) + 1 2√1 − −4 +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ¤ 67 28.  √1 − 2 =  √1 1− 2      == ,,  =   35 = − 1 1 ln  1 + √1 − 2   +  = −ln  1 + √1 − 2   +  = −ln1 + √1− 2  +  29. Let  = . Then  = ln,  = , so  2 − 1 =  √2− 1  =41 2 − 1 − cos−1(1) +  = 2 − 1 − cos−1(−) + . 30. Let  =  − 3 and assume that  6= 0. Then  =   and   sin( − 3) = 1  (+3) sin  = 1 3  (1) sin  98 = 1  3 (1) (1)2 + 12 1 sin − cos +  = 1 3(1) 1 +22 1 sin − cos  +  = 1 1 + 2 (+3)(sin − cos ) +  = 1 +12  [sin( − 3) − cos( − 3)] +  31.  √10 4  − 2 =  (45) 2 − 2 = 1 5  √ 2 − 2 = = 55,4  43 = 1 5 ln  + √2 − 2  +  = 1 5 ln 5 + √10 − 2  +  32. Let  = tan and  = 3. Then  = sec2   and  sec √92− tan tan22  =  √2−2 2  =34 −2√2 − 2 + 22 sin−1  +  = − 1 2 tan √9 − tan2  + 9 2 sin−1tan 3   +  33. Use disks about the -axis:  = 0 (sin2 )2  =  0 sin4   73 = − 1 4 sin3  cos 0 + 3 4 0 sin2   63 =  0 + 3 4 1 2 − 1 4 sin 2 0  =   3 4 1 2 − 0 = 3 82 34. Use shells about the -axis:  = 01 2arcsin  90 = 2224− 1 sin−1  +  √14− 2 1 0 = 214 · 2 + 0 − 0 = 142 35. (a)   13  +  −  +2 − 2ln| + | +  = 13  + ( + 2 )2 − (2+) = 1 3 ( + )2 + ( +2−)2( + )2 = 1 3 (+3 2 )2  = ( +2)2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.68 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (b) Let  =  +  ⇒  =  . Note that  =  −   and  = 1 .  (+2  )2 = 13  ( −2)2  = 13  2 − 2 2 + 2  = 13  1 − 2 + 22  = 1 3  − 2ln|| − 2  +  = 13  +  −  +2 − 2ln| + | +  36. (a)   8(22 − 2)√2 − 2 + 84 sin−1  +  =  8 (22 − 2) √−2 − 2 + √2 − 2 8(4) + (22 − 2) 1 8 + 84 1 −1 22 = − 2(22 − 2) 8√2 − 2 + √2 − 2 22 + 228− 2  + 8√24− 2 = 1 2 (2 − 2)−12−42 (22 − 2) + 2(2 − 2) + 1 4(2 − 2)(22 − 2) + 44  = 1 2 (2 − 2)−12[222 − 24] = √2(22 −−22) = 2 √2 − 2 (b) Let  = sin ⇒  = cos . Then  2 √2 − 2  =  2 sin2   1 − sin2  cos  = 4  sin2  cos2   = 4  1 2(1 + cos 2) 1 2(1 − cos 2) = 1 4 4  (1 − cos2 2) = 1 4 4  1 − 1 2(1 + cos 4)  = 1 4 4 1 2  − 1 8 sin 4 +  = 1 4 4 1 2  − 1 8 · 2sin 2 cos 2 +  = 1 4 4 1 2  − 1 2 sin cos(1 − 2sin2 ) +  = 4 8 sin−1  −  √2− 2 1 − 222  +  = 84 sin−1  −  √2− 2 2 −222  +  =  8 (22 − 2)√2 − 2 + 4 8 sin−1   +  37. Maple and Mathematica both give  sec4   = 2 3 tan + 1 3 tan sec2 , while Derive gives the second term as sin 3cos3  = 1 3 sin cos 1 cos2  = 1 3 tan sec2 . Using Formula 77, we get  sec4   = 1 3 tan sec2  + 2 3  sec2   = 1 3 tan sec2  + 2 3 tan + . 38. Derive gives  csc5   = 3 8 lntan2 − cos 8sin 32  + 4sin 14  and Maple gives − 1 4 cos sin4  − 3 8 cos sin2  + 3 8 ln(csc − cot). Using a half-angle identity for tangent, tan  2 = 1 − cos sin , we have ln tan  2 = ln 1 − cos sin = lnsin1 − cos sin = ln(csc − cot), so those two answers are equivalent. Mathematica gives °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ¤ 69  = − 3 32 csc2  2 − 1 64 csc4 2 − 3 8 log cos 2 + 3 8 log sin 2 + 32 3 sec2 2 + 64 1 sec4 2 = 3 8 log sin 2 − log cos 2 + 32 3 sec2 2 − csc2 2 + 64 1 sec4 2 − csc4 2 = 3 8 log sin(2) cos(2) + 3 32cos2(12) − sin2(12) + 64 1 cos4(12) − sin4(12) = 3 8 log tan  2 + 3 32sin cos 2(2 ( 2)2) sin − cos 22( (2) 2) + 64 1 sin cos 4(4 ( 2)2) sin − cos 44( (2) 2) Now sin2(2) − cos2(2) cos2(2) sin2(2) = 1 − cos 2 − 1 + cos 2 1 + cos 2 · 1 − cos  2 = − 2cos  2 1 − cos2  4 = −4cos sin2  and sin4(2) − cos4(2) cos4(2) sin4(2) = sin2(2) − cos2(2) cos2(2) sin2(2) sin2(2) + cos2(2) cos2(2) sin2(2) = −4cos  sin2  1 1 + cos 2 · 1 − cos 2 = − 4cos sin2  4 1 − cos2  = − 16 cos sin4  Returning to the expression for , we get  = 3 8 log tan  2 + 3 32−sin 4cos 2  + 64 1 −16 cos sin4   = 3 8 log tan 2 − 3 8 sin cos2 − 1 4 sin cos4, so all are equivalent. Now use Formula 78 to get  csc5   = −41 cot csc3  + 34  csc3   = −1 4 cos sin sin13  + 3 4−21 cotcsc + 12  csc  = − 1 4 cos sin4  − 3 8 cos sin 1 sin + 3 8  csc  = −14 sin cos4 − 3 8 sin cos2 + 3 8 ln|csc − cot| +  39. Derive gives  2 √2 + 4 = 1 4 (2 + 2)√2 + 4 − 2ln√2 + 4 + . Maple gives 1 4 (2 + 4)32 − 1 2  √2 + 4 − 2arcsinh 1 2 . Applying the command convert(%,ln); yields 1 4 (2 + 4)32 − 1 2  √2 + 4 − 2ln 1 2  + 1 2 √2 + 4 = 1 4 (2 + 4)12(2 + 4) − 2 − 2ln + √2 + 42 = 1 4 (2 + 2)√2 + 4 − 2ln√2 + 4 +  + 2ln2 Mathematica gives 1 4 (2 + 2)√3 + 2 − 2arcsinh(2). Applying the TrigToExp and Simplify commands gives 1 4 (2 + 2)√4 + 2 − 8log 1 2 + √4 + 2  = 1 4 (2 + 2)√2 + 4 − 2ln + √4 + 2 + 2ln2, so all are equivalent (without constant). Now use Formula 22 to get  2 22 + 2  = 8(22 + 22)√22 + 2 − 284 ln + √22 + 2  +  =  8 (2)(2 + 2)√4 + 2 − 2ln + √4 + 2  +  = 1 4 (2 + 2)√2 + 4 − 2ln√2 + 4 +  +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.70 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 40. Derive gives  (3  + 2) = −−2 + 3ln(34 + 2) − 34, Maple gives 3 4 ln(3 + 2) − 21 − 34 ln(), and Mathematica gives − − 2 + 3 4 log(3 + 2−) = −− 2 + 3 4 log3+ 2 = −−2 + 34 ln(3ln+ 2) = −−2 + 34 ln(3 + 2) − 3 4, so all are equivalent. Now let  = , so  =   and  = . Then  (31 + 2)  =  (31+ 2)   =  2(2 + 3 1 )  =50 −21 + 232 ln   2 + 3      +  = − 1 2 + 3 4 ln(2 + 3) − 3 4 ln +  = −21 + 34 ln(3 + 2) − 3 4 +  41. Derive and Maple give  cos4   = sin 4cos3  + 3sin8 cos + 38, while Mathematica gives 3 8 + 1 4 sin(2) + 1 32 sin(4) = 3 8 + 1 4 (2sin cos) + 1 32 (2sin2 cos2) = 3 8 + 1 2 sin cos + 1 16 [2sin cos(2cos2  − 1)] = 3 8 + 1 2 sin cos + 1 4 sin cos3  − 18 sin cos, so all are equivalent. Using tables,  cos4   74 = 1 4 cos3  sin + 3 4  cos2   64 = 1 4 cos3  sin + 3 4 1 2  + 1 4 sin2 +  = 1 4 cos3  sin + 3 8  + 16 3 (2sin cos) +  = 1 4 cos3  sin + 3 8  + 3 8 sin cos +  42. Derive gives  21 − 2  = arcsin 8  + √1 − 28(22 − 1), Maple gives −  4 (1 − 2)32 +  8 √1 − 2 + 1 8 arcsin =  8 (1 − 2)12[−2(1 − 2) + 1] + 1 8 arcsin =  8 (1 − 2)12(22 − 1) + 1 8 arcsin and Mathematica gives 1 8√1 − 2(−1 + 22) + arcsin, so all are equivalent. Now use Formula 31 to get  21 − 2  = 8(22 − 1)1 − 2 + 18 sin−1  +  43. Maple gives  tan5   = 1 4 tan4  − 1 2 tan2  + 1 2 ln(1 + tan2 ), Mathematica gives  tan5   = 1 4[−1 − 2cos(2)]sec4  − ln(cos), and Derive gives  tan5   = 1 4 tan4  − 1 2 tan2  − ln(cos ). These expressions are equivalent, and none includes absolute value bars or a constant of integration. Note that Mathematica’s and Derive’s expressions suggest that the integral is undefined where cos   0, which is not the case. Using Formula 75,  tan5   = 5 −1 1 tan5−1  −  tan5−2   = 1 4 tan4  −  tan3  . Using Formula 69,  tan3   = 1 2 tan2  + ln|cos| + , so  tan5   = 1 4 tan4  − 1 2 tan2  − ln|cos| + . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.DISCOVERY PROJECT PATTERNS IN INTEGRALS ¤ 71 44. Derive, Maple, and Mathematica all give  1 +1 √3   = 2 5  √3  + 13 √3 2 − 4 √3  + 8. [Maple adds a constant of − 16 5 . We’ll change the form of the integral by letting  = √3 , so that 3 =  and 32  = . Then  1 +1 √3   =  √31 + 2  = 3 56 15(1) 2 3 8(1)2 + 3(1)22 − 4(1)(1) √1 +   +  = 2 5 (8 + 32 − 4)√1 +  +  = 2 58 + 3 √3 2 − 4 √3   1 + √3  +  45. (a) () =  () =   √11− 2  35 = −1 1 ln   1 + √1 − 2     +  = −ln   1 + √1 − 2     + .  has domain  |  6= 0, 1 − 2  0 = { |  6= 0, ||  1} = (−10) ∪ (01).  has the same domain. (b) Derive gives () = ln√1 − 2 − 1 − ln and Mathematica gives () = ln − ln1 + √1 − 2 . Both are correct if you take absolute values of the logarithm arguments, and both would then have the same domain. Maple gives () = −arctanh1√1 − 2 . This function has domain    ||  1 −1  1√1 − 2  1 =    ||  1, 1√1 − 2  1 =    ||  1 √1 − 2  1 = ∅, the empty set! If we apply the command convert(%,ln); to Maple’s answer, we get − 1 2 ln√1 1− 2 + 1 + 12 ln1 − √1 1− 2 , which has the same domain, ∅. 46. None of Maple, Mathematica and Derive is able to evaluate  (1 + ln)1 + (ln)2 . However, if we let  = ln, then  = (1 + ln) and the integral is simply  √1 + 2 , which any CAS can evaluate. The antiderivative is 1 2 lnln + 1 + (ln)2  + 1 2ln1 + (ln)2 + . DISCOVERY PROJECT Patterns in Integrals 1. (a) The CAS results are listed. Note that the absolute value symbols are missing, as is the familiar “ +  ”. (i)  ( + 2)( 1 + 3)  = ln( + 2) − ln( + 3) (ii)  ( + 1)( 1 + 5)  = ln(4+ 1) − ln(4+ 5) (iii)  ( + 2)( 1 − 5)  = ln(7− 5) − ln(7+ 2) (iv)  ( + 2) 1 2  = − + 2 1 (b) If  6= , it appears that ln( + ) is divided by  −  and ln( + ) is divided by  − , so we guess that  ( + )( 1  + )  = ln(−+) + ln( −+) + . If  = , as in part (a)(iv), it appears that  ( +1)2  = − +1  + . (c) The CAS verifies our guesses. Now 1 ( + )( + ) =   +  +   +  ⇒ 1 = ( + ) + ( + ) Setting  = − gives  = 1( − ) and setting  = − gives  = 1( − ). So  ( + )( 1  + )  =  1(+−) + 1(+−)  = ln|−+| + ln|−+| +  [continued] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.72 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION and our guess for  6=  is correct. If  = , then 1 ( + )( + ) = 1 ( + )2 = ( + )−2. Letting  =  +  ⇒  = , we have  ( + )−2  =  −2  = −1 +  = − +1  + , and our guess for  =  is also correct. 2. (a) (i)  sin cos 2  = cos 2  − cos 3 6  (ii)  sin 3 cos 7  = cos 4 8  − cos 10 20  (iii)  sin 8 cos 3  = −cos 11 22  − cos 5 10  (b) Looking at the sums and differences of  and  in part (a), we guess that  sin cos   = cos(( 2(−−))) − cos(( 2(++))) +  Note that cos(( − )) = cos(( − )). (c) The CAS verifies our guess. Again, we can prove that the guess is correct by differentiating:   cos(( 2(−−))) − cos(( 2(++))) = 2(1− ) [−sin(( − ))]( − ) − 2(1+ ) [−sin(( + ))]( + ) = 1 2 sin( − ) + 1 2 sin( + ) = 1 2 (sin cos − cos sin) + 1 2(sin cos + cos  sin) = sin cos Our formula is valid for  6= . 3. (a) (i)  ln  = ln −  (ii)  ln  = 1 22 ln − 1 42 (iii)  2 ln  = 1 33 ln − 1 93 (iv)  3 ln  = 1 44 ln − 16 1 4 (v)  7 ln  = 1 88 ln − 64 1 8 (b) We guess that   ln  = 1  + 1 +1 ln − 1 ( + 1)2 +1. (c) Let  = ln,  =   ⇒  =   ,  = 1  + 1 +1. Then   ln  =  + 1 1 +1 ln −  + 1 1    =  + 1 1 +1 ln −  + 1 1 ·  + 1 1 +1, which verifies our guess. We must have  + 1 6= 0 ⇔  6= −1. 