C H a P T E R 8 Integration Techniques, L'hôpital's Rule, And

Home , Trapezoidal rule

CHAPTER 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

Section 8.1 Basic Integration Rules ...... 95

Section 8.2 Integration by Parts ...... 106

Section 8.3 Trigonometric Integrals ...... 128

Section 8.4 Trigonometric Substitution ...... 141

Section 8.5 Partial Fractions ...... 161

Section 8.6 Integration by Tables and Other Integration Techniques . . 173

Section 8.7 Indeterminate Forms and L’Hôpital’s Rule ...... 184

Section 8.8 Improper Integrals ...... 199

Review Exercises ...... 212

Problem Solving ...... 223 CHAPTER 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

Section 8.1 Basic Integration Rules

d 1 d 1 2x x 1. (a) 2x2 1 C 2 x2 1 1 22x 2. (a) lnx2 1 C dx 2 dx 2 x2 1 x2 1 2x d 2x x2 122 2x2x2 12x (b) C x2 1 dx x2 12 x2 14

2 d 1 x 21 3x (b) x2 1 C x2 1 1 22x dx 2 x2 1 x2 13

d 1 1 1 d 1 (c) x2 1 C x2 1 1 22x (c) arctan x C dx 2 2 2 dx 1 x2 x d 2x (d) lnx2 1 C 2x2 1 dx x2 1 d 2x x (d) lnx2 1 C dx matches (a). dx x2 1 x2 1 x dx matches (b). x2 1

d 1 2x x 3. (a) lnx2 1 C dx 2 x2 1 x2 1 d 2x x2 122 2x2x2 12x 21 3x2 (b) C dx x2 12 x2 14 x2 13 d 1 (c) arctan x C dx 1 x2 d 2x (d) lnx2 1 C dx x2 1 1 dx matches (c). x2 1

d 4. (a) 2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1 dx d 1 1 (b) sinx2 1 C cosx2 12x x cosx2 1 dx 2 2 d 1 1 (c) sinx2 1 C cosx2 12x x cosx2 1 dx 2 2 d (d) 2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1 dx

x cosx2 1 dx matches (c).

95 96 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

2t 1 1 5. 3x 24 dx 6. dt 7. dx t2 t 2 x1 2x u 3x 2, du 3 dx, n 4 u t2 t 2, du 2t 1 dt 1 u 1 2x, du dx du x Use un du. Use . u du Use . u

2 3 2x 8. dt 9. dt 10. dx 2t 12 4 1 t2 x2 4 u 2t 1, du 2 dt, a 2 u t, du dt, a 1 1 u x2 4, du 2x dx, n 2 du du Use . Use . u2 a2 2 2 a u Use un du.

11. t sin t 2 dt 12. sec 3x tan 3x dx 13. cos xe sin x dx

u t 2, du 2t dt u 3x, du 3 dx u sin x, du cos x dx

Use sin u du. Use sec u tan u du. Use eu du.

1 14. dx 15. Let u x 4, du dx. xx2 4 x 46 u x, du dx, a 2 6x 45 dx 6x 45 dx 6 C 6 du Use . x 46 C uu2 a2

16. Let u t 9, du dt. 17. Let u z 4, du dz. 2 2 5 z 44 dt 2t 92 dt C dz 5z 45 dz 5 C t 92 t 9 z 45 4 5 C 4z 44

1 1 18. Let u t3 1, du 3t2 dt. 19. v dv v dv 3v 133 dv 3v 13 3 1 t 23 t3 1 dt t 3 1133t 2 dt 1 1 3 v2 C 2 63v 12 1 t 3 143 C 3 43 t 3 143 C 4

3 3 20. x dx x dx 2x 322 dx 21. Let u t3 9t 1, du 3t2 9 dt 2x 32 2 3t2 3 dt. x2 3 2x 31 C t 2 3 1 3t 2 3 2 2 1 dt dt t3 9t 1 3 t3 9t 1 x2 3 C 1 2 22x 3 lnt3 9t 1 C 3 Section 8.1 Basic Integration Rules 97

x2 1 22. Let u x2 2x 4, du 2x 1 dx. 23. dx x 1 dx dx x 1 x 1 x 1 1 dx x2 2x 4 1 22x 1 dx 1 x2 2x 4 2 x2 x lnx 1 C 2 x2 2x 4 C

2x 8 24. dx 2 dx dx 25. Let u 1 ex, du ex dx. x 4 x 4 ex 2x 8 lnx 4 C dx ln1 ex C 1 ex

1 1 1 1 1 1 26. dx 3 dx 3 dx 3x 1 3x 1 3 3x 1 3 3x 1 1 1 1 3x 1 ln3x 1 ln3x 1 C ln C 3 3 3 3x 1

4 4 x 27. 1 2x22 dx 4x 4 4x2 1 dx x5 x3 x C 12x 4 20x2 15 C 5 3 15

1 3 3 3 1 3 1 1 1 28. x1 x1 dx x 3 dx x2 3x 3 lnx C x x x2 x3 x x2 2 x

1 29. Let u 2x2, du 4x dx. 30. sec 4x dx sec4x4 dx 4 1 xcos 2 x2 dx cos 2 x24x dx 1 4 lnsec 4x tan 4x C 4 1 sin 2x2 C 4

31. Let u x, du dx. 32. Let u cos x, du sin x dx. 1 sin x cscx cotx dx cscx cotx dx dx cos x12sin x dx cos x 1 2 cos x C csc x C

33. Let u 5x, du 5 dx. 34. Let u cot x, du csc2 x dx. 1 1 e5x dx e5x5 dx e5x C csc2 xecot x dx ecot xcsc2 x dx ecot x C 5 5

5 1 ex 35. Let u 1 e x, du e x dx. 36. dx 5 dx 3ex 2 3ex 2 ex 2 1 e x dx 2 dx ex ex 1 ex 1 e x 5 dx 3 2ex e x 2 dx 5 1 1 e x 2ex dx 2 3 2ex 2 ln1 e x C 5 ln3 2ex C 2 98 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

ln x2 1 ln x2 sin x 37. dx 2ln x dx 2 C ln x2 C 38. Let u lncos x, du dx tan x dx. x x 2 cos x

tan xln cos x dx ln cos xtan x dx

lncos x2 C 2

1 sin x 1 sin x 1 sin x 39. dx dx Alternate Solution: cos x cos x 1 sin x 1 sin x 1 sin2 x dx sec x tan x dx dx cos x cos x1 sin x lnsec x tan x lnsec x C cos2 x dx cos x1 sin x lnsec xsec x tan x C cos x dx 1 sin x ln1 sin x C, u 1 sin x

1 cos 40. d csc d cot d sin lncsc cot lnsin C

1 1 cos 1 cos 1 2 2 1 sec x 1 41. 42. dx dx cos 1 cos 1 cos 1 cos2 1 3sec x 1 3 sec x 1 sec x 1 cos 1 2 sec x 1 csc cot csc2 dx sin2 3 tan2 x 1 2 sec x 2 d csc cot csc2 d dx cot2 x dx cos 1 3 tan2 x 3 csc cot C 2 cos x 2 dx csc2 x 1 dx 3 sin2 x 3 1 cos C sin sin 2 1 2 2 cot x x C 3 sin x 3 3 1 cos C sin 2 csc x cot x x C 3

43. Let u 2t 1, du 2 dt. 44. Let u 3x, du 3 dx. 1 1 2 1 1 3 dt dt dx dx 1 2t 12 2 1 2t 12 4 3x2 3 4 3x 2 1 1 3x arcsin2t 1 C arctan C 2 23 2

2 2 sin2t 1 1 45. Let u cos , du dt. 46. Let u , du dt. t t 2 t t2 tan2t 1 1 2 sin2t e1t 1 dt dt dt e1t dt e1t C t 2 2 cos2t t 2 t2 t2 1 2 ln cos C 2 t Section 8.1 Basic Integration Rules 99

3 1 x 3 47. dx 3 dx 3 arcsin C 6x x2 9 x 32 3

1 2 48. dx dx arcsec2x 1 C x 14x2 8x 3 2x 12x 12 1

4 1 1 x 12 1 2x 1 49. dx dx arctan C arctan C 4x2 4x 65 x 122 16 4 4 4 8

1 1 1 x 2 50. dx dx dx arcsin C, a 5 1 4x x2 5 x2 4x 4 5 x 22 5

ds t 1 51. , 0, dt 1 t4 2

(a) s (b) u t2, du 2t dt 0.8 1 t 1 2t dt dt − 1 t 4 2 1 t 22 1.2 1.2

t −1 1 1 2 arcsin t C − 2 0.8 1 1 1 1 −1 0, : arcsin 0 C ⇒ C 2 2 2 2 1 1 s arcsin t2 2 2

dy 52. tan22x, 0, 0 dx

y (a) (b) tan22x dx sec22x 1 dx 1.2 1 1 tan2x x C −1.2 1.2 2 x −0.6 0.6 0, 0 : 0 C −1.2 1 −1 y tan2x x 2

2 53. (a) y (b) y sec x tan x dx 9 8

sec2 x 2 sec x tan x tan2 x dx −9 9

x 8 sec2 x 2 sec x tan x sec2 x 1 dx −9

−8 2 sec2 x 2 sec x tan x 1 dx

2 tan x 2 sec x x C At 0, 1: 1 0 2 0 C ⇒ C 1 y 2 tan x 2 sec x x 1 100 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 54. (a) y (b) y dx 2 4x x2 2 1 1 dx 04 4 x2 4x 4 x 4 1 dx −2 −1 4 x 22 −2 x 2 arcsin C 2 1 1 1 At 2, : arcsin0 C ⇒ C 2 2 2 x 2 1 y arcsin 2 2

x 2 55. 10 56. (0, 1) 5 57. y 1 e dx

e2x 2ex 1 dx −2 2 −10 10

− 1 2 −2 e2x 2ex x C 2 y 3e0.2x y 5 4ex

1 et2 1 2et e2t dy sec2 x 58. r dt dt 59. et et dx 4 tan2 x Let u tan x, du sec2 x dx. et 2 et dt et 2t et C sec2 x 1 tan x y dx arctan C 4 tan2 x 2 2