4. (a) (i)    = ( − 1) (ii)  2  = (2 − 2 + 2) (iii)  3  = (3 − 32 + 6 − 6) (iv)  4  = (4 − 43 + 122 − 24 + 24) (v)  5  = 5 − 54 + 203 − 602 + 120 − 120 (b) Notice from part (a) that we can write  4  = (4 − 43 + 4 · 32 − 4 · 3 · 2 + 4 · 3 · 2 · 1) and  5  = (5 − 54 + 5 · 43 − 5 · 4 · 32 + 5 · 4 · 3 · 2 − 5 · 4 · 3 · 2 · 1) °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 73 So we guess that  6  = (6 − 65 + 6 · 54 − 6 · 5 · 43 + 6 · 5 · 4 · 32 − 6 · 5 · 4 · 3 · 2 + 6 · 5 · 4 · 3 · 2 · 1) = (6 − 65 + 304 − 1203 + 3602 − 720 + 720) The CAS verifies our guess. (c) From the results in part (a), as well as our prediction in part (b), we speculate that    =  − −1 + ( − 1)−2 − ( − 1)( − 2)−3 + · · · ± ! ∓ ! =   =0 (−1)− ! ! . (We have reversed the order of the polynomial’s terms.) (d) Let  be the statement that    =   =0 (−1)− ! ! . 1 is true by part (a)(i). Suppose  is true for some , and consider +1. Integrating by parts with  = +1,  =   ⇒  = ( + 1) ,  = , we get  +1  = +1 − ( + 1)   = +1 − ( + 1) =0  (−1)− !!  = +1 − ( + 1) =0  (−1)− !!  = +1 + =0  (−1)−+1 ( + 1) ! !  =  +1   =0 (−1)(+1)− ( + 1)! !  This verifies  for  =  + 1. Thus, by mathematical induction,  is true for all , where  is a positive integer. 7.7 Approximate Integration 1. (a) ∆ = ( − ) = (4 − 0)2 = 2 2 = 2 =1 (−1)∆ = (0) · 2 + (1) · 2 = 2 [(0) + (2)] = 2(05 + 25) = 6 2 = 2 =1 ()∆ = (1) · 2 + (2) · 2 = 2 [(2) + (4)] = 2(25 + 35) = 12 2 = 2 =1 ()∆ = (1) · 2 + (2) · 2 = 2 [(1) + (3)] ≈ 2(16 + 32) = 96 (b) 2 is an underestimate, since the area under the small rectangles is less than the area under the curve, and 2 is an overestimate, since the area under the large rectangles is greater than the area under the curve. It appears that 2 is an overestimate, though it is fairly close to . See the solution to Exercise 47 for a proof of the fact that if  is concave down on [ ], then the Midpoint Rule is an overestimate of  (). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.74 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (c) 2 =  1 2 ∆[(0) + 2(1) + (2)] = 2 2[(0) + 2(2) + (4)] = 05 + 2(25) + 35 = 9. This approximation is an underestimate, since the graph is concave down. Thus, 2 = 9  . See the solution to Exercise 47 for a general proof of this conclusion. (d) For any , we will have         . 2. The diagram shows that 4  4  02 ()  4, and it appears that 4 is a bit less than 02 (). In fact, for any function that is concave upward, it can be shown that     02 ()    . (a) Since 09540  08675  08632  07811, it follows that  = 09540,  = 08675,  = 08632, and  = 07811. (b) Since   02 ()  , we have 08632  02 ()  08675. 3. () = cos2, ∆ = 1 −4 0 = 1 4 (a) 4 = 41· 2 (0) + 2 1 4  + 2 2 4  + 2 3 4  + (1) ≈ 0895759 (b) 4 = 1 4  1 8  +  3 8  +  5 8  +  7 8  ≈ 0908907 The graph shows that  is concave down on [0 1]. So 4 is an underestimate and 4 is an overestimate. We can conclude that 0895759  01 cos2   0908907. 4. (a) Since  is increasing on [01], 2 will underestimate  (since the area of the darkest rectangle is less than the area under the curve), and 2 will overestimate . Since  is concave upward on [01], 2 will underestimate  and 2 will overestimate  (the area under the straight line segments is greater than the area under the curve). (b) For any , we will have         . (c) 5 = 5 =1 (−1) ∆ = 1 5[(00) + (02) + (04) + (06) + (08)] ≈ 01187 5 = 5 =1 () ∆ = 1 5[(02) + (04) + (06) + (08) + (1)] ≈ 02146 5 = 5 =1 () ∆ = 1 5[(01) + (03) + (05) + (07) + (09)] ≈ 01622 5 =  1 2 ∆[(0) + 2(02) + 2(04) + 2(06) + 2(08) + (1)] ≈ 01666 From the graph, it appears that the Midpoint Rule gives the best approximation. (This is in fact the case, since  ≈ 016371405.) °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 75 5. (a) () =  1 + 2 , ∆ =  −  = 2 10 − 0 = 15 10 = 1 5   10 1  +  10 3  +  10 5  + · · · +   19 10  ≈ 0806598 (b) 10 = 51· 3 (0) + 4 1 5  + 2 2 5  + 4 3 5  + 2 4 5  + · · · + 4 9 5  + (2) ≈ 0804779 Actual:  = 02 1 +2  =  1 2 ln 1 + 2 2 0 [ = 1 + 2,  = 2 ] = 1 2 ln5 − 1 2 ln1 = 1 2 ln5 ≈ 0804719 Errors:  = actual − 10 =  − 10 ≈ −0001879  = actual − 10 =  − 10 ≈ −0000060 6. (a) () = cos, ∆ =  −   =  − 0 4 =  4 4 =  4  8  +  38 +  58 +   78 ≈ −1945744 (b) 4 = 4· 3 (0) + 4 4  + 2 24 + 4 34 + () ≈ −1985611 Actual:  = 0 cos  = sin + cos 0 [use parts with  =  and  = cos ] = (0 + (−1)) − (0 + 1) = −2 Errors:  = actual − 4 =  − 4 ≈ −0054256  = actual − 4 =  − 4 ≈ −0014389 7. () = √3 − 1, ∆ =  −   = 2 − 1 10 = 1 10 (a) 10 = 101· 2[(1) + 2(11) + 2(12) + 2(13) + 2(14) + 2(15) + 2(16) + 2(17) + 2(18) + 2(19) + (2)] ≈ 1506361 (b) 10 = 10 1 [(105) + (115) + (125) + (135) + (145) + (155) + (165) + (175) + (185) + (195)] ≈ 1518362 (c) 10 = 101· 3[(1) + 4(11) + 2(12) + 4(13) + 2(14) + 4(15) + 2(16) + 4(17) + 2(18) + 4(19) + (2)] ≈ 1511519 8. () = 1 1 + 6 , ∆ =  −  = 2 −8 0 = 14 (a) 8 = 41· 2[(0) + 2(025) + 2(05) + 2(075) + 2(1) + 2(125) + 2(15) + 2(175) + (2)] ≈ 1040756 (b) 8 = 1 4[(0125) + (0375) + (0625) + (0875) + (1125) + (1375) + (1625) + (1875)] ≈ 1041109 (c) 8 = 41· 3[(0) + 4(025) + 2(05) + 4(075) + 2(1) + 4(125) + 2(15) + 4(175) + (2)] ≈ 1042172 9. () =  1 + 2 , ∆ =  −  = 210 − 0 = 1 5 (a) 10 = 51· 2[(0) + 2(02) + 2(04) + 2(06) + 2(08) + 2(1) + 2(12) + 2(14) + 2(16) + 2(18) + (2)] ≈ 2660833 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.76 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (b) 10 = 1 5[(01) + (03) + (05) + (07) + (09) + (11) + (13) + (15) + (17) + (19)] ≈ 2664377 (c) 10 = 51· 3[(0) + 4(02) + 2(04) + 4(06) + 2(08) + 4(1) + 2(12) + 4(14) + 2(16) + 4(18) + (2)] ≈ 2663244 10. () = √3 1 + cos, ∆ = 24− 0 = 8 (a) 4 = 8· 2 (0) + 2 8  + 2 28  + 2 38  +  2  ≈ 1838967 (b) 4 = 8  16   +  316   +  516   +  716   ≈ 1845390 (c) 4 = 8· 3 (0) + 4 8  + 2 28  + 4 38  +  2  ≈ 1843245 11. () = 3 sin, ∆ = 4 −8 0 = 1 2 (a) 8 = 21· 2 (0) + 2 1 2  + 2(1) + 2 3 2  + 2(2) + 2 5 2  + 2(3) + 2 7 2  + (4) ≈ −7276910 (b) 8 = 1 2  1 4  +  3 4  +  5 4  +  7 4  +  9 4  +  11 4  +  13 4  +  15 4  ≈ −4818251 (c) 8 = 21· 3 (0) + 4 1 2  + 2(1) + 4 3 2  + 2(2) + 4 5 2  + 2(3) + 4 7 2  + (4) ≈ −5605350 12. () = 1, ∆ = 3 −8 1 = 1 4 (a) 8 = 41· 2 (1) + 2 5 4  + 2 3 2  + 2 7 4  + 2(2) + 2 9 4  + 2 5 2  + 2 11 4  + (3) ≈ 3534934 (b) 8 = 1 4  9 8  +  11 8  +  13 8  +  15 8  +  17 8  +  19 8  +  21 8  +  23 8  ≈ 3515248 (c) 8 = 41· 3 (1) + 4 5 4  + 2 3 2  + 4 7 4  + 2(2) + 4 9 4  + 2 5 2  + 4 11 4  + (3) ≈ 3522375 13. () = √ cos, ∆ = 4 −8 0 = 1 2 (a) 8 = 21· 2 (0) + 2 1 2  + 2(1) + 2 3 2  + 2(2) + 2 5 2  + 2(3) + 2 7 2  + (4) ≈ −2364034 (b) 8 = 1 2  1 4  +  3 4  +  5 4  +  7 4  +  9 4  +  11 4  +  13 4  +  15 4  ≈ −2310690 (c) 8 = 21· 3 (0) + 4 1 2  + 2(1) + 4 3 2  + 2(2) + 4 5 2  + 2(3) + 4 7 2  + (4) ≈ −2346520 14. () = 1 ln, ∆ = 310 − 2 = 10 1 (a) 10 = 101· 2{(2) + 2[(21) + (22) + · · · + (29)] + (3)} ≈ 1119061 (b) 10 = 10 1 [(205) + (215) + · · · + (285) + (295)] ≈ 1118107 (c) 10 = 101· 3[(2) + 4(21) + 2(22) + 4(23) + 2(24) + 4(25) + 2(26) + 4(27) + 2(28) + 4(29) + (3)] ≈ 1118428 15. () = 2 1 + 4 , ∆ = 110 − 0 = 10 1 (a) 10 = 101· 2{(0) + 2[(01 + (02) + · · · + (09)] + (1)} ≈ 0243747 (b) 10 = 10 1 [(005) + (015) + · · · + (085) + (095)] ≈ 0243748 (c) 10 = 101· 3[(0) + 4(01) + 2(02) + 4(03) + 2(04) + 4(05) + 2(06) + 4(07) + 2(08) + 4(09) + (1)] ≈ 0243751 Note: 01 () ≈ 024374775. This is a rare case where the Trapezoidal and Midpoint Rules give better approximations than Simpson’s Rule. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 77 16. () = sin  , ∆ = 3 − 1 4 = 1 2 (a) 4 = 21· 2[(1) + 2(15) + 2(2) + 2(25) + (3)] ≈ 0901645 (b) 4 = 1 2[(125) + (175) + (225) + (275)] ≈ 0903031 (c) 4 = 21· 3[(1) + 4(15) + 2(2) + 4(25) + (3)] ≈ 0902558 17. () = ln(1 + ), ∆ = 4 −8 0 = 1 2 (a) 8 = 21· 2 {(0) + 2[(05) + (1) + · · · + (3) + (35)] + (4)} ≈ 8814278 (b) 8 = 1 2[(025) + (075) + · · · + (325) + (375)] ≈ 8799212 (c) 8 = 21· 3[(0) + 4(05) + 2(1) + 4(15) + 2(2) + 4(25) + 2(3) + 4(35) + (4)] ≈ 8804229 18. () = √ + 3, ∆ = 110 − 0 = 10 1 (a) 10 = 21· 2 {(0) + 2[(01) + (02) + · · · + (08) + (09)] + (1)} ≈ 0787092 (b) 10 = 1 2[(005) + (015) + · · · + (085) + (095)] ≈ 0793821 (c) 10 = 21· 3[(0) + 4(01) + 2(02) + 4(03) + 2(04) + 4(05) + 2(06) + 4(07) + 2(08) + 4(09) + (1)] ≈ 0789915 19. () = cos(2), ∆ = 1 −8 0 = 1 8 (a) 8 = 81· 2(0) + 2 1 8 +  2 8 + · · · +  7 8 + (1) ≈ 0902333 8 = 1 8 16 1  +  16 3  +  16 5  + · · · +  15 16 = 0905620 (b) () = cos(2),  0() = −2sin(2),  00() = −2sin(2) − 42 cos(2). For 0 ≤  ≤ 1, sin and cos are positive, so | 00()| = 2 sin(2) + 42 cos(2) ≤ 2 · 1 + 4 · 1 · 1 = 6 since sin(2) ≤ 1 and cos2 ≤ 1 for all , and 2 ≤ 1 for 0 ≤  ≤ 1. So for  = 8, we take  = 6,  = 0, and  = 1 in Theorem 3, to get | | ≤ 6 · 13(12 · 82) = 128 1 = 00078125 and || ≤ 256 1 = 000390625. [A better estimate is obtained by noting from a graph of  00 that | 00()| ≤ 4 for 0 ≤  ≤ 1.] (c) Take  = 6 [as in part (b)] in Theorem 3. | | ≤ ( − )3 122 ≤ 00001 ⇔ 6(112−20)3 ≤ 10−4 ⇔ 1 22 ≤ 10 14 ⇔ 22 ≥ 104 ⇔ 2 ≥ 5000 ⇔  ≥ 71. Take  = 71 for . For , again take  = 6 in Theorem 3 to get || ≤ 10−4 ⇔ 42 ≥ 104 ⇔ 2 ≥ 2500 ⇔  ≥ 50. Take  = 50 for . 20. () = 1, ∆ = 210 − 1 = 10 1 (a) 10 = 101· 2[(1) + 2(11) + 2(12) + · · · + 2(19) + (2)] ≈ 2021976 10 = 10 1 [(105) + (115) + (125) + · · · + (195)] ≈ 2019102 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.78 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (b) () = 1,  0() = − 1 2 1,  00() = 2+ 1 4 1. Now  00 is decreasing on [12], so let  = 1 to take  = 3. | | ≤ 3(2 − 1)3 12(10)2 =  400 ≈ 0006796. || ≤ |2 | = 800  ≈ 0003398. (c) Take  = 3 [as in part (b)] in Theorem 3. | | ≤ ( − )3 122 ≤ 00001 ⇔ 3(2 12−21)3 ≤ 10−4 ⇔  42 ≤ 10 14 ⇔ 2 ≥ 1044 ⇔  ≥ 83. Take  = 83 for . For , again take  = 3 in Theorem 3 to get || ≤ 10−4 ⇔ 2 ≥ 104 8 ⇔  ≥ 59. Take  = 59 for . 21. () = sin, ∆ = 10 − 0 = 10  (a) 10 = 10· 2 (0) + 2 10   + 2 210   + · · · + 2 910   + () ≈ 1983524 10 = 10   20   +  320   +  520   + · · · +  19 20  ≈ 2008248 10 = 10· 3 (0) + 4 10   + 2 210   + 4 310   + · · · + 4 910   + () ≈ 2000110 Since  = 0 sin  = −cos 0 = 1 − (−1) = 2,  =  − 10 ≈ 0016476,  =  − 10 ≈ −0008248, and  =  − 10 ≈ −0000110. (b) () = sin ⇒   ()()   ≤ 1, so take  = 1 for all error estimates. | | ≤ ( − )3 122 = 1( − 0)3 12(10)2 = 3 1200 ≈ 0025839. || ≤ | | 2 = 3 2400 ≈ 0012919. || ≤ ( − )5 1804 = 1( − 0)5 180(10)4 = 5 1,800,000 ≈ 0000170. The actual error is about 64% of the error estimate in all three cases. (c) | | ≤ 000001 ⇔ 3 122 ≤ 10 15 ⇔ 2 ≥ 1012 53 ⇒  ≥ 5083. Take  = 509 for . || ≤ 000001 ⇔ 3 242 ≤ 10 15 ⇔ 2 ≥ 1024 53 ⇒  ≥ 3594. Take  = 360 for . || ≤ 000001 ⇔ 5 1804 ≤ 10 15 ⇔ 4 ≥ 10180 55 ⇒  ≥ 203. Take  = 22 for  (since  must be even). 22. From Example 7(b), we take  = 76 to get || ≤ 76(1)5 1804 ≤ 000001 ⇒ 4 ≥ 180(076 00001)  ⇒  ≥ 184. Take  = 20 (since  must be even). 23. (a) Using a CAS, we differentiate () = cos  twice, and find that  00() = cos (sin2  − cos). From the graph, we see that the maximum value of | 00()| occurs at the endpoints of the interval [02]. Since  00(0) = −, we can use  =  or  = 28. (b) A CAS gives 10 ≈ 7954926518. (In Maple, use Student[Calculus1][RiemannSum] or Student[Calculus1][ApproximateInt].) °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 79 (c) Using Theorem 3 for the Midpoint Rule, with  = , we get || ≤ (2 − 0)3 24 · 102 ≈ 0280945995. With  = 28, we get || ≤ 28(2 − 0)3 24 · 102 = 0 289391916. (d) A CAS gives  ≈ 7954926521. (e) The actual error is only about 3 × 10−9, much less than the estimate in part (c). (f) We use the CAS to differentiate twice more, and then graph  (4)() = cos (sin4  − 6sin2  cos + 3 − 7sin2  + cos). From the graph, we see that the maximum value of   (4)()   occurs at the endpoints of the interval [02]. Since  (4)(0) = 4, we can use  = 4 or  = 109. (g) A CAS gives 10 ≈ 7953789422. (In Maple, use Student[Calculus1][ApproximateInt].) (h) Using Theorem 4 with  = 4, we get || ≤ 4(2 − 0)5 180 · 104 ≈ 0059153618. With  = 109, we get || ≤ 109(2 − 0)5 180 · 104 ≈ 0059299814. (i) The actual error is about 7954926521 − 7953789422 ≈ 000114. This is quite a bit smaller than the estimate in part (h), though the difference is not nearly as great as it was in the case of the Midpoint Rule. (j) To ensure that || ≤ 00001, we use Theorem 4: || ≤ 4(2)5 180 · 4 ≤ 00001 ⇒ 180 4·(200001 )5 ≤ 4 ⇒ 4 ≥ 5,915,362 ⇔  ≥ 493. So we must take  ≥ 50 to ensure that | − | ≤ 00001. ( = 109 leads to the same value of .) 24. (a) Using the CAS, we differentiate () = √4 − 3 twice, and find that  00() = − 94 4(4 − 3)32 − 3 (4 − 3)12 . From the graph, we see that | 00()|  22 on [−11]. (b) A CAS gives 10 ≈ 3995804152. (In Maple, use Student[Calculus1][RiemannSum] or Student[Calculus1][ApproximateInt].) (c) Using Theorem 3 for the Midpoint Rule, with  = 22, we get || ≤ 22[1 − (−1)]3 24 · 102 ≈ 000733. (d) A CAS gives  ≈ 3995487677. (e) The actual error is about −00003165, much less than the estimate in part (c). (f) We use the CAS to differentiate twice more, and then graph  (4)() = 9 16 2(6 − 2243 − 1280) (4 − 3)72 . From the graph, we see that   (4)()    181 on [−1 1]. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.80 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (g) A CAS gives 10 ≈ 3995449790. (In Maple, use Student[Calculus1][ApproximateInt].) (h) Using Theorem 4 with  = 181, we get || ≤ 181[1 − (−1)]5 180 · 104 ≈ 0000322. (i) The actual error is about 3995487677 − 3995449790 ≈ 00000379. This is quite a bit smaller than the estimate in part (h). (j) To ensure that || ≤ 00001, we use Theorem 4: || ≤ 181(2)5 180 · 4 ≤ 00001 ⇒ 180 18·1(2) 00001 5 ≤ 4 ⇒ 4 ≥ 32,178 ⇒  ≥ 134. So we must take  ≥ 14 to ensure that | − | ≤ 00001. 25.  = 01  = [( − 1)]1 0 [parts or Formula 96] = 0 − (−1) = 1, () = , ∆ = 1  = 5: 5 = 1 5[(0) + (02) + (04) + (06) + (08)] ≈ 0742943 5 = 1 5[(02) + (04) + (06) + (08) + (1)] ≈ 1286599 5 = 51· 2[(0) + 2(02) + 2(04) + 2(06) + 2(08) + (1)] ≈ 1014771 5= 1 5[(01) + (03) + (05) + (07) + (09)] ≈ 0992621  =  − 5 ≈ 1 − 0742943 = 0257057  ≈ 1 − 1286599 = −0286599  ≈ 1 − 1014771 = −0014771  ≈ 1 − 0992621 = 0007379  = 10: 10 = 10 1 [(0) + (01) + (02) + · · · + (09)] ≈ 0867782 10 = 10 1 [(01) + (02) + · · · + (09) + (1)] ≈ 1139610 10 = 101· 2{(0) + 2[(01) + (02) + · · · + (09)] + (1)} ≈ 1003696 10= 10 1 [(005) + (015) + · · · + (085) + (095)] ≈ 0998152  =  − 10 ≈ 1 − 0867782 = 0132218  ≈ 1 − 1139610 = −0139610  ≈ 1 − 1003696 = −0003696  ≈ 1 − 0998152 = 0001848  = 20: 20 = 20 1 [(0) + (005) + (010) + · · · + (095)] ≈ 0932967 20 = 20 1 [(005) + (010) + · · · + (095) + (1)] ≈ 1068881 20 = 201· 2{(0) + 2[(005) + (010) + · · · + (095)] + (1)} ≈ 1000924 20= 20 1 [(0025) + (0075) + (0125) + · · · + (0975)] ≈ 0999538  =  − 20 ≈ 1 − 0932967 = 0067033  ≈ 1 − 1068881 = −0068881  ≈ 1 − 1000924 = −0000924  ≈ 1 − 0999538 = 0000462 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 81      5 0742943 1286599 1014771 0992621 10 0867782 1139610 1003696 0998152 20 0932967 1068881 1000924 0999538      5 0257057 −0286599 −0014771 0007379 10 0132218 −0139610 −0003696 0001848 20 0067033 −0068881 −0000924 0000462 Observations: 1.  and  are always opposite in sign, as are  and . 2. As  is doubled,  and  are decreased by about a factor of 2, and  and  are decreased by a factor of about 4. 3. The Midpoint approximation is about twice as accurate as the Trapezoidal approximation. 4. All the approximations become more accurate as the value of  increases. 5. The Midpoint and Trapezoidal approximations are much more accurate than the endpoint approximations. 26.  = 12 12  = −12 1 = −12 − (−1) = 12 () = 12  ∆ = 1  = 5: 5 = 1 5[(1) + (12) + (14) + (16) + (18)] ≈ 0580783 5 = 1 5[(12) + (14) + (16) + (18) + (2)] ≈ 0430783 5 = 51· 2[(1) + 2(12) + 2(14) + 2(16) + 2(18) + (2)] ≈ 0505783 5= 1 5[(11) + (13) + (15) + (17) + (19)] ≈ 0497127  =  − 5 ≈ 1 2 − 0580783 = −0080783  ≈ 1 2 − 0430783 = 0069217  ≈ 1 2 − 0505783 = −0005783  ≈ 1 2 − 0497127 = 0002873  = 10: 10 = 1 10[(1) + (11) + (12) + · · · + (19)] ≈ 0538955 10 = 10 1 [(11) + (12) + · · · + (19) + (2)] ≈ 0463955 10 = 101· 2{(1) + 2[(11) + (12) + · · · + (19)] + (2)} ≈ 0501455 10= 10 1 [(105) + (115) + · · · + (185) + (195)] ≈ 0499274  =  − 10 ≈ 1 2 − 0538955 = −0038955  ≈ 1 2 − 0463955 = 0036049  ≈ 1 2 − 0501455 = −0001455  ≈ 1 2 − 0499274 = 0000726  = 20: 20 = 1 20[(1) + (105) + (110) + · · · + (195)] ≈ 0519114 20 = 20 1 [(105) + (110) + · · · + (195) + (2)] ≈ 0481614 20 = 201· 2{(1) + 2[(105) + (110) + · · · + (195)] + (2)} ≈ 0500364 20= 20 1 [(1025) + (1075) + (1125) + · · · + (1975)] ≈ 0499818 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.82 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION  =  − 20 ≈ 1 2 − 0519114 = −0019114  ≈ 1 2 − 0481614 = 0018386  ≈ 1 2 − 0500364 = −0000364  ≈ 1 2 − 0499818 = 0000182      5 0580783 0430783 0505783 0497127 10 0538955 0463955 0501455 0499274 20 0519114 0481614 0500364 0499818      5 −0080783 0069217 −0005783 0002873 10 −0038955 0036049 −0001455 0000726 20 −0019114 0018386 −0000364 0000182 Observations: 1.  and  are always opposite in sign, as are  and . 2. As  is doubled,  and  are decreased by about a factor of 2, and  and  are decreased by a factor of about 4. 3. The Midpoint approximation is about twice as accurate as the Trapezoidal approximation. 4. All the approximations become more accurate as the value of  increases. 5. The Midpoint and Trapezoidal approximations are much more accurate than the endpoint approximations. 27.  = 02 4  =  1 552 0 = 32 5 − 0 = 64, () = 4, ∆ = 2 − 0 = 2  = 6: 6 = 62· 2 (0) + 2 1 3  +  2 3  +  3 3  +  4 3  +  5 3  + (2) ≈ 6695473 6 = 2 6  1 6  +  3 6  +  5 6  +  7 6  +  9 6  +  11 6  ≈ 6252572 6 = 62· 3 (0) + 4 1 3  + 2 2 3  + 4 3 3  + 2 4 3  + 4 5 3  + (2) ≈ 6403292  =  − 6 ≈ 64 − 6695473 = −0295473  ≈ 64 − 6252572 = 0147428  ≈ 64 − 6403292 = −0003292  = 12: 12 = 2 12 · 2 (0) + 2 1 6  +  2 6  +  3 6  + · · · +  11 6  + (2) ≈ 6474023 6 = 12 2  12 1  +  12 3  +  12 5  + · · · +  23 12  ≈ 6363008 6 = 122· 3 (0) + 4 1 6  + 2 2 6  + 4 3 6  + 2 4 6  + · · · + 4 11 6  + (2) ≈ 6400206  =  − 12 ≈ 64 − 6474023 = −0074023  ≈ 64 − 6363008 = 0036992  ≈ 64 − 6400206 = −0000206     6 6695473 6252572 6403292 12 6474023 6363008 6400206     6 −0295473 0147428 −0003292 12 −0074023 0036992 −0000206 Observations: 1.  and  are opposite in sign and decrease by a factor of about 4 as  is doubled. 2. The Simpson’s approximation is much more accurate than the Midpoint and Trapezoidal approximations, and  seems to decrease by a factor of about 16 as  is doubled. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 83 28.  = 14 √1  = 2√ 4 1 = 4 − 2 = 2, () = √1 , ∆ = 4 − 1 = 3  = 6: 6 = 63· 2(1) + 2 3 2 +  4 2 +  5 2 +  6 2 +  7 2 + (4) ≈ 2008966 6 = 3 6 5 4 +  7 4 +  9 4 +  11 4  +  13 4  +  15 4  ≈ 1995572 6 = 63· 3(1) + 4 3 2 + 2 4 2 + 4 5 2 + 2 6 2 + 4 7 2 + (4) ≈ 2000469  =  − 6 ≈ 2 − 2008966 = −0008966,  ≈ 2 − 1995572 = 0004428,  ≈ 2 − 2000469 = −0000469  = 12: 12 = 3 12 · 2(1) + 2 5 4 +  6 4 +  7 4 + · · · +  15 4  + (4) ≈ 2002269 12 = 3 12 9 8 +  11 8  +  13 8  + · · · +  31 8  ≈ 1998869 12 = 3 12 · 3(1) + 4 5 4 + 2 6 4 + 4 7 4 + 2 8 4 + · · · + 4 15 4  + (4) ≈ 2000036  =  − 12 ≈ 2 − 2002269 = −0002269  ≈ 2 − 1998869 = 0001131  ≈ 2 − 2000036 = −0000036     6 2008966 1995572 2000469 12 2002269 1998869 2000036     6 −0008966 0004428 −0000469 12 −0002269 0001131 −0000036 Observations: 1.  and  are opposite in sign and decrease by a factor of about 4 as  is doubled. 2. The Simpson’s approximation is much more accurate than the Midpoint and Trapezoidal approximations, and  seems to decrease by a factor of about 16 as  is doubled. 