60. Let u 2x, du 2 dx. 61. Let u 2x, du 2 dx. 1 2 4 1 4 y dx dx cos 2x dx cos 2x2 dx 2 2 x4x 1 2x2x 1 0 2 0 arcsec2x C 1 4 1 sin 2x 2 0 2

62. Let u sin t, du cos t dt. 63. Let u x2, du 2x dx.

1 1 1 1 2 1 2 1 2 sin2 t cos t dt sin3 t 0 xe x dx e x 2x dx e x 0 3 0 0 2 0 2 0 1 1 e1 0.316 2

1 64. Let u 1 ln x, du dx. 65. Let u x2 9, du 2x dx. x 4 2x 4 e 1 ln x e 1 dx x2 9122x dx 2 dx 1 ln x dx 0 x 9 0 1 x 1 x 4 1 e 1 2x2 9 4 1 ln x2 0 2 1 2 Section 8.1 Basic Integration Rules 101

2 x 2 2 2 66. dx 1 dx 67. Let u 3x, du 3 dx. x x 1 1 2 3 1 1 2 3 3 2 dx dx x 2 ln x 1 ln 4 0.386 4 9x2 3 4 3x2 1 0 0 1 3x 23 arctan 6 2 0 0.175 18

4 1 x 4 4 68. dx arcsin arcsin 0.927 2 0 25 x 5 0 5

52 2 69. A 2x 532 dx 70. A x8 2x2 dx 0 0 1 5 2 1 2 5 2x322 dx 8 2x2124x dx 2 0 4 0 1 52 1 2 5 2x52 8 2x232 5 0 6 0 1 1 0 552 532 0 832 5 6 55 11.1803 82 3.7712 3

5 3x 2 3 3 71. A dx 72. A dx 2 2 0 x 9 3 x 1 5 3x 5 2 3 3 dx dx 2 dx 2 2 2 0 x 9 0 x 9 0 x 1 3 2 x 5 3 lnx2 9 arctan 6 arctanx 2 3 3 0 0 3 2 5 3 6 arctan 3 ln34 arctan ln 9 2 3 3 2 7.4943 3 34 2 5 ln arctan 2 9 3 3 2.6806

2 73. y2 x21 x2 74. A sin 2x dx 0 y ±x21 x2 1 2 1 cos 2x A 4 x1 x2 dx 2 0 0 1 1 1 1 1 2 1 x2122x dx 2 0 4 1 1 x32 3 0 4 4 0 1 3 3 102 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 1 x 2 75. dx arctan C 1 x2 4x 13 3 3 C = 0 The antiderivatives are vertical translations of each other. −7 5

C = −0.2

−1

6 x 2 1 4 x 2 76. dx lnx2 4x 13 arctan C x2 4x 13 2 3 3 −10 10 The antiderivatives are vertical translations of each other.

−6

1 2 ex ex 3 1 77. d tan sec C or 78. dx e3x 9ex 9ex e3x C 1 sin 1 tan2 2 24 The antiderivatives are vertical translations of each other. The antiderivatives are vertical translations of each other.

6 5

C = 2 − 7 −55 2 2 C = 0

−6 −5

un1 79. Power Rule: un du C, n 1 80. sec u tan u du sec u C n 1 u x2 1, n 3

du du 1 u 81. Log Rule: lnu C, u x2 1 82. Arctan Rule: arctan C u a2 u2 a a

83. They are equivalent because 84. They differ by a constant. x C1 x C1 x C1 2 2 2 e e e Ce , C e . sec x C1 tan x 1 C1 tan x C

85. sin x cos x a sinx b sin x cos x a sin x cos b a cos x sin b sin x cos x a cos b sin x a sin b cos x Equate coefficients of like terms to obtain the following. 1 a cos b and 1 a sin b Thus,a 1cos b. Now, substitute for a in 1 a sin b. 1 1 sin b cos b 1 tan b ⇒ b 4 1 Since b , a 2. Thus, sin x cos x 2 sinx . 4 cos4 4 dx dx 1 1 cscx dx ln cscx cotx C sin x cos x 2 sinx 4 2 4 2 4 4 Section 8.1 Basic Integration Rules 103

1 a 1 a 1a 1 1 86. x ax2 dx x2 x3 87. f x x3 7x2 10x 2 5 0 2 3 0 6a 5 1 2 1 2 f x dx < 0 because more area is below the x- axis Let 2 , 12a 3, a . 6a 3 2 0 than above. y

5 11, 2 ()aa yx=

05 1 yax= 2

−5 x 12

2 4x 2 4 88. No. When u x2, it does not 89. dx 3 90. dx 4 x2 1 x2 1 follow that x u since x is 0 0 negative on 1, 0. Matches (a). Matches (d).

y y

3 2 2 1 1

x x 123 1234

91. (a) y 2x2, 0 ≤ x ≤ 2 (b) y 2x, 0 ≤ x ≤ 2 (c) y x, 0 ≤ x ≤ 2

y y y y = 2x 25 3 3 2 20 2 1 15 x − − − 10 3 2 1 1 2 3 y = x −1 x 5 −2 −2 2 −1 x −3 −3123−2 −1

1 92. (a) x y, 0 ≤ y ≤ 4 (b) x y, 0 ≤ y ≤ 4 (c) x , 0 ≤ y ≤ 4 2 1 y x2, 0 ≤ x ≤ 2 y x, 0 ≤ x ≤ 4 4 1 2 y dy y 0 2 y 4 y 8 4 6

4 2 2 y = x x = 1 x 2 π 234π ππ −2

−4 x x −2 2 −2 −1 1 2

−4 104 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

93. (a) Shell Method: y (b) Shell Method: Let u x2, du 2x dx. b 1 V 2 xex2 dx 1 0 x2 1 V 2 xe dx b 2 2 0 ex 1 0 2 x x 2 4 e 2x dx b 1 1 1 e 0 2 3 1 x2 e b2 3 4 0 e 3 1 e1 1.986 3 b ln 0.743 3 4

94. y f x lnsin x 95. y 2x cos x 1 fx y sin x x 2 cos2 x 2 sin2 x cos2 x 1 x 1 2 s 1 2 dx 2 dx 1 y 1 4 sin x 4 sin x x x 2 1 2 9 x 1 dx csc x dx S 2 2x dx y 4 sin x 4 0 x 2 9 12 lncsc x cot x 2 2 x 1 dx 9 4 0 6 ln1 ln 2 1 9 2 32 4 x 1 3 ln2 1 0.8814 3 0 x 8 36912 1010 1 3 256.545

4 5 x 4 4 96. A dx 5 arcsin 5 arcsin y 2 0 25 x 5 0 5 4 1 4 5 x x dx 3 2 A 0 25 x 2 (2.157,y ) 1 5 4 212 25 x 2x dx 1 5 arcsin 4 5 2 0 x 1 4 1234 525 x212 5 arcsin45 0 1 2 3 5 2.157 arcsin45 arcsin45

b b 1 1 97. Average value f x dx 98. Average value f x dx b a a b a a 1 3 1 1 n dx 2 sin nx dx 3 3 3 1 x n 0 0 1 3 n 1 n arctanx cosnx 6 3 n 0 1 1 arctan3 arctan3 cos cos0 6 1 2 arctan3 0.4163 3 Section 8.1 Basic Integration Rules 105

99. y tanx 100. y x23 y sec2x 2 y 3x13 1 y2 1 2 sec4x 4 1 4 1 y2 1 s 1 2 sec4x dx 9x23 0 8 4 1.0320 s 1 23 dx 7.6337 1 9x

101. (a) cos3 x dx 1 sin2 x cos x dx (c) cos7 x dx 1 sin2 x3 cos x dx

sin3 x sin x C 1 3 sin2 x 3 sin4 x sin6 x cos x dx 3 3 1 (b) cos5 x dx 1 sin2 x2 cos x dx sin x sin3 x sin5 x sin7 x C 5 7

1 2 sin2 x sin4 x cos x dx (d) cos15 x dx 1 cos2 x7 cos x dx

2 sin5 x You would expand 1 cos2 x7. sin x sin3 x C 3 5

102. (a) tan3 x dx sec2 x 1 tan x dx (c) tan2k1 x dx sec2 x 1 tan2k1 x dx

tan2kx sec2 x tan x dx tan x dx tan2k1 x dx 2k tan2 x (d) You would use these formulas recursively. tan x dx 2 tan2 x tan3 x dx lncos x C 2

(b) tan5 x dx sec2 x 1 tan3 x dx

tan4 x tan3 x dx 4

1 103. Let f x xx2 1 ln x x2 1 C. 2 1 1 1 1 fx x x2 1122x x2 1 1 x2 1122x 2 2 x x2 1 2 1 x2 1 x x2 1 1 2 x2 1 x x2 1 x2 1 1 x2 x2 1 1 x2 1 x 2 x2 1 x x2 1 x2 1 1 2x2 1 1 1 2x2 1 x2 1 2 x2 1 x2 1 2 x2 1 1 Thus, x2 1 dx xx2 1 ln x x2 1 C. 2

—CONTINUED— 106 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

103. —CONTINUED— 1 Let gx xx2 1 arcsinhx. 2 1 1 1 gx x x2 1122x x2 1 2 2 x2 1 1 x2 1 x2 1 2 x2 1 x2 1 1 x2 x2 1 1 2 x2 1 1 2x2 1 x2 1 2 x2 1 1 Thus, x2 1 dx xx2 1 arcsinhx C. 2

4 ln9 x 104. Let I dx. 2 ln9 x lnx 3 I is defined and continuous on 2, 4. Note the symmetry: as x goes from 2 to 4, 9 x goes from 7 to 5 and x 3 goes from 5 to 7. So, let y 6 x, dy dx. 2 ln3 y 4 ln3 y I dy dy 4 ln3 y ln9 y 2 ln3 y ln9 y Adding: 4 ln9 x 4 ln3 x 4 2I dx dx dx 2 ⇒ I 1 2 ln9 x lnx 3 2 ln3 x ln9 x 2 You can easily check this result numerically.