29. (a) ∆ = ( − ) = (6 − 0)6 = 1 6 = 1 2[(0) + 2(1) + 2(2) + 2(3) + 2(4) + 2(5) + (6)] ≈ 1 2[2 + 2(1) + 2(3) + 2(5) + 2(4) + 2(3) + 4] = 1 2(38) = 19 (b) 6 = 1[(05) + (15) + (25) + (35) + (45) + (55)] ≈ 13 + 15 + 46 + 47 + 33 + 32 = 186 (c) 6 = 1 3[(0) + 4(1) + 2(2) + 4(3) + 2(4) + 4(5) + (6)] ≈ 1 3[2 + 4(1) + 2(3) + 4(5) + 2(4) + 4(3) + 4] = 1 3(56) = 186 30. If  = distance from left end of pool and  = () = width at , then Simpson’s Rule with  = 8 and ∆ = 2 gives Area = 016   ≈ 2 3[0 + 4(62) + 2(72) + 4(68) + 2(56) + 4(50) + 2(48) + 4(48) + 0] ≈ 84 m2. 31. (a) 15 () ≈ 4 = 5 −4 1[(15) + (25) + (35) + (45)] = 1(29 + 36 + 40 + 39) = 144 (b) −2 ≤  00() ≤ 3 ⇒ | 00()| ≤ 3 ⇒  = 3, since | 00()| ≤ . The error estimate for the Midpoint Rule is || ≤ ( − )3 242 = 3(5 − 1)3 24(4)2 = 1 2 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.84 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 32. (a) 016 () ≈ 8 = 186·−3 0[(0) + 4(02) + 2(04) + 4(06) + 2(08) + 4(10) + 2(12) + 4(14) + (16)] = 1 15[121 + 4(116) + 2(113) + 4(111) + 2(117) + 4(122) + 2(126) + 4(130) + 132] = 1 15(2881) = 2881 150 ≈ 192 (b) −5 ≤ (4)() ≤ 2 ⇒   (4)()   ≤ 5 ⇒  = 5, since   (4)()   ≤ . The error estimate for Simpson’s Rule is || ≤ ( − )5 1804 = 5(16 − 0)5 180(8)4 = 2 28,125 = 71 × 10−5. 33. We use Simpson’s Rule with  = 12 and ∆ = 2412 − 0 = 2. 12 = 2 3[(0) + 4(2) + 2(4) + 4(6) + 2(8) + 4(10) + 2(12) + 4(14) + 2(16) + 4(18) + 2(20) + 4(22) + (24)] ≈ 2 3[666 + 4(654) + 2(644) + 4(617) + 2(673) + 4(721) + 2(749) + 4(774) + 2(791) + 4(754) + 2(756) + 4(714) + 675] = 2 3(25503) = 17002. Thus, 024 () ≈ 12 and ave = 241− 0 024 () ≈ 7084◦F. 34. We use Simpson’s Rule with  = 10 and ∆ = 1 2: distance = 05 () ≈ 10 = 21· 3[(0) + 4(05) + 2(1) + · · · + 4(45) + (5)] = 1 6 [0 + 4(467) + 2(734) + 4(886) + 2(973) + 4(1022) + 2(1051) + 4(1067) + 2(1076) + 4(1081) + 1081] = 1 6 (26841) = 44735 m 35. By the Net Change Theorem, the increase in velocity is equal to 06 (). We use Simpson’s Rule with  = 6 and ∆ = (6 − 0)6 = 1 to estimate this integral: 06 () ≈ 6 = 1 3[(0) + 4(1) + 2(2) + 4(3) + 2(4) + 4(5) + (6)] ≈ 1 3[0 + 4(05) + 2(41) + 4(98) + 2(129) + 4(95) + 0] = 1 3(1132) = 3773 fts 36. By the Net Change Theorem, the total amount of water that leaked out during the first six hours is equal to 06 (). We use Simpson’s Rule with  = 6 and ∆ = 6 −6 0 = 1 to estimate this integral: 06 () ≈ 6 = 1 3[(0) + 4(1) + 2(2) + 4(3) + 2(4) + 4(5) + (6)] ≈ 1 3[4 + 4(3) + 2(24) + 4(19) + 2(14) + 4(11) + 1] = 1 3(366) = 122 liters 37. By the Net Change Theorem, the energy used is equal to 06 (). We use Simpson’s Rule with  = 12 and ∆ = 6 − 0 12 = 1 2 to estimate this integral: 06 () ≈ 12 = 132[(0) + 4(05) + 2(1) + 4(15) + 2(2) + 4(25) + 2(3) + 4(35) + 2(4) + 4(45) + 2(5) + 4(55) + (6)] = 1 6 [1814 + 4(1735) + 2(1686) + 4(1646) + 2(1637) + 4(1609) + 2(1604) + 4(1611) + 2(1621) + 4(1666) + 2(1745) + 4(1886) + 2052] = 1 6 (61,064) = 10,1773 megawatt-hours °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.7 APPROXIMATE INTEGRATION ¤ 85 38. By the Net Change Theorem, the total amount of data transmitted is equal to 08 () × 3600 [since () is measured in megabits per second and  is in hours]. We use Simpson’s Rule with  = 8 and ∆ = (8 − 0)8 = 1 to estimate this integral: 08 () ≈ 8 = 1 3[(0) + 4(1) + 2(2) + 4(3) + 2(4) + 4(5) + 2(6) + 4(7) + (8)] ≈ 1 3[035 + 4(032) + 2(041) + 4(050) + 2(051) + 4(056) + 2(056) + 4(083) + 088] = 1 3 (1303) = 4343 Now multiply by 3600 to obtain 15,636 megabits. 39. (a) Let  = () denote the curve. Using disks,  = 210 [()]2  =  210 () = 1. Now use Simpson’s Rule to approximate 1: 1 ≈ 8 = 10 − 2 3(8) [(2) + 4(3) + 2(4) + 4(5) + 2(6) + 4(7) + (8)] ≈ 1 3[02 + 4(15)2 + 2(19)2 + 4(22)2 + 2(30)2 + 4(38)2 + 2(40)2 + 4(31)2 + 02] = 1 3 (18178) Thus,  ≈  · 1 3(18178) ≈ 1904 or 190 cubic units. (b) Using cylindrical shells,  = 210 2() = 2 210 () = 21. Now use Simpson’s Rule to approximate 1: 1 ≈ 8 = 10 − 2 3(8) [2(2) + 4 · 3(3) + 2 · 4(4) + 4 · 5(5) + 2 · 6(6) + 4 · 7(7) + 2 · 8(8) + 4 · 9(9) + 10(10)] ≈ 1 3[2(0) + 12(15) + 8(19) + 20(22) + 12(30) + 28(38) + 16(40) + 36(31) + 10(0)] = 1 3 (3952) Thus,  ≈ 2 · 1 3(3952) ≈ 8277 or 828 cubic units. 40. Work = 018 () ≈ 6 = 18 6 ·−30 [(0) + 4(3) + 2(6) + 4(9) + 2(12) + 4(15) + (18)] = 1 · [98 + 4(91) + 2(85) + 4(80) + 2(77) + 4(75) + 74] = 148 joules 41. The curve is  = () = 1(1 + −). Using disks,  = 010 [()]2  =  010 () = 1. Now use Simpson’s Rule to approximate 1: 1 ≈ 10 = 10 10−· 30[(0) + 4(1) + 2(2) + 4(3) + 2(4) + 4(5) + 2(6) + 4(7) + 2(8) + 4(9) + (10)] ≈ 880825 Thus,  ≈ 1 ≈ 277 or 28 cubic units. 42. Using Simpson’s Rule with  = 10, ∆ =  102,  = 1, 0 = 42 180  radians,  = 98 ms2, 2 = sin2 1 20, and () = 11 − 2 sin2 , we get  = 4 02 1 − 2 sin2  ≈ 4 10 = 4 918  10 ·23 (0) + 4 20   + 2 220   + · · · + 4 920   +  2  ≈ 207665 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.86 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 43. () =  2 sin2  2 , where  = sin ,  = 10,000,  = 10−4, and  = 6328 × 10−9. So () = (104)22sin2  , where  = (104)(10−4) sin 6328 × 10−9 . Now  = 10 and ∆ = 10−6 −10 (−10−6) = 2 × 10−7, so 10 = 2 × 10−7[(−00000009) + (−00000007) + · · · + (00000009)] ≈ 594. 44. () = cos(), ∆ = 2010 − 0 = 2 ⇒ 10 = 2 2{(0) + 2[(2) + (4) + · · · + (18)] + (20)} = 1[cos 0 + 2(cos 2 + cos 4 + · · · + cos 18) + cos 20] = 1 + 2(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + 1 = 20 The actual value is 020 cos() = 1 sin20 0 = 1 (sin 20 − sin 0) = 0. The discrepancy is due to the fact that the function is sampled only at points of the form 2, where its value is (2) = cos(2) = 1. 45. Consider the function  whose graph is shown. The area 02 () is close to 2. The Trapezoidal Rule gives 2 = 2 − 0 2 · 2 [(0) + 2(1) + (2)] = 1 2 [1 + 2 · 1 + 1] = 2. The Midpoint Rule gives 2 = 2 −2 0 [(05) + (15)] = 1[0 + 0] = 0, so the Trapezoidal Rule is more accurate. 46. Consider the function () = | − 1|, 0 ≤  ≤ 2. The area 02 () is exactly 1. So is the right endpoint approximation: 2 = (1)∆ + (2)∆ = 0 · 1 + 1 · 1 = 1. But Simpson’s Rule approximates  with the parabola  = ( − 1)2, shown dashed, and 2 = ∆ 3 [(0) + 4(1) + (2)] = 1 3 [1 + 4 · 0 + 1] = 2 3 . 47. Since the Trapezoidal and Midpoint approximations on the interval [ ] are the sums of the Trapezoidal and Midpoint approximations on the subintervals [−1 ],  = 12     , we can focus our attention on one such interval. The condition  00()  0 for  ≤  ≤  means that the graph of  is concave down as in Figure 5. In that figure,  is the area of the trapezoid ,  () is the area of the region , and  is the area of the trapezoid , so    ()  . In general, the condition  00  0 implies that the graph of  on [ ] lies above the chord joining the points ( ()) and ( ()). Thus,  ()  . Since  is the area under a tangent to the graph, and since  00  0 implies that the tangent lies above the graph, we also have    (). Thus,    ()  . 48. Let  be a polynomial of degree ≤ 3; say () = 3 + 2 +  + . It will suffice to show that Simpson’s estimate is exact when there are two subintervals ( = 2), because for a larger even number of subintervals the sum of exact estimates is exact. As in the derivation of Simpson’s Rule, we can assume that 0 = −, 1 = 0, and 2 = . Then Simpson’s approximation is °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 87 − () ≈ 1 3 [(−) + 4(0) + ()] = 1 3 −3 + 2 −  +  + 4 + 3 + 2 +  +  = 1 3 [22 + 6] = 2 3 3 + 2 The exact value of the integral is −(3 + 2 +  + ) = 20(2 + ) [by Theorem 5.5.7(a) and (b)] = 2 1 3 3 +  0 = 2 3 3 + 2 Thus, Simpson’s Rule is exact. 49.  = 1 2 ∆[(0) + 2(1) + · · · + 2(−1) + ()] and  = ∆[(1) + (2) + · · · + (−1) + ()], where  = 1 2(−1 + ). Now 2 = 1 2  1 2∆[(0) + 2(1) + 2(1) + 2(2) + 2(2) + · · · + 2(−1) + 2(−1) + 2() + ()] so 1 2 ( + ) = 1 2  + 1 2  = 1 4 ∆[(0) + 2(1) + · · · + 2(−1) + ()] + 1 4∆[2(1) + 2(2) + · · · + 2(−1) + 2()] = 2 50.  = ∆ 2 (0) + 2 =1 −1 () + () and  = ∆ =1   − ∆2 , so 1 3  + 2 3  = 1 3( + 2) = ∆ 3 · 2(0) + 2 =1 −1 () + () + 4 =1   − ∆2  where ∆ =  −   . Let  =  −  2 . Then ∆ = 2, so 1 3  + 2 3  =  3 (0) + 2 =1 −1 () + () + 4 =1  ( − ) = 1 3 [(0) + 4(1 − ) + 2(1) + 4(2 − ) + 2(2) + · · · + 2(−1) + 4( − ) + ()] Since 0 1 −  1 2 −  2     −1  −   are the subinterval endpoints for 2, and since  =  −  2 is the width of the subintervals for 2, the last expression for 1 3  + 2 3  is the usual expression for 2. Therefore, 1 3  + 2 3  = 2. 7.8 Improper Integrals 1. (a) Since  =   − 1 has an infinite discontinuity at  = 1, 12  − 1  is a Type 2 improper integral. (b) Since 0∞ 1 +13  has an infinite interval of integration, it is an improper integral of Type 1. (c) Since −∞ ∞ 2−2  has an infinite interval of integration, it is an improper integral of Type 1. (d) Since  = cot has an infinite discontinuity at  = 0, 04 cot  is a Type 2 improper integral. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.88 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 2. (a) Since  = tan is defined and continuous on 0 4 , 04 tan  is proper. (b) Since  = tan has an infinite discontinuity at  = 2 , 0 tan  is a Type 2 improper integral. (c) Since  = 1 2 −  − 2 = 1 ( − 2)( + 1) has an infinite discontinuity at  = −1, −11 2 −  − 2 is a Type 2 improper integral. (d) Since 0∞ −3  has an infinite interval of integration, it is an improper integral of Type 1. 3. The area under the graph of  = 13 = −3 between  = 1 and  =  is () = 1 −3  = − 1 2 −2 1 = − 1 2−2 − − 1 2  = 1 2 − 122. So the area for 1 ≤  ≤ 10 is (10) = 05 − 0005 = 0495, the area for 1 ≤  ≤ 100 is (100) = 05 − 000005 = 049995, and the area for 1 ≤  ≤ 1000 is (1000) = 05 − 00000005 = 04999995. The total area under the curve for  ≥ 1 is lim →∞ () = lim →∞  1 2 − 1(22) = 1 2. 4. (a) (b) The area under the graph of  from  = 1 to  =  is () = 1 () = 1 −11  = − 011 −01 1 = −10(−01 − 1) = 10(1 − −01) and the area under the graph of  is () = 1 () = 1 −09  =  011 01 1 = 10(01 − 1).  () () 10 206 259 100 369 585 104 602 1512 106 749 2981 1010 9 90 1020 99 990 (c) The total area under the graph of  is lim →∞ () = lim →∞ 10(1 − −01) = 10. The total area under the graph of  does not exist, since lim →∞ () = lim →∞ 10(01 − 1) = ∞. 5. 3∞ ( −12)32  = lim →∞ 3( − 2)−32  = lim →∞ −2( − 2)−12  3 [ =  − 2,  = ] = lim →∞ √− −2 2 + √21 = 0 + 2 = 2. Convergent 6. 0∞ √4 1 + 1   = lim →∞ 0(1 + )−14  = lim →∞  4 3(1 + )34  0 [ = 1 + ,  = ] = lim →∞  4 3(1 + )34 − 4 3  = ∞. Divergent °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 89 7. −∞ 0 3 −14  = lim →−∞  0 3 −14  = lim →−∞ − 1 4 ln|3 − 4| 0  = lim →−∞ − 1 4 ln 3 + 1 4 ln|3 − 4|  = ∞ Divergent 8. 1∞ (2 + 1) 1 3  = lim →∞ 1 (2 + 1) 1 3  = lim →∞ −4(21+ 1)2  1 = lim →∞ −4(21+ 1)2 + 36 1  = 0 + 36 1 . Convergent 9. 2∞ −5  = lim →∞ 2 −5  = lim →∞ − 1 5−5 2 = lim →∞ − 1 5−5 + 1 5−10 = 0 + 1 5−10 = 1 5−10. Convergent 10. −∞ 0 2  = lim →−∞  0 2  = lim →−∞ ln 2 2 0  = lim →−∞ ln 2 1 − ln 2 2  = ln 2 1 − 0 = ln 2 1 . Convergent 11. 0∞ √1 + 2 3  = lim →∞ 0 √1 + 2 3  = lim →∞ 2 31 + 3 0 = lim →∞  2 31 + 3 − 2 3 = ∞. Divergent 12.  = −∞ ∞ (3 − 32) = 1 + 2 = −∞ 0 (3 − 32) + 0∞(3 − 32), but 1 = lim →−∞  1 44 − 30  = lim →−∞(3 − 1 44) = −∞. Since 1 is divergent,  is divergent, and there is no need to evaluate 2. Divergent 13. −∞ ∞ −2  = −∞ 0 −2  + 0∞ −2 . −∞ 0 −2  = lim →−∞ − 1 2−20  = lim →−∞− 1 21 − −2 = − 1 2 · 1 = − 1 2, and 0∞ −2  = lim →∞ − 1 2−2 0 = lim →∞− 1 2−2 − 1 = − 1 2 · (−1) = 1 2. Therefore, −∞ ∞ −2  = − 1 2 + 1 2 = 0. Convergent 14. 1∞ −12  = lim →∞ 1 −12  = lim →∞ −1 1 = lim →∞(−1 − −1) = 1 − 1 . Convergent 15. 0∞ sin2   = lim →∞ 0 1 2(1 − cos 2) = lim →∞  1 2 − 1 2 sin 2 0 = lim →∞  1 2 − 1 2 sin 2 − 0 = ∞. Divergent 16. 0∞ sin cos   = lim →∞ 0 sin cos   = lim →∞ −cos  0 = lim →∞(−cos  + ) This limit does not exist since cos  oscillates in value between −1 and 1, so cos  oscillates in value between −1 and 1. Divergent 17. 1∞ 2 1+   = lim →∞ 1 (1+ 1)  = lim →∞ 1 1 −  + 1 1   [partial fractions] = lim →∞ ln|| − ln| + 1|  1 = lim →∞ ln    + 1       1 = lim →∞ ln  + 1  − ln 12 = 0 − ln 1 2 = ln 2. Convergent °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.90 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 18. 2∞ 2 + 2  − 3 = lim →∞2 ( + 3)(  − 1) = lim →∞2 −+ 3 1 4 +  −1 4 1  = lim →∞− 1 4 ln| + 3| + 1 4 ln| − 1|  2 = 1 4 lim →∞ln  −+ 3 1 2 = 1 4 lim →∞ln  −+ 3 1 − ln 15 = 1 4(0 + ln 5) = 1 4 ln 5. Convergent 19. −∞ 0 2  = lim →−∞ 0 2  = lim →−∞  1 2 2 − 1 4 20  integration by parts with  = ,  = 2   = lim →−∞ 0 − 1 4 −  1 2 2 − 1 4 2 = − 1 4 − 0 + 0 [by l’Hospital’s Rule] = − 1 4 . Convergent 20. 2∞ −3  = lim →∞ 2 −3  = lim →∞ − 1 3 −3 − 1 9 −3 2 integration by parts with = ,  = −3   = lim →∞ − 1 3 −3 − 1 9 −3 − − 2 3 −6 − 1 9 −6 = 0 − 0 + 7 9 −6 [by l’Hospital’s Rule] = 7 9 −6. Convergent 21. 1∞ ln  = lim →∞(ln2)2  1 by substitution with  = ln ,  =  = lim →∞ (ln2)2 = ∞. Divergent 22. 1∞ ln 2  = lim →∞1 ln2  = lim →∞−ln − 1 1  = ln integration by parts with ,  = (12)   = lim →∞−ln  − 1 + 1 = lim H →∞−11 − lim →∞ 1 + lim →∞1 = 0 − 0 + 1 = 1. Convergent 23. −∞ 0 4 + 4  = lim →−∞ 0 4 + 4  = lim →−∞ 1 21 2 tan−122 0    == 2 2  ,  = lim →−∞0 − 14 tan−122  = −142  = −8 . Convergent 24. ∞ (ln1)2  = lim →∞ (ln1)2  = lim →∞−ln1    = ln = (1, )  = lim →∞−ln1 + 1 = 0 + 1 = 1. Convergent 25. 0∞ −√  = lim →∞0 −√  = lim →∞0√ − (2 )   == 1 √(2 , √ )  = lim →∞−2−√ 0  + 0√ 2−    = 2 = 2 , ,   == −−−   = lim →∞−2√ −√ + −2−√ 0   = lim →∞−2√√  − √2  + 2 = 0 − 0 + 2 = 2. Convergent Note: lim →∞ √ √ H = lim →∞ 2√ 2√ √ = lim →∞ √1  = 0 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 91 26. 1∞ √ +√  = lim →∞1 √(1 +  ) = lim →∞1√ (1 +1 2) (2 )   == 1 √(2 , √ )  = lim →∞1√ 1 +22  = lim →∞2tan−1 √ 1  = lim →∞2(tan−1 √ − tan−1 1) = 2( 2 − 4 ) = 2 . Convergent 27. 01 1  = lim →0+  1 1  = lim →0+ ln|| 1  = lim →0+(−ln) = ∞. Divergent 28. 05 √3 51−   = lim →5− 0(5 − )−13  = lim →5− − 3 2(5 − )23 0 = lim →5− − 3 2[(5 − )23 − 523] = 3 2 523 Convergent 29. −14 2 √4 + 2 = lim →−2+  14( + 2)−14  = lim →−2+ 4 3( + 2)3414  = 4 3 →− lim2+ 1634 − ( + 2)34 = 4 3 (8 − 0) = 32 3 . Convergent 30. −21 ( + 1)  2  = lim →−1+  2 ( + 1)  2  = lim →−1+  2  + 1 1 − ( + 1) 1 2   [partial fractions] = lim →−1+ ln| + 1| +  + 1 1 2  = lim →−1+ ln 3 + 1 3 − ln( + 1) +  + 1 1  = −∞. Divergent Note: To justify the last step, lim →−1+ ln( + 1) +  + 1 1  = lim →0+ ln + 1 substitute for  + 1  = lim →0+ ln + 1 = ∞ since lim →0+ (ln) = lim →0+ ln 1 H = lim →0+ 1 −12 = lim →0+(−) = 0. 31. −32  4 = −02  4 + 03  4 , but −02  4 = lim →0− −−3 3  −2 = lim →0− −313 − 24 1  = ∞. Divergent 32. 01 √1 − 2 = lim →1− 0 √1 − 2 = lim →1− sin−1  0 = lim →1− sin−1  = 2 . Convergent 33. There is an infinite discontinuity at  = 1. 09 √3 1− 1  = 01( − 1)−13  + 19( − 1)−13 . Here 01( − 1)−13  = lim →1− 0( − 1)−13  = lim →1−  3 2( − 1)23 0 = lim →1−  3 2( − 1)23 − 3 2 = − 3 2 and 19( − 1)−13  = lim →1+ 9( − 1)−13  = lim →1+  3 2( − 1)239  = lim →1+ 6 − 3 2( − 1)23 = 6. Thus, 09 √3 1− 1  = −32 + 6 = 92. Convergent 34. There is an infinite discontinuity at  = 2. 02  − 2  = lim →2− 0 1 +  −2 2  = lim →2−  + 2ln| − 2|  0 = lim →2− ( + 2ln| − 2| − 2ln2) = −∞, so 02  − 2  diverges, and hence, 05  − 2  diverges. Divergent °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.92 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 35. 02 tan2   = lim →(2)− 0 tan2   = lim →(2)− 0(sec2  − 1) = lim →(2)− tan −  0 = lim →(2)− (tan − ) = ∞ since tan → ∞ as  → 2 −. Divergent 36. 04 2 − − 2 = 04 ( − 2)(  + 1) = 02 ( − 2)(  + 1) + 24 ( − 2)(  + 1) Considering only 02 ( − 2)(  + 1) and using partial fractions, we have 02 ( − 2)(  + 1) = lim →2− 0   −1 3 2 −  + 1 1 3   = lim →2−  1 3 ln| − 2| − 1 3 ln| + 1|  0 = lim →2−  1 3 ln| − 2| − 1 3 ln| + 1| − 1 3 ln 2 + 0 = −∞ since ln| − 2| → −∞ as  → 2−. Thus, 02 2 − − 2 is divergent, and hence, 04 2 − − 2 is divergent as well. 37. 01  ln  = lim →0+ 1  ln  = lim →0+  1 2 2 ln − 1 4 21    = ln = (1 , ) ,   ==   1 2 2  = lim →0+ 0 − 1 4  −  1 2 2 ln − 1 4 2 = − 1 4 − 0 = − 1 4 since lim →0+ 2 ln = lim →0+ ln 12 H = lim →0+ 1 −23 = lim →0+(− 1 2 2) = 0. Convergent 38. 02 √cos sin  = lim →0+  2 √cos sin  = lim →0+ 2√sin  2   = sin = cos  ,  = lim →0+ (2 − 2√sin) = 2 − 0 = 2. Convergent 39. −01 1 3  = lim →0− −1 11 · 12  = lim →0− −11  (−)   = 1 = −  , 2  = lim →0− ( − 1)− 1 1 or Formula 96 use parts  = lim →0− −2−1 − 1 − 11 = − 2  − lim →−∞ ( − 1) [ = 1] = −2  − lim →−∞  − 1 − H= − 2  − lim →−∞ 1 −− = − 2  − 0 = −2  . Convergent 40. 01 1 3  = lim →0+  1 11 · 12  = lim →0+ 1 1  (−)   = 1 = −  , 2  = lim →0+ ( − 1)1 1 or Formula 96 use parts  = lim →0+ 1 − 11 − 0 = lim →∞ ( − 1) [ = 1] = ∞. Divergent °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 93 41. Area = 1∞ −  = lim →∞ 1 −  = lim →∞−− 1 = lim →∞ (−− + −1) = 0 + −1 = 1 42. Area = −∞ 0   = lim →−∞ 0   = lim →−∞0  = lim →−∞ (0 − ) = 1 − 0 = 1 43. Area = 1∞ 3 1+   = lim →∞1 (21+ 1)  = lim →∞1  1 − 2 + 1  [partial fractions] = lim →∞ln|| − 12 ln 2 + 1  1 = lim →∞ln √2+ 1 1 = lim →∞ln √2+ 1 − ln √12 = ln 1 − ln 2−12 = 1 2 ln 2 44. Area = 0∞ −  = lim →∞ 0 −  = lim →∞−− − − 0 [use parts wtih  =  and  = − ] = lim →∞ (−− − −) − (−1) = 0 [use l’Hospital’s Rule] − 0 + 1 = 1 45. Area = 02 sec2   = lim →(2)− 0 sec2   = lim →(2)− tan 0 = lim →(2)− (tan − 0) = ∞ Infinite area 46. Area = −02 √1+ 2  = lim →−2+  0 √1+ 2  = lim →−2+ 2√ + 20  = lim →−2+ 2√2 − 2√ + 2 = 2√2 − 0 = 2√2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.94 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 47. (a)  1 () 2 0447453 5 0577101 10 0621306 100 0668479 1000 0672957 10,000 0673407 () = sin2  2 . It appears that the integral is convergent. (b) −1 ≤ sin ≤ 1 ⇒ 0 ≤ sin2  ≤ 1 ⇒ 0 ≤ sin2  2 ≤ 12 . Since 1∞ 12  is convergent [Equation 2 with  = 2  1], 1∞ sin22   is convergent by the Comparison Theorem. (c) Since 1∞ () is finite and the area under () is less than the area under () on any interval [1 ], 1∞ () must be finite; that is, the integral is convergent. 48. (a)  2 () 5 3830327 10 6801200 100 23328769 1000 69023361 10,000 208124560 () = √1− 1. It appears that the integral is divergent. (b) For  ≥ 2, √  √ − 1 ⇒ √1  √1− 1. Since 2∞ √1  is divergent [Equation 2 with  = 1 2 ≤ 1], 2∞ √1− 1  is divergent by the Comparison Theorem. (c) Since 2∞ () is infinite and the area under () is greater than the area under () on any interval [2 ], 2∞ () must be infinite; that is, the integral is divergent. 49. For   0,  3 + 1   3 = 1 2 . 1∞ 12  is convergent by Equation 2 with  = 2  1, so 1∞ 3 + 1  is convergent by the Comparison Theorem. 01 3+ 1  is a constant, so 0∞ 3 + 1  = 01 3 + 1  + 1∞ 3+ 1  is also convergent. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 95 50. For  ≥ 1, 1 + sin √ 2  ≥ √1. 1∞ √1  is divergent by Equation 2 with  = 1 2 ≤ 1, so 1∞ 1 + sin √ 2   is divergent by the Comparison Theorem. 51. For   1, () = √4+ 1 −   √+ 1 4  2 = 1, so 2∞ () diverges by comparison with 2∞ 1 , which diverges by Equation 2 with  = 1 ≤ 1. Thus, 1∞ () = 12 () + 2∞ () also diverges. 52. For  ≥ 0, arctan   2  2, so arctan 2 +   2 2 +   2  = 2−. Now  = 0∞ 2−  = lim →∞0 2−  = lim →∞−2− 0 = lim →∞−2 + 2 = 2, so  is convergent, and by comparison, 0∞ arctan 2 +   is convergent. 53. For 0   ≤ 1, sec2   √  1 32 . Now  = 01 −32  = lim →0+  1 −32  = lim →0+  − 2−121  = lim →0+ −2 + √2 = ∞, so  is divergent, and by comparison, 01 sec  √2 is divergent. 54. For 0   ≤ 1, sin √2 ≤ √1. Now  = 0 √1  = lim →0+   −12  = lim →0+ 212  = lim →0+2 − 2√  = 2 − 0 = 2, so  is convergent, and by comparison, 0 sin √2  is convergent. 55. 0∞ √(1 +  ) = 01 √(1 +  ) + 1∞ √(1 +  ) = lim →0+  1 √(1 +  ) + lim →∞1 √(1 +  ). Now  √(1 +  ) =  (1 + 2 2)  =√= 2 ,   = 2,  = 2 1 +2 = 2 tan−1  +  = 2 tan−1√ + , so 0∞ √(1 +  ) = lim →0+ 2tan−1√ 1  + lim →∞2tan−1√  1 = lim →0+ 2 4  − 2tan−1√  + lim →∞ 2tan−1√ − 2 4  = 2 − 0 + 2 2  − 2 = . 56. 