Section 8.2 Integration by Parts

d 1. sin x x cos x cos x x sin x cos x x sin x dx Matches (b)

d 2. x2 sin x 2x cos x 2 sin x x2 cos x 2x sin x 2x sin x 2 cos x 2 cos x x2 cos x dx Matches (d)

d d 1 3. x2ex 2xex 2ex x2ex 2xex 2xex 2ex 2ex 4. x x ln x 1 x ln x ln x dx dx x x2ex Matches (a) Matches (c)

5. xe2x dx 6. x2e2x dx 7. ln x2 dx

u x, dv e2x dx u x2, dv e2x dx u ln x2, dv dx Section 8.2 Integration by Parts 107

8. ln 3x dx 9. x sec2 x dx 10. x2 cos x dx

u ln 3x, dv dx u x, dv sec2 x dx u x2, dv cos x dx

1 11. dv e2x dx ⇒ v e2x dx e2x 12. dv ex dx ⇒ v ex dx ex 2 u x ⇒ du dx u x ⇒ du dx 1 1 x xe2x dx xe2x e2x dx 2 dx 2xex dx 2 2 ex 1 1 xe2x e2x C 2 xex ex dx 2 4 1 2xex ex C 2x 1 C 4e2x 2xex 2ex C

13. Use integration by parts three times.

(1) dv ex dx ⇒ v ex dx ex (2) dv ex dx ⇒ v ex dx ex (3) dv ex dx ⇒ v ex dx ex

u x3 ⇒ du 3x2 dx u x2 ⇒ du 2x dx u x ⇒ du dx

x3ex dx x3ex 3x2ex dx x3ex 3x2ex 6xex dx

x3ex 3x2ex 6xex 6ex C exx3 3x2 6x 6 C

1t e 1 3 1 3 1 3 14. dt e1t dt e1t C 15. x2ex dx ex 3x2 dx ex C t 2 t 2 3 3

x5 t2 16. dv x4 dx ⇒ v 17. dv t dt ⇒ v t dt 5 2 1 1 u ln x ⇒ du dx u lnt 1 ⇒ du dt x t 1 x5 x5 1 t2 1 t2 x4 ln x dx ln x dx t lnt 1 dt lnt 1 dt 5 5 x 2 2 t 1 x5 1 t 2 1 1 ln x x4 dx lnt 1 t 1 dt 5 5 2 2 t 1 x5 1 t2 1 t2 ln x x5 C lnt 1 t lnt 1 C 5 25 2 2 2 x5 1 5 ln x 1 C 2t 2 1 lnt 1 t 2 2t C 25 4

1 1 18. Let u ln x, du dx. 19. Let u ln x, du dx. x x 1 1 1 ln x2 1 ln x3 dx ln x3 dx C dx ln x2 dx C xln x3 x 2ln x2 x x 3 108 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 1 1 1 20. dv dx ⇒ v dx 21. dv dx ⇒ v 2x 12 dx x2 x2 x 2x 12 1 1 u ln x ⇒ du dx x 22x 1 ln x ln x 1 ln x 1 u xe2x ⇒ du 2xe2x e2x dx dx dx C x2 x x2 x x e2x2x 1 dx xe2x xe2x e2x dx dx 2x 12 22x 1 2 xe2x e2x e2x C C 22x 1 4 42x 1

x 1 22. dv dx ⇒ v x2 12 x dx x2 12 2x2 1 u x2ex2 ⇒ du 2x3ex2 2xex2 dx 2xex2x2 1 dx

3 x2 2 x2 2 x2 x2 x2 x e x e 2 x e e e dx xex dx C C x2 12 2x2 1 2x2 1 2 2x2 1

23. Use integration by parts twice.

(1) dv ex dx ⇒ v ex dx ex (2) dv ex dx ⇒ v ex dx ex

u x2 ⇒ du 2x dx u x ⇒ du dx

x2 1ex dx x2ex dx ex dx x2ex 2xex dx ex

x2ex 2xex ex dx ex x2ex 2xex ex C x 12ex C

1 1 1 2 24. dv dx ⇒ v dx 25. dv x 1 dx ⇒ v x 112 dx x 132 x2 x2 x 3 1 u x ⇒ du dx u ln 2x ⇒ du dx x 2 2 xx 1 dx xx 132 x 132 dx ln2x ln2x 1 ln2x 1 3 3 dx dx C x2 x x2 x x 2 4 xx 132 x 152 C ln2x 1 3 15 C x 2x 132 3x 2 C 15

1 2 26. dv dx ⇒ v 2 3x12 dx 2 3x 2 3x 3 u x ⇒ du dx x 2x2 3x 2 dx 2 3x dx 2 3x 3 3 2x2 3x 4 22 3x 22 3x 2 3x32 C 9x 22 3x C 3x 4 C 3 27 27 27 Section 8.2 Integration by Parts 109