2∞  √2 − 4 = 23 √2 − 4 + 3∞  √2 − 4 = lim →2+  3 √2 − 4 + lim →∞3  √2 − 4. Now   √2 − 4 =  2sec 2sectan 2tan    0 ≤     = 2 sec 2 or , where ≤   32  = 1 2 +  = 1 2 sec−1 1 2 + , so 2∞  √2 − 4 = lim →2+  1 2 sec−1 1 23  + lim →∞ 1 2 sec−1 1 2 3 = 1 2 sec−1 3 2 − 0 + 1 2 2  − 1 2 sec−1 3 2 = 4 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.96 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 57. If  = 1, then 01   = lim →0+  1   = lim →0+ [ln]1  = ∞. Divergent If  6= 1, then 01   = lim →0+  1   [note that the integral is not improper if   0] = lim →0+ −−+ 1 +1 1  = lim →0+ 1 −1  1 − 1−1  If   1, then  − 1  0, so 1 −1 → ∞ as  → 0+, and the integral diverges. If   1, then  − 1  0, so 1 −1 → 0 as  → 0+ and 01   = 1 −1  →lim0+ 1 − 1− = 1 −1 . Thus, the integral converges if and only if   1, and in that case its value is 1 1 −  . 58. Let  = ln. Then  =  ⇒ ∞ (ln ) = 1∞   . By Example 4, this converges to  −1 1 if   1 and diverges otherwise. 59. First suppose  = −1. Then 01  ln  = 01 ln  = lim →0+  1 ln  = lim →0+  1 2(ln)21  = − 1 2 →lim0+(ln)2 = −∞, so the integral diverges. Now suppose  6= −1. Then integration by parts gives   ln  = + 1 +1 ln −  + 1   = + 1 +1 ln − (+ 1) +1 2 + . If   −1, then  + 1  0, so 01  ln  = lim →0+ + 1 +1 ln − (+ 1) +1 2 1  = ( −+ 1) 1 2 −  + 1 1  →lim0+ +1 ln −  + 1 1  = ∞. If   −1, then  + 1  0 and 01  ln  = −1 ( + 1)2 −  + 1 1  →lim0+ ln −−1(+1) ( + 1) =H ( −+ 1) 1 2 −  + 1 1  →lim0+ −( + 1) 1−(+2) = −1 ( + 1)2 + 1 ( + 1)2 →lim0+ +1 = ( −+ 1) 1 2 Thus, the integral converges to − 1 ( + 1)2 if   −1 and diverges otherwise. 60. (a)  = 0: 0∞ −  = lim →∞0 −  = lim →∞−− 0 = lim →∞−− + 1 = 0 + 1 = 1  = 1: 0∞ −  = lim →∞0 − . To evaluate  − , we’ll use integration by parts with  = ,  = −  ⇒  = ,  = −−. So  −  = −− −  −−  = −− − − +  = (− − 1)− +  and °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 97 lim →∞0 −  = lim →∞(− − 1)− 0 = lim →∞(− − 1)− + 1 = lim →∞ − − − − + 1 = 0 − 0 + 1 [use l’Hospital’s Rule] = 1  = 2: 0∞ −  = lim →∞0 2− . To evaluate  2− , we could use integration by parts again or Formula 97. Thus, lim →∞0 2−  = lim →∞−2− 0 + 2 lim →∞0 −  = 0 + 0 + 2(1) [use l’Hospital’s Rule and the result for  = 1] = 2  = 3: 0∞ − = lim →∞0 3−  = lim 97 →∞−3− 0 + 3 lim →∞0 2−  = 0 + 0 + 3(2) [use l’Hospital’s Rule and the result for  = 2] = 6 (b) For  = 1, 2, and 3, we have 0∞ −  = 1, 2, and 6. The values for the integral are equal to the factorials for , so we guess 0∞ −  = !. (c) Suppose that 0∞ −  = ! for some positive integer . Then 0∞ +1−  = lim →∞ 0 +1− . To evaluate  +1− , we use parts with  = +1,  = −  ⇒  = ( + 1) ,  = −−. So  +1− = −+1− −  −( + 1)−  = −+1− + ( + 1) −  and lim →∞ 0 +1−  = lim →∞ −+1− 0 + ( + 1) lim →∞0 −  = lim →∞ −+1− + 0 + ( + 1)! = 0 + 0 + ( + 1)! = ( + 1)!, so the formula holds for  + 1. By induction, the formula holds for all positive integers. (Since 0! = 1, the formula holds for  = 0, too.) 61. (a)  = −∞ ∞   = −∞ 0   + 0∞  , and 0∞   = lim →∞ 0   = lim →∞  1 22 0 = lim →∞ 1 22 − 0 = ∞, so  is divergent. (b) −    =  1 22 − = 1 22 − 1 22 = 0, so lim →∞−    = 0. Therefore, −∞ ∞   6= lim →∞−   . 62. Let  =  2 so that  = √4 32 0∞ 3−2 . Let  denote the integral and use parts to integrate . Let  = 2,  = −2  ⇒  = 2 ,  = − 1 2 −2:  = lim →∞−21 2−2 0 + 1 0∞ −2 0 = −21 lim →∞2−2 + 1 lim →∞−21 −2 H= − 1 2 · 0 − 212 (0 − 1) = 21 2 Thus,  = √4 32 · 21 2 = (2)12 = [(22)]12 = 2√√2 √ = 8  . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.98 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 63. Volume = 1∞ 12  =  lim →∞1  2 =  lim →∞−1 1 =  lim →∞1 − 1  =   ∞. 64. Work = ∞  2  = lim →∞  2  = lim →∞−1  =  lim →∞−1 + 1  =   , where  = mass of the earth = 598 × 1024 kg,  = mass of satellite = 103 kg,  = radius of the earth = 637 × 106 m, and  = gravitational constant = 667 × 10−11 N·m2kg. Therefore, Work = 667 × 10−11 · 598 × 1024 · 103 637 × 106 ≈ 626 × 1010 J. 65. Work = ∞   = lim →∞  2  = lim →∞1 − 1  =   . The initial kinetic energy provides the work, so 1 202 =   ⇒ 0 = 2  . 66. () =  √22− 2 () and () = 1 2( − )2 ⇒ () = lim →+   √(2−−)22  = lim →+   3 −√2 2 −2 +22  = lim →+    √23 −2 − 2   √22 −2 + 2   √  2 − 2  = lim →+1 − 22 + 23 =  For 1: Let  = √2 − 2 ⇒ 2 = 2 − 2, 2 = 2 + 2, 2  = 2 , so, omitting limits and constant of integration, 1 =  (2 +2)  =  (2 + 2) = 1 33 + 2 = 1 3(2 + 32) = 1 3 √2 − 2 (2 − 2 + 32) = 1 3√2 − 2 (2 + 22) For 2: Using Formula 44, 2 =  2 √2 − 2 + 2 2 ln  + √2 − 2  . For 3: Let  = 2 − 2 ⇒  = 2 . Then 3 = 1 2  √ = 1 2 · 2√ = 2 − 2. Thus,  = lim →+  1 3√2 − 2 (2 + 22) − 22 √2 − 2 + 22 ln  + √2 − 2    + 2√2 − 2   = lim →+  1 3√2 − 2(2 + 22) − 22 √2 − 2 + 22 ln  + √2 − 2    + 2√2 − 2  − lim →+  1 3√2 − 2 2 + 22 − 22 √2 − 2 + 22 ln  + √2 − 2    + 2√2 − 2  =  1 3√2 − 2 (2 + 22) − 2 ln  + √2 − 2    − −2 ln|| = 1 3 √2 − 2 (2 + 22) − 2 ln + √ 2 − 2  67. We would expect a small percentage of bulbs to burn out in the first few hundred hours, most of the bulbs to burn out after close to 700 hours, and a few overachievers to burn on and on. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 99 (a) (b) () =  0() is the rate at which the fraction () of burnt-out bulbs increases as  increases. This could be interpreted as a fractional burnout rate. (c) 0∞ () = lim →∞ () = 1, since all of the bulbs will eventually burn out. 68.  = 0∞   = lim →∞ 12 ( − 1) 0 [Formula 96, or parts] = lim →∞ 1 − 12  − −12 . Since   0 the first two terms approach 0 (you can verify that the first term does so with l’Hospital’s Rule), so the limit is equal to 12. Thus,  = − = −12 = −1 = −1(−0000121) ≈ 82645 years. 69.  = 0∞ (1 − −) −  =   lim →∞0 − − (−−)  =   lim →∞−1 − − −1−  (−−) 0 =   lim →∞− 1  + ( + )1(+) − −1 +  +1   =    1 −  +1   =    +(+−)  = (+ ) 70. 0∞ () = lim →∞0  0 −  =  0 lim →∞ −−   0 =  0 −  lim →∞ − − 1 = −0(0 − 1) = 0 0∞ () represents the total amount of urea removed from the blood if dialysis is continued indefinitely. The fact that 0∞ () = 0 means that, in the limit, as  → ∞, all the urea in the blood at time  = 0 is removed. The calculation says nothing about how rapidly that limit is approached. 71.  = ∞ 21+ 1  = lim →∞ 21+ 1  = lim →∞tan−1   = lim →∞tan−1  − tan−1  = 2 − tan−1 .   0001 ⇒  2 − tan−1   0001 ⇒ tan−1   2 − 0001 ⇒   tan 2 − 0001 ≈ 1000. 72. () = −2 and ∆ = 4 −8 0 = 1 2. 04 () ≈ 8 = 21· 3[(0) + 4(05) + 2(1) + · · · + 2(3) + 4(35) + (4)] ≈ 1 6(531717808) ≈ 08862 Now   4 ⇒ − ·   − · 4 ⇒ −2  −4 ⇒ 4∞ −2   4∞ −4 . 4∞ −4  = lim →∞ − 1 4 −4 4 = − 1 4 0 − −16 = 1(416) ≈ 00000000281  00000001, as desired. 73. (a) () = 0∞ ()−  = 0∞ −  = lim →∞−−  0 = lim →∞ −−  + 1 . This converges to 1 only if   0. Therefore () = 1  with domain { |   0}. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.100 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION (b) () = 0∞ ()−  = 0∞ −  = lim →∞0 (1−)  = lim →∞1 −1 (1−) 0 = lim →∞1(1−−) − 1 −1  This converges only if 1 −   0 ⇒   1, in which case () = 1  − 1 with domain { |   1}. (c) () = 0∞ ()−  = lim →∞ 0 − . Use integration by parts: let  = ,  = −  ⇒  = ,  = − −  . Then () = lim →∞− − − 12 − 0 = lim →∞ −  − 21 + 0 + 12  = 12 only if   0. Therefore, () = 1 2 and the domain of  is { |   0}. 74. 0 ≤ () ≤  ⇒ 0 ≤ ()− ≤ − for  ≥ 0. Now use the Comparison Theorem: 0∞ −  = lim →∞ 0 (−)  =  · lim →∞ −1 (−) 0 =  · lim →∞  −1  (−) − 1 This is convergent only when  −   0 ⇒   . Therefore, by the Comparison Theorem, () = 0∞ ()−  is also convergent for   . 75. () = 0∞  0()− . Integrate by parts with  = −,  =  0() ⇒  = −−,  = (): () = lim →∞ ()− 0 +  0∞ ()−  = lim →∞ ()− − (0) + () But 0 ≤ () ≤  ⇒ 0 ≤ ()− ≤ − and lim →∞ (−) = 0 for   . So by the Squeeze Theorem, lim →∞ ()− = 0 for    ⇒ () = 0 − (0) + () = () − (0) for   . 76. Assume without loss of generality that   . Then −∞  () + ∞ () = lim →−∞  () + lim →∞  () = lim →−∞  () + lim →∞ () +  () = lim →−∞  () +  () + lim →∞  () = lim →−∞ () +  () + ∞ () = lim →−∞  () + ∞ () = −∞  () + ∞ () 77. We use integration by parts: let  = ,  = −2  ⇒  = ,  = − 1 2−2. So 0∞ 2−2  = lim →∞−1 2−2 0 + 1 2 0∞ −2  = lim →∞−22  + 12 0∞ −2  = 12 0∞ −2  (The limit is 0 by l’Hospital’s Rule.) °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.SECTION 7.8 IMPROPER INTEGRALS ¤ 101 78. 0∞ −2  is the area under the curve  = −2 for 0 ≤   ∞ and 0   ≤ 1. Solving  = −2 for , we get  = −2 ⇒ ln = −2 ⇒ −ln = 2 ⇒  = ±√−ln. Since  is positive, choose  = √−ln, and the area is represented by 01 √−ln . Therefore, each integral represents the same area, so the integrals are equal. 79. For the first part of the integral, let  = 2 tan ⇒  = 2 sec2  .  √21+ 4  =  2sec 2sec2  =  sec  = ln|sec + tan|. From the figure, tan =  2 , and sec = √2 + 4 2 . So  = 0∞√21+ 4 −  + 2  = lim →∞ln   √22+ 4 + 2    −  ln| + 2| 0 = lim →∞ln √2 + 4 + 2  −  ln( + 2) − (ln 1 −  ln 2) = lim →∞ln√2( 2+ 4 + + 2)  + ln 2 = lnlim →∞  +(√+ 2) 2 + 4   + ln 2−1 Now  = lim →∞  + √2 + 4 ( + 2) H = lim →∞ 1 + √2 + 4  ( + 2)−1 = 2  lim →∞ ( + 2)−1 . If   1,  = ∞ and  diverges. If  = 1,  = 2 and  converges to ln 2 + ln 20 = ln 2. If   1,  = 0 and  diverges to −∞. 80.  = 0∞  2 + 1 − 3+ 1  = lim →∞ 1 2 ln(2 + 1) − 1 3  ln(3 + 1) 0 = lim →∞ln(2 + 1)12 − ln(3 + 1)3 = lim →∞ln (3 (2 + 1) + 1) 123  = lnlim →∞ (3√ + 1) 2 + 1 3  For  ≤ 0, the integral diverges. For   0, we have  = lim →∞ √2 + 1 (3 + 1)3 H = lim →∞  √2 + 1 (3 + 1)(3)−1 = 1  lim →∞ 1 (3 + 1)(3)−1 For 3  1 ⇔   3,  = ∞ and  diverges. For  = 3,  = 1 3 and  = ln 1 3. For   3,  = 0 and  diverges to −∞. 81. No,  = 0∞ () must be divergent. Since lim →∞ () = 1, there must exist an  such that if  ≥ , then () ≥ 1 2. Thus,  = 1 + 2 = 0 () + ∞ (), where 1 is an ordinary definite integral that has a finite value, and 2 is improper and diverges by comparison with the divergent integral ∞ 1 2 . 82. As in Exercise 55, we let  = 0∞ 1 +  = 1 + 2, where 1 = 01 1 +  and 2 = 1∞ 1 + . We will show that 1 converges for   −1 and 2 converges for    + 1, so that  converges when   −1 and    + 1. [continued] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.102 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 1 is improper only when   0. When 0 ≤  ≤ 1, we have 1 1 +  ≤ 1 ⇒ −(1 + 1 ) ≤ −1 . The integral 01 −1  converges for −  1 [or   −1] by Exercise 57, so by the Comparison Theorem, 01 −(1 + 1 )  converges for −1    0. 