27. dv cos x dx ⇒ v cos x dx sin x 28. dv sin x dx ⇒ v cos x u x ⇒ du dx u x ⇒ du dx x sin dx x cos x cos x dx x cos x dx x sin x sin x dx x sin x cos x C x cos x sin x C

29. Use integration by parts three times. (1) u x3, du 3x2 dx, dv sin x dx, v cos x (2) u x2, du 2x dx, dv cos x dx, v sin x

x3 sin dx x3 cos x 3x2 cos x dx x3 sin x dx x3 cos x 3x2 sin x 2x sin x dx

x3 cos x 3x2 sin x 6x sin x dx

(3) u x, du dx, dv sin x dx, v cos x

x3 sin x dx x3 cos x 3x2 sin x 6x cos x cos x dx

x3 cos x 3x2 sin x 6x cos x 6 sin x C

30. Use integration by parts twice. (1) u x2, du 2x dx, dv cos x dx, v sin x (2) u x, du dx, dv sin x dx, v cos x

x2 cos x dx x2 sin x 2x sin x dx x2 cos x dx x2 sin x 2x cos x cos x dx

x2 sin x 2x cos x 2 sin x C

31. u t, du dt, dv csc t cot dt, v csc t 32. dv sec tan d ⇒ v sec tan d sec

t csc t cot t dt t csc t csc t dt u ⇒ du d

t csc t lncsc t cot t C sec tan d sec sec d

sec lnsec tan C

33. dv dx ⇒ v dx x 34. dv dx ⇒ v dx x

1 1 u arctan x ⇒ du dx u arccos x ⇒ du dx 1 x2 1 x2

x x arctan x dx x arctan x 2 dx 4 arccos x dx 4x arccos x dx 1 x 1 x2

1 2 x arctan x ln1 x2 C 4x arccos x 1 x C 2

35. Use integration by parts twice. 1 1 (1) dv e2x dx ⇒ v e2x dx e2x (2) dv e2x dx ⇒ v e2x dx e2x 2 2 u sin x ⇒ du cos x dx u cos x ⇒ du sin x dx —CONTINUED— 110 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

35. —CONTINUED— 1 1 1 1 1 1 e2x sin x dx e2x sin x e2x cos x dx e2x sin x e2x cos x e2x sin x dx 2 2 2 2 2 2 5 1 1 e2x sin x dx e2x sin x e2x cos x 4 2 4 1 e2x sin x dx e2x2 sin x cos x C 5

36. Use integration by parts twice.

(1) dv e x dx ⇒ v e x dx e x (2) dv e x dx ⇒ v e x dx e x

u cos 2x ⇒ du 2 sin 2x dx u sin 2x ⇒ du 2 cos 2x dx

ex cos 2x dx ex cos 2x 2ex sin 2x dx ex cos 2x 2ex sin 2x 2ex cos 2x dx

5e x cos 2x dx ex cos 2x 2ex sin 2x

ex e x cos 2x dx cos 2x 2 sin 2x C 5

37. y xex2 38. dv dx ⇒ v x

2 1 2 1 y xex dx ex C u ln x ⇒ du dx 2 x y ln x 1 y ln x dx x ln x x dx x x ln x x C x1 ln x C

39. Use integration by parts twice. 1 2 (1) dv dt ⇒ v 2 3t12 dt 2 3t 2 3t 3 u t 2 ⇒ du 2t dt 2 (2) dv 2 3t dt ⇒ v 2 3t12 dt 2 3t32 9 u t ⇒ du dt t2 2t22 3t 4 y dt t2 3t dt 2 3t 3 3 2t 22 3t 4 2t 2 2 3t32 2 3t32 dt 3 3 9 9 2t 22 3t 8t 16 2 3t32 2 3t52 C 3 27 405 22 3t 27t2 24t 32 C 405 Section 8.2 Integration by Parts 111

40. Use integration by parts twice. 2 (1) dv x 1 dx ⇒ v x 112 dx x 132 3 u x2 ⇒ du 2x dx 2 (2) dv x 132 dx ⇒ v x 132 dx x 152 5 u x ⇒ du dx

y x2x 1 dx

2 4 2 4 2 2 x2x 132 xx 132 dx x2x 132 xx 152 x 152 dx 3 3 3 3 5 5 2 8 16 2x 132 x2x 132 xx 152 x 172 C 15x2 12x 8 C 3 15 105 105

41. cos yy 2x

cos y dy 2x dx

sin y x2 C

42. dv dx ⇒ v dx x

x 1 1 2 u arctan ⇒ du dx dx 2 1 x22 2 4 x2 x x 2x x y arctan dx x arctan dx x arctan ln4 x2 C 2 2 4 x2 2

43. (a) y

x −4 −2 2 4

dy (b) xy cos x, 0, 4 6 dx dy x cos x dx y − 66

y12 dy x cos x dx u x, du dx, dv cos x dx, v sin x −2

2 y12 x sin x sin x dx

x sin x cos x C 0, 4: 2412 0 1 C ⇒ C 3 2y x sin x cos x 3 112 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