1 is not improper when  ≥ 0, so it has a finite real value in that case. Therefore, 1 has a finite real value (converges) when   −1. 2 is always improper. When  ≥ 1,  1 +  = 1 −(1 + ) = 1 − + −  1 − . By (2), 1∞ 1−  converges for  −   1 (or    + 1), so by the Comparison Theorem, 1∞ 1 +  converges for    + 1. Thus,  converges if   −1 and    + 1. 7 Review 1. False. Since the numerator has a higher degree than the denominator, 2 + 4 2 − 4 =  + 8 2 − 4 =  +   + 2 +   − 2 . 2. True. In fact,  = −1,  =  = 1. 3. False. It can be put in the form   +  2 +   − 4. 4. False. The form is   +  +  2 + 4 . 5. False. This is an improper integral, since the denominator vanishes at  = 1. 04 2 − 1  = 01 2 − 1  + 14 2 − 1  and 01 2 − 1  = lim →1− 0 2 − 1  = lim →1−  1 2 ln 2 − 1   0 = lim →1− 1 2 ln 2 − 1  = ∞ So the integral diverges. 6. True by Theorem 7.8.2 with  = √2  1. 7. False. See Exercise 61 in Section 7.8. 8. False. For example, with  = 1 the Trapezoidal Rule is much more accurate than the Midpoint Rule for the function in the diagram. 9. (a) True. See the end of Section 7.5. (b) False. Examples include the functions () = 2, () = sin(2), and () = sin  . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 103 10. True. If  is continuous on [0 ∞), then 01 () is finite. Since 1∞ () is finite, so is 0∞ () = 01 () + 1∞ (). 11. False. If () = 1, then  is continuous and decreasing on [1 ∞) with lim →∞ () = 0, but 1∞ () is divergent. 12. True. ∞ [() + ()] = lim →∞  [() + ()] = lim →∞ () +  () = lim →∞  () + lim →∞  () since both limits in the sum exist  = ∞ () + ∞ () Since the two integrals are finite, so is their sum. 13. False. Take () = 1 for all  and () = −1 for all . Then ∞ () = ∞ [divergent] and ∞ () = −∞ [divergent], but ∞ [() + ()]  = 0 [convergent]. 14. False. 0∞ () could converge or diverge. For example, if () = 1, then 0∞ () diverges if () = 1 and converges if () = 0. 1. 12 ( + 1)  2  = 12 2 + 2  + 1  = 12  + 2 + 1  = 122 + 2 + ln||2 1 = (2 + 4 + ln 2) −  1 2 + 2 + 0 = 7 2 + ln 2 2. 12 ( + 1)  2  = 23 −2 1    == + 1  = 23 1 − 12   = ln|| + 13 2 = ln 3 + 13 − ln 2 + 1 2 = ln 3 2 − 16 3.  sec sin  =  cos sin   =      = sin = cos  ,  =  +  = sin  +  4. 06 sin 2  = − 1 2 cos 2 0 6 − 06 − 1 2 cos 2    ==  , ,   == sin 2 − 1 2 cos 2    = (− 12  · 1 2) − (0) +  1 4 sin 2 0 6 = − 24  + 1 8√3 5.  22 + 3  + 1 =  (2 + 1)( 1  + 1)  =  2 2+ 1 −  + 1 1   [partial fractions] = ln|2 + 1| − ln| + 1| +  6. 12 5 ln  =  1 6 6 ln2 1 − 12 1 6 5    = ln = 1 , ,   == 1 65 6  = 64 6 ln 2 − 0 −  36 1 62 1 = 32 3 ln 2 −  64 36 − 36 1  = 32 3 ln 2 − 7 4 7. 02 sin3  cos2   = 02(1 − cos2 )cos2  sin  = 10(1 − 2)2 (−)  = cos = − sin ,   = 01(2 − 4) =  1 3 3 − 1 5 51 0 =  1 3 − 1 5  − 0 = 15 2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.104 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 8. Let  = √ − 1, so that 2 =  − 1, 2  =  , and  = 2 + 1. Then  √1− 1  =  1 22  + 1 = 2 21+ 1  = 2 tan−1  +  = 2 tan−1 √ − 1 + . 9. Let  = ln,  = . Then  sin(ln  )  =  sin  = −cos +  = −cos(ln) + . 10. Let  = arctan,  = (1 + 2). Then 01 √1 + arctan 2   = 04 √  = 2 332 0 4 = 2 343322 − 0 = 2 3 · 1832 = 12 1 32. 11. Let  = sec. Then 12 √2− 1  = 03 tan sec sec tan  = 03 tan2   = 03(sec2  − 1) = tan −  0 3 = √3 − 3 . 12.  1 +2 4  =  1 +12  1 2    == 2 22   = 1 2 tan−1  +  = 1 2 tan−1 2 +  13. Let  = √3 . Then 3 =  and 32  = , so   √3   =   · 32  = 3. To evaluate , let  = 2,  =   ⇒  = 2 ,  = , so  =  2  = 2 −  2 . Now let  = ,  =   ⇒  = ,  = . Thus,  = 2 − 2 −    = 2 − 2 + 2 + 1, and hence 3 = 3(2 − 2 + 2) +  = 3 √3 (23 − 213 + 2) + . 14.  2+ 2 + 2  =   − 2 +  + 2 6   = 1 22 − 2 + 6 ln| + 2| +  15.  − 1 2 + 2 =  − 1 ( + 2) =   +   + 2 ⇒  − 1 = ( + 2) + . Set  = −2 to get −3 = −2, so  = 3 2. Set  = 0 to get −1 = 2, so  = − 1 2. Thus,  2 + 2 − 1  =  −1 2 +  + 2 3 2   = − 1 2 ln|| + 3 2 ln| + 2| + . 16.  tan sec62   =  (tan2 tan + 1) 2 2 sec2   = tan = − sec ,2   =  (2+ 1) 2 2  =  4 + 2 2 2 + 1  =  2 + 2 + 12   = 33 + 2 − 1 +  = 1 3 tan3  + 2 tan − cot +  17.  cosh  = sinh −  sinh    ==  , ,   = sinh = cosh   = sinh − cosh +  18. 2 + 8 − 3 3 + 32 = 2 + 8 − 3 2( + 3) =   +  2 +   + 3 ⇒ 2 + 8 − 3 = ( + 3) + ( + 3) + 2. Taking  = 0, we get −3 = 3, so  = −1. Taking  = −3, we get −18 = 9, so  = −2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 105 Taking  = 1, we get 6 = 4 + 4 +  = 4 − 4 − 2, so 4 = 12 and  = 3. Now  23+ 8 + 3 −2 3  =  3 − 12 −  + 3 2   = 3ln|| + 1 − 2ln| + 3| + . 19.  92 + 6 + 1  + 5  =  (92 + 6  + 1 + 1) + 4  =  (3 + 1)  + 12 + 4  = 3 = 3+ 1, =   1 3(−2 + 4 1) + 1 13  = 13 · 13  (−2 1) + 3 + 4  = 1 9  2+ 4  + 19  2 + 2 2 2  = 1 9 · 12 ln(2 + 4) + 29 · 12 tan−11 2 +  = 1 18 ln(92 + 6 + 5) + 1 9 tan−1 1 2(3 + 1) +  20.  tan5  sec3   =  tan4  sec2  sec tan  =  (sec2  − 1)2 sec2  sec tan   = sec = sec  ,tan   =  (2 − 1)22  =  (6 − 24 + 2) = 1 7 7 − 2 55 + 1 33 +  = 1 7 sec7  − 2 5 sec5  + 1 3 sec3  +  21.  √2− 4 =  (2 − 4  + 4) − 4 =  ( − 2)2 − 22 =  2sec2tan  tan  = 2 sec − 2 = 2 sec  tan  ,  =  sec  = ln|sec + tan| + 1 = ln   − 2 2 + √2 − 4 2  + 1 = ln  − 2 + √2 − 4   + , where  = 1 − ln 2 22.  cos√  =  2 cos   2 == √, ,2  =   = 2 sin −  2sin    ==  , ,   = sin = cos   = 2 sin + 2 cos +  = 2√ sin√ + 2 cos√ +  23. Let  = tan, so that  = sec2  . Then  √2 + 1 =  tan sec2   sec =  tan sec  =  csc  = ln|csc − cot| +  = ln  √2 + 1  − 1   +  = ln  √2 + 1 − 1   +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.106 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 24. Let  = cos,  =   ⇒  = −sin ,  = : (∗)  =   cos   =  cos +   sin . To integrate   sin , let  = sin,  =   ⇒  = cos ,  = . Then   sin  =  sin −   cos  =  sin − . By substitution in (∗),  =  cos +  sin −  ⇒ 2 = (cos + sin) ⇒  = 1 2(cos + sin) + . 25. 33 − 2 + 6 − 4 (2 + 1)(2 + 2) =  +  2 + 1 +  +  2 + 2 ⇒ 33 − 2 + 6 − 4 = ( + )2 + 2 + ( + )2 + 1. Equating the coefficients gives  +  = 3,  +  = −1, 2 +  = 6, and 2 +  = −4 ⇒  = 3,  = 0,  = −3, and  = 2. Now  3(32−+ 1)( 2 + 6 2 + 2) − 4  = 3 2−+ 1 1  + 2 2 + 2 = 3 2 ln2 + 1 − 3tan−1  + √2tan−1√2 + . 26.  sin cos  =  1 2sin 2   = = 1 2 1 2  ,  = sin 2 = − 1 2 cos 2  ,  = − 1 4 cos 2 +  1 4 cos 2  = − 1 4cos 2 + 1 8 sin 2 +  27. 02 cos3  sin 2  = 02 cos3 (2 sin cos)  = 02 2cos4  sin  = − 2 5 cos5  0 2 = 2 5 28. Let  = √3 . Then  = 3,  = 32 ⇒  √ √3 3   + 1 − 1  =   + 1 − 1 32 = 3 2 + 2 + 2 +  −2 1  = 3 + 32 + 6 + 6 ln| − 1| +  =  + 323 + 6√3  + 6 ln|√3  − 1| +  29. The integrand is an odd function, so −33 1 +||  = 0 [by 5.5.7(b)]. 30. Let  = −,  = −− . Then   √1 − −2 =  1−−( −)2 =  √1−−2 = −sin−1  +  = −sin−1(−) + . 31. Let  = √ − 1. Then 2 =  − 1 and 2  =  . Also,  + 8 = 2 + 9. Thus, 0ln 10 √+ 8  − 1  = 03 ·22+ 9   = 203 2+ 9 2  = 203 1 − 29+ 9  = 2 − 9 3 tan−133 0 = 2(3 − 3tan−1 1) − 0 = 23 − 3 · 4  = 6 − 32 32. 04 cos sin3   = 04 tan sec2    = =  ,  = tan = 1 2 tan  2sec  2  , = 2 tan2  0 4 − 12 04 tan2   = 8 · 12 − 0 − 1 2 04(sec2  − 1) =  8 − 1 2 tan −  0 4 = 8 − 121 − 4  = 4 − 12 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 107 33. Let  = 2sin ⇒ 4 − 232 = (2 cos)3,  = 2 cos , so  (4 −22)32  =  8cos 4sin23  2cos  =  tan2   =  sec2  − 1  = tan −  +  = √4− 2 − sin−12 +  34. Integrate by parts twice, first with  = (arcsin)2,  = :  =  (arcsin)2  = (arcsin)2 −  2arcsin √1 − 2  Now let  = arcsin,  = √1− 2  ⇒  = √1 1− 2 ,  = −√1 − 2. So  = (arcsin)2 − 2arcsin −√1 − 2  +   = (arcsin)2 + 2√1 − 2 arcsin − 2 +  35.  √ +132  =  (1 +  √) =  √ 1 + √   == 1 + 2 √√,  =  2√  =  2−12  = 4√ +  = 41 + √ +  36.  1 1 + tan − tan   =  cos  cos  − sin cos  cos  cos  + sin cos   =  cos cos  − + sin sin  = ln|cos + sin| +  37.  (cos + sin)2 cos 2  =  cos2  + 2 sincos + sin2 cos 2  =  (1 + sin 2)cos 2  =  cos 2  + 1 2  sin 4  = 1 2 sin 2 − 1 8 cos 4 +  Or:  (cos + sin)2 cos2  =  (cos + sin)2(cos2  − sin2 ) =  (cos + sin)3(cos  − sin) = 1 4(cos + sin)4 + 1 38.  2√√  =  2 (2)   == 1 √ (2√)   = 2 · 2 ln2 +  = 2√ln2 +1 +  39. We’ll integrate  =  (1 + 2 2)2  by parts with  = 2 and  = (1 + 2 )2 . Then  = ( · 22 + 2 · 1) and  = −1 2 · 1 1 + 2 , so  = −1 2 · 2 1 + 2 −  −12 · 21 + 2 (2 + 1)    = −4  + 2 2 + 12 · 1 22 +  = 21 4 − 4+ 2 +  Thus, 012 (1 + 2 2)2  = 21 4 − 4+ 21 02 = 14 − 18 − 114 − 0 = 18 − 1 4. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.108 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 40.   43 √sin2 tan  =   43 2sin  cos sincos    =   43 1 2(sin)−12(cos )−32  =   43 1 2cos sin −12 (cos)−2 =   43 1 2(tan)−12 sec2   = √tan  3 4 = √3 − √1 = √4 3 − 1 41. 1∞ (2 + 1) 1 3  = lim →∞ 1 (2 + 1) 1 3  = lim →∞ 1 1 2(2 + 1)−3 2 = lim →∞ −4(21+ 1)2  1 = − 1 4 lim →∞ (2 + 1) 1 2 − 19 = −1 40 − 19 = 36 1 42. 1∞ ln 4  = lim →∞ 1 ln4   = ln =  ,  ==  −1(3 4,3) = lim →∞ −ln 33  1 + 1 314  = lim →∞−ln 33 + 0 + 9−13  1 = lim H →∞−913 + 9−13 + 19 = 0 + 0 + 1 9 = 1 9 43.   ln  = ln =  ,  =    = ln|| +  = ln|ln| +  so 2∞  ln = lim →∞ 2  ln = lim →∞ ln|ln|  2 = lim →∞[ln(ln) − ln(ln 2)] = ∞, so the integral is divergent. 44. Let  = √ − 2. Then  = 2 + 2 and  = 2 , so  √   − 2 =  2 + 22  = 2 2 + 2  = 2 1 3 3 + 2 +  Thus, 26 √   − 2= lim →2+  6 √   − 2 = lim →2+  2 3( − 2)32 + 4 − 26  = lim →2+  16 3 + 8 − 2 3( − 2)32 − 4√ − 2 = 40 3 . 45. 04 ln √  = lim →0+  4 ln √  = lim  →0+ 2√ ln − 4√ 4  = lim →0+ (2 · 2ln4 − 4 · 2) − 2√ ln − 4√   = (4 ln 4 − 8) − (0 − 0) = 4 ln 4 − 8 () Let  = ln,  = √1  ⇒  = 1 ,  = 2√. Then  ln √  = 2√ ln − 2 √ = 2√ ln − 4√ +  () lim →0+ 2√ ln = lim →0+ 2ln −12 H = lim →0+ 2 − 1 2 −32 = lim →0+ −4√  = 0 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 109 46. Note that () = 1(2 − 3) has an infinite discontinuity at  = 2 3. Now 023 2 −13  = lim →(23)− 0 2 −13  = lim →(23)− − 1 3 ln|2 − 3|  0 = − 1 3 →lim (23)− ln|2 − 3| − ln 2 = ∞ Since 023 2 −13  diverges, so does 01 2 −13 . 