44. (a) y 4

x −6 4

−4

dy 18 (b) ex3 sin 2x, 0, dx 37

y ex3 sin 2x dx

Use integration by parts twice. (1) u sin 2x, du 2 cos 2x dv ex3 dx, v 3ex3

ex3 sin 2x dx 3ex3 sin 2x 6ex3 cos 2x dx

(2) u cos 2x, du 2 sin 2x dv ex3 dx, v 3ex3

ex3 sin 2x dx 3ex3 sin 2x 63ex3 cos 2x 6ex3 sin 2x dx C

37ex3 sin 2x dx 3ex3 sin 2x 18ex3 cos 2x C

1 y ex3 sin 2x dx 3ex3 sin 2x 18ex3 cos 2x C 37 18 18 1 0, : 0 18 C ⇒ C 0 37 37 37 1 y 3ex3 sin 2x 18ex3 cos 2x 37

dy x dy x 45. ex 8, y0 2 10 46. sin x, y0 4 8 (0, 4) dx y dx y

− −10 10 5 10

−2 −2

47. u x, du dx, dv ex2 dx, v 2ex2

xex2 dx 2xex2 2ex2 dx 2xex2 4ex2 C

4 4 T hus, xex2 dx 2xex2 4ex2 0 0 8e2 4e2 4 12e2 4 2.376. Section 8.2 Integration by Parts 113

48. See Exercise 3. 49. See Exercise 27. 1 1 2 2 x2e x dx x2ex 2xex 2e x e 2 0.718 x cos x dx x sin x cos x 1 0 0 0 0 2

1 1 50. dv sin 2x dx ⇒ v sin 2x dx cos 2x 51. u arccos x, du dx, dv dx, v x 2 1 x2 u x ⇒ du dx x arccos x dx x arccos x dx 1 x2 1 1 x sin 2x dx x cos 2x cos 2x dx 2 2 x arccos x 1 x2 C 1 1 1 2 12 x cos 2x sin 2x C Thus, arccos x x arccos x 1 x2 2 4 0 0 1 1 1 3 sin 2x 2x cos 2x C arccos 1 4 2 2 4 1 3 Thus, x sin 2x dx sin 2x 2x cos 2x . 1 0.658. 0 4 0 2 6 2

x2 52. dv x dx ⇒ v x dx 2 2x u arcsin x2 ⇒ du dx 1 x 4 x2 x3 x arcsin x2 dx arcsin x2 dx 2 1 x 4 x2 1 arcsin x2 21 x 412 C 2 4 1 x2 arcsin x2 1 x 4 C 2

1 1 1 1 Thus, x arcsin x2 dx x2 arcsin x2 1 x 4 2. 0 2 0 4

53. Use integration by parts twice.

(1) dv e x dx ⇒ v e x dx e x (2) dv ex dx ⇒ v ex dx ex

u sin x ⇒ du cos x dx u cos x ⇒ du sin x dx

e x sin x dx e x sin x e x cos x dx e x sin x e x cos x e x sin x dx

2e x sin x dx e xsin x cos x

e x e x sin x dx sin x cos x C 2

1 e x 1 e 1 esin 1 cos 1 1 Thus, e x sin x dx sin x cos x sin 1 cos 1 0.909. 0 2 0 2 2 2 114 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

54. Use integration by parts twice. (1) dv ex, v ex, u cos x, du sin x dx

ex cos x dx ex cos x ex sin x dx

(2) dv ex dx, v ex, u sin x, du cos x dx

ex cos x dx ex cos x ex sin x ex cos x dx ⇒ 2ex cos x dx ex sin x ex cos x

2 ex sin x ex cos x 2 e2 1 Thus, ex cos x dx sin 2 cos 2 . 0 2 0 2 2

x3 1 55. dv x2 dx, v , u ln x, du dx 56. dv dx ⇒ v dx x 3 x x3 x3 1 2x x2 ln x dx ln x dx u ln1 x2 ⇒ du dx 3 3 x 1 x2 x3 1 2x2 ln x x2 dx ln1 x2 dx x ln1 x2 dx 3 3 1 x2 2 x3 1 2 1 Hence, x2 ln x dx ln x x3 x ln1 x2 2 1 dx 2 1 3 9 1 1 x 8 8 1 ln 2 x ln1 x2 2x 2 arctan x C 3 9 9 8 7 Thus, ln 2 1.071. 3 9 1 1 ln1 x2 dx x ln1 x2 2x 2 arctan x 0 0 ln 2 2 . 2

x2 1 57. dv x dx, v , u arcsec x, du dx 58. u x, du dx, dv sec2 x dx, v tan x 2 xx2 1 x2 x22 x sec2 x dx x tan x tan x dx x arcsec x dx arcsec x dx 2 xx2 1 Hence, x2 1 2x arcsec x dx 4 2 4 2 4 x 1 x sec2 x dx x tan x lncos x x2 1 0 0 arcsec x x2 1 C 2 2 2 ln 0 4 2 Hence, 1 4 x2 1 4 ln 2. x arcsec x dx arcsec x x2 1 4 2 2 2 2 2 15 2 3 8 arcsec 4 2 3 2 15 3 2 8 arcsec 4 2 2 3 7.380. Section 8.2 Integration by Parts 115