47. 01 √−1  = lim →0+  1√ − √1 = lim →0+  1(12 − −12) = lim →0+ 2 3 32 − 2121  = lim →0+ 2 3 − 2 −  2 3 32 − 212 = − 4 3 − 0 = − 4 3 48.  = −11 2  − 2 = −11 ( − 2) = −01 ( − 2) + 01 ( − 2) = 1 + 2. Now 1 ( − 2) =   +   − 2 ⇒ 1 = ( − 2) + . Set  = 2 to get 1 = 2, so  = 1 2. Set  = 0 to get 1 = −2,  = − 1 2. Thus, 2 = lim →0+  1−1 2 +  −1 2 2  = lim →0+ − 1 2 ln|| + 1 2 ln| − 2|1  = lim →0+ (0 + 0) − − 1 2 ln + 1 2 ln| − 2| = − 1 2 ln 2 + 1 2 lim →0+ ln = −∞. Since 2 diverges,  is divergent. 49. Let  = 2 + 1. Then −∞ ∞ 42 + 4  + 5 = −∞ ∞ 122  + 4 = 1 2 −∞ 0 2 + 4 + 1 2 0∞ 2 + 4 = 1 2 lim →−∞  1 2 tan−1 1 2 0  + 1 2 lim →∞ 1 2 tan−1 1 2  0 = 1 40 − − 2  + 1 4 2 − 0 = 4 . 50. 1∞ tan−21   = lim →∞1 tan−21  . Integrate by parts:  tan−21   = −tan −1  +  1 1 +2 = −tan −1  +  1 − 2+ 1  = −tan−1   + ln|| − 1 2 ln(2 + 1) +  = −tan−1   + 1 2 ln 2 2 + 1 +  Thus, 1∞ tan−21   = lim →∞−tan−1  + 12 ln 2+ 1 2  1 = lim →∞−tan−1  + 12 ln 2+ 1 2 + 4 − 12 ln 12 = 0 + 1 2 ln 1 + 4 + 1 2 ln 2 = 4 + 1 2 ln 2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.110 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 51. We first make the substitution  =  + 1, so ln(2 + 2 + 2) = ln( + 1)2 + 1 = ln(2 + 1). Then we use parts with  = ln(2 + 1),  = :  ln(2 + 1) =  ln(2 + 1) −  (2 2 + 1 ) =  ln(2 + 1) − 2 22+ 1  =  ln(2 + 1) − 2 1 − 2 1+ 1  =  ln(2 + 1) − 2 + 2 arctan +  = ( + 1) ln(2 + 2 + 2) − 2 + 2 arctan( + 1) + , where  =  − 2 [Alternatively, we could have integrated by parts immediately with  = ln(2 + 2 + 2).] Notice from the graph that  = 0 where  has a horizontal tangent. Also,  is always increasing, and  ≥ 0. 52. Let  = 2 + 1. Then 2 =  − 1 and   = 1 2 , so  √23+ 1  =  (√−1)  1 2  = 12  (12 − −12) = 1 2  2 3 32 − 212 +  = 1 3(2 + 1)32 − (2 + 1)12 +  = 1 3 (2 + 1)12 (2 + 1) − 3 +  = 1 3√2 + 1 (2 − 2) +  53. From the graph, it seems as though 02 cos2  sin3   is equal to 0. To evaluate the integral, we write the integral as  = 02 cos2 (1 − cos2 ) sin  and let  = cos ⇒  = −sin . Thus,  = 11 2(1 − 2)(−) = 0. 54. (a) To evaluate  5−2  by hand, we would integrate by parts repeatedly, always taking  = −2 and starting with  = 5. Each time we would reduce the degree of the -factor by 1. (b) To evaluate the integral using tables, we would use Formula 97 (which is proved using integration by parts) until the exponent of  was reduced to 1, and then we would use Formula 96. (d) (c)  5−2  = − 1 8 −245 + 104 + 203 + 302 + 30 + 15 +  55.  √42 − 4 − 3 =  (2 − 1)2 − 4  = 2 = 2− 1, =  √2 − 22 1 2  39 = 1 2  2 √2 − 22 − 222 ln  + √2 − 22  +  = 1 4 √2 − 4 − ln  + √2 − 4  +  = 1 4 (2 − 1)√42 − 4 − 3 − ln 2 − 1 + √42 − 4 − 3  +  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 111 56.  csc5   =78 − 1 4 cot csc3  + 3 4  csc3   =72 − 1 4 cot csc3  + 3 4 − 1 2 csc cot + 1 2 ln|csc − cot| +  = − 1 4 cot csc3  − 3 8 csc cot + 3 8 ln|csc − cot| +  57. Let  = sin, so that  = cos . Then  cos 4 + sin2   =  √22 + 2  21 =  2 √22 + 2 + 22 2 ln + √22 + 2  +  = 1 2 sin 4 + sin2  + 2 lnsin + 4 + sin2  +  58. Let  = sin. Then  = cos , so  √1 + 2 sin cot   =  √1 + 2   =1 57 with = ln , =2    √√1 + 2 1 + 2  − + 1 1    +  = ln   √ √1 + 2 sin 1 + 2 sin  − + 1 1    +  59. (a)   −1 √2 − 2 − sin−1  +  = 12 √2 − 2 + √21− 2 − 1 −122 · 1 = 2 − 2−12  12 2 − 2 + 1 − 1 = √2−2 2 (b) Let  = sin ⇒  = cos  , 2 − 2 = 21 − sin2  = 2 cos2 .  √2−2 2  =  22 cos sin22   =  1 −sinsin 2 2   =  (csc2  − 1) = −cot −  +  = − √2 − 2  − sin−1   +  60. Work backward, and use integration by parts with  = −(−1) and  = ( + )−12  ⇒  = −( − 1)  and  = 2  √ + , to get  −1 √ +  =    =  −    = 2√+−1 + 2(− 1)  √+   = 2√ +  −1 + 2( − 1)   √+  +   = 2√ +  −1 + 2( − 1) −1 √ +  + 2(− 1)   √  +  Rearranging the equation gives 2( − 1)    √  +  = −2√+−1 − (2 − 3)  −1 √ +  ⇒   √  +  = (−√−1) + −1 − 2(2 ( −−3) 1)  −1 √ +  61. For  ≥ 0, 0∞   = lim →∞ +1( + 1) 0 = ∞. For   0, 0∞   = 01   + 1∞  . Both integrals are improper. By (7.8.2), the second integral diverges if −1 ≤   0. By Exercise 7.8.57, the first integral diverges if  ≤ −1. Thus, 0∞   is divergent for all values of . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.112 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 62.  = 0∞  cos  = lim →∞0  cos  99 with = lim =1 →∞2 + 1 (cos + sin) 0 = lim →∞2 + 1 (cos + sin) − 2 1+ 1 () = 2 1+ 1 lim →∞(cos + sin) − . For  ≥ 0, the limit does not exist due to oscillation. For   0, lim →∞ (cos + sin) = 0 by the Squeeze Theorem, because  (cos + sin)  ≤ (|| + 1), so  = 1 2 + 1(−) = −2 + 1. 63. () = 1 ln , ∆ =  −  = 410 − 2 = 1 5 (a) 10 = 51· 2{(2) + 2[(22) + (24) + · · · + (38)] + (4)} ≈ 1925444 (b) 10 = 1 5[(21) + (23) + (25) + · · · + (39)] ≈ 1920915 (c) 10 = 51· 3[(2) + 4(22) + 2(24) + · · · + 2(36) + 4(38) + (4)] ≈ 1922470 64. () = √cos, ∆ =  −   = 4 − 1 10 = 3 10 (a) 10 = 103· 2{(1) + 2[(13) + (16) + · · · + (37)] + (4)} ≈ −2835151 (b) 10 = 10 3 [(115) + (145) + (175) + · · · + (385)] ≈ −2856809 (c) 10 = 103· 3[(1) + 4(13) + 2(16) + · · · + 2(34) + 4(37) + (4)] ≈ −2849672 65. () = 1 ln ⇒  0() = −(ln1)2 ⇒  00() = 2 + ln 2(ln)3 = 2(ln 2 )3 + 2(ln 1 )2 . Note that each term of  00() decreases on [24], so we’ll take  =  00(2) ≈ 2022. | | ≤ ( − )3 122 ≈ 2022(4 − 2)3 12(10)2 = 001348 and || ≤ ( − )3 242 = 000674. | | ≤ 000001 ⇔ 212 022(8) 2 ≤ 10 15 ⇔ 2 ≥ 105(212 022)(8) ⇒  ≥ 3672. Take  = 368 for . || ≤ 000001 ⇔ 2 ≥ 105(2022)(8) 24 ⇒  ≥ 2596. Take  = 260 for . 66. 14   ≈ 6 = (4 −31)6 [(1) + 4(15) + 2(2) + 4(25) + 2(3) + 4(35) + (4)] ≈ 17739438 67. ∆ =  10 60 − 010 = 60 1 . Distance traveled = 010   ≈ 10 = 1 60 · 3[40 + 4(42) + 2(45) + 4(49) + 2(52) + 4(54) + 2(56) + 4(57) + 2(57) + 4(55) + 56] = 1 180(1544) = 857 mi 68. We use Simpson’s Rule with  = 6 and ∆ = 246− 0 = 4: Increase in bee population = 024 () ≈ 6 = 4 3 [(0) + 4(4) + 2(8) + 4(12) + 2(16) + 4(20) + (24)] = 4 3 [0 + 4(300) + 2(3000) + 4(11,000) + 2(4000) + 4(400) + 0] = 4 3 (60,800) ≈ 81,067 bees °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 113 69. (a) () = sin(sin). A CAS gives  (4)() = sin(sin)[cos4  + 7cos2  − 3] + cos(sin)6cos2 sin + sin From the graph, we see that   (4)()    38 for  ∈ [0 ]. (b) We use Simpson’s Rule with () = sin(sin) and ∆ = 10  : 0 () ≈ 10· 3 (0) + 4 10   + 2 210   + · · · + 4 910   + () ≈ 1786721 From part (a), we know that   (4)()    38 on [0 ], so we use Theorem 7.7.4 with  = 38, and estimate the error as || ≤ 38( − 0)5 180(10)4 ≈ 0000646. (c) If we want the error to be less than 000001, we must have || ≤ 385 1804 ≤ 000001, so 4 ≥ 385 180(000001) ≈ 646,0416 ⇒  ≥ 2835. Since  must be even for Simpson’s Rule, we must have  ≥ 30 to ensure the desired accuracy. 70. With an -axis in the normal position, at  = 7 we have  = 2 = 45 ⇒ (7) = 245  . Using Simpson’s Rule with  = 4 and ∆ = 7, we have  = 028 [()]2  ≈ 4 = 7 3 0 + 4 245  2 + 2 253  2 + 4 245  2 + 0 = 7 3  214,818   ≈ 4051 cm3. 71. (a) 2 + sin √  ≥ √1 for  in [1 ∞). 1∞ √1  is divergent by (7.8.2) with  = 1 2 ≤ 1. Therefore, 1∞ 2 + sin √   is divergent by the Comparison Theorem. (b) √1 + 1 4  √14 = 12 for  in [1 ∞). 1∞ 12  is convergent by (7.8.2) with  = 2  1. Therefore, 1∞ √1 + 1 4  is convergent by the Comparison Theorem. 72. The line  = 3 intersects the hyperbola 2 − 2 = 1 at two points on its upper branch, namely −2√23 and 2√23. The desired area is  = −22√√22 3 − 2 + 1  = 202√2 3 − 2 + 1  = 2 21 3 − 1 2  2 + 1 − 1 2 ln + 2 + 120√2 = 6 −  √2 + 1 − ln + √2 + 12 0√2 = 12√2 − 2√2 · 3 − ln2√2 + 3 = 6√2 − ln3 + 2√2 Another method:  = 213 2 − 1 and use Formula 39. 73. For  in 0 2 , 0 ≤ cos2  ≤ cos. For  in  2  , cos  ≤ 0 ≤ cos2 . Thus, area = 02(cos − cos2 ) +   2(cos2  − cos) = sin − 1 2  − 1 4 sin2 0 2 +  1 2  + 1 4 sin2 − sin 2 = 1 − 4  − 0 +  2 −  4 − 1 = 2 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.114 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 74. The curves  = 1 2 ± √ are defined for  ≥ 0. For   0, 2 −1√  2 +1√. Thus, the required area is 01 2 −1√ − 2 +1√  = 01 2 −1  − 2 +1 2   =   = 201 − − 2 −  + 2    = 201 −1 −  −2 2 − 1 +  + 2 2   = 22ln     + 2 − 2    − 21 0 = 4 ln 3 − 4. 75. Using the formula for disks, the volume is  = 02  [()]2  =  02(cos2 )2  =  02  1 2(1 + cos 2)2  = 4 02(1 + cos2 2 + 2 cos 2) = 4 02 1 + 1 2(1 + cos 4) + 2 cos 2  = 4  3 2 + 1 2 1 4 sin 4 + 2 1 2 sin 2 0 2 = 4  34 + 1 8 · 0 + 0 − 0 = 16 3 2 76. Using the formula for cylindrical shells, the volume is  = 02 2() = 2 02 cos2   = 2 02  1 2(1 + cos 2)  = 2 1 2  02( + cos 2) =  1 2 2 0 2 +  1 2 sin 2 0 2 − 02 1 2 sin 2  parts with  = cos 2   = ,  =  1 2 2 2 + 0 − 1 2− 1 2 cos 2 0 2 = 83 + 4 (−1 − 1) = 1 8(3 − 4) 77. By the Fundamental Theorem of Calculus, 0∞  0() = lim →∞ 0  0() = lim →∞ [() − (0)] = lim →∞ () − (0) = 0 − (0) = −(0). 78. (a) (tan−1 )ave = lim →∞ 1  − 0 0 tan−1   = lim 89 →∞1 tan−1  − 1 2 ln(1 + 2) 0 = lim →∞1  tan−1  − 1 2 ln(1 + 2) = lim →∞tan−1  − ln1 + 2 2 H=  2 − lim →∞ 2(1 + 2) 2 =  2 − 0 =  2 (b) () ≥ 0 and ∞ () is divergent ⇒ lim →∞  () = ∞. ave = lim →∞  ()  −   = lim H →∞ () 1 [by FTC1] = lim →∞ (), if this limit exists. (c) Suppose ∞ () converges; that is, lim →∞  () =   ∞. Then ave = lim →∞ −1   () = lim →∞  −1  · lim →∞ () = 0 ·  = 0. (d) (sin)ave = lim →∞ 1  0 sin  = lim →∞1 −cos 0 = lim →∞−cos   + 1  = lim →∞ 1 −cos = 0 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.CHAPTER 7 REVIEW ¤ 115 79. Let  = 1 ⇒  = 1 ⇒  = −(12). 0∞ 1 + ln2  = ∞0 1 + 1 ln (1 )2 − 2  = ∞0 −2ln + 1  (−) = ∞0 1 + ln2  = − 0∞ 1 + ln2  Therefore, 0∞ 1 + ln2  = − 0∞ 1 + ln2  = 0. 80. If the distance between  and the point charge is , then the potential  at  is  =  = ∞   = ∞ 402  = lim →∞ 4  0 −1   = 4  0 lim →∞−1 + 1  = −4  0. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.116 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. [Show More]

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