1 1 1 59. x2e2x dx x2 e2x 2x e2x 2 e2x C Alternate u and itsv and its 2 4 8 signs derivatives antiderivatives 1 1 1 x2 e2x x2e2x xe2x e2x C 2 2 4 1 2x 2x 2 e 1 e2x2x2 2x 1 C 2 1 e2x 4 4 1 2x 0 8 e

1 1 1 1 60. x3e2x dx x3 e2x 3x2 e2x 6x e2x 6 e2x C Alternate u and itsv and its 2 4 8 16 signs derivatives antiderivatives

1 e2x4x3 6x2 6x 3 C x3 e 2x 8 2 1 2x 3x 2e 1 2x 6x 4e 1 2x 6 8e 1 2x 0 16e

61. x3 sin x dx x3cos x 3x2sin x 6x cos x 6 sin x C Alternate u and itsv and its signs derivatives antiderivatives x3 cos x 3x2 sin x 6x cos x 6 sin x C x3 sin x 3x2 6 sin x x3 6x cos x C 3x2 cos x 6x sin x 6 cos x 0 sin x

1 1 1 1 62. x3 cos 2x dx x3 sin 2x 3x2 cos 2x 6x sin 2x 6 cos 2x C 2 4 8 16 1 3 3 3 Alternate u and itsv and its x3 sin 2x x2 cos 2x x sin 2x cos 2x C 2 4 4 8 signs derivatives antiderivatives 1 x3 cos 2x 4x3 sin 2x 6x2 cos 2x 6x sin 2x 3 cos 2x C 8 2 1 3x 2 sin 2x 1 6x 4 cos 2x 1 6 8 sin 2x 1 0 16 cos 2x

63. x sec2 x dx x tan x lncos x C Alternate u and itsv and its signs derivatives antiderivatives x sec2 x 1 tan x 0 lncos x 116 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

2 8 16 64. x2x 232 dx x2x 252 xx 272 x 292 C Alternate u and itsv and its 5 35 315 signs derivatives antiderivatives

2 2 32 x 25 235x2 40x 32 C x x 2 315 2 52 2x 5 x 2 4 72 2 35 x 2 8 92 0 315 x 2

65. u x ⇒ u2 x ⇒ 2u du dx 66. u x2, du 2x dx

sin x dx sin u2u du 2u sin u du 2x3 cosx2 dx x2 cosx22x dx u cos u du

Integration by parts: w u, dw du, dv sin u du, Integration by parts: w u, dw du, dv cos u du, v cos u v sin u

2 u sin u du 2u cos u cos u du u cos u du u sin u sin u du

2u cos u sin u C u sin u cos u C 2x cos x sin x C x2 sinx2 cosx2 C

67. Let u 4 x, du dx, x 4 u. 68. Let u 2x, u2 2x, 2u du 2 dx.

4 0 2 2 x4 x dx 4 uu12du e2x dx euu du 0 4 0 0 4 2 4u12 u32 du ueu eu (Integration by parts) 0 0 2 2 8 2 4 2e e 0 1 u32 u52 3 5 0 e2 1 8 2 128 8 32 3 5 15

1 69. Let w ln x, dw dx, x ew, dx ew dw. x

cosln x dx cos wew dw

Now use integration by parts twice.

cos wew dw cos wew sin wew dw u cos w, dv ew dw

cos wew sin wew cos wew dw u sin w, dv ew dw

2 cos wew dw cos wew sin wew

1 cos wew dw ewcos w sin w C 2 1 cosln x dx xcosln x sinln x C 2 Section 8.2 Integration by Parts 117

70. Let w 1 x2, dw 2x dx, x2 w 1, x w 1. dw lnx2 1 dx lnw 2w 1 1 1 Integration by parts: u ln w, du dw, dv dw, v w 1 w 2w 1 w 1 lnx2 1 dx lnww 1 dw 2 Substitution: z w 1, z2 w 1, 2z dz dw z lnx2 1 dx lnww 1 2z dz z2 1 1 lnww 1 21 dz z2 1 lnww 1 2z 2 arctanz C ln1 x2x 2x 2 arctanx C

71. Integration by parts is 72. Answers will vary. 73. No 74. Yes based on the Product Substitution u ln x, dv x dx Rule.

75. Yes 76. No 77. Yes. Let u x and 78. No u x2, dv e2x dx Substitution Substitution 1 du dx. x 1 Substitution also works. Let u x 1.

e4t 79. (a) t 3e4t dt 32t 3 24t 2 12t 3 C 128

(b) 5

C = 2

C = 1 −2 4

−1

1 80. (a) 4 sin d 4 cos 43 sin 122 cos 24 sin 24 cos C 5

(b) 35

C = 10

−4 4 C = 0

−10

(c) The graphs are vertical translations of each other. 118 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

e2x 81. (a) e2x sin 3x dx 2 sin 3x 3 cos 3x C 13 2 1 e2x sin 3x dx 2e 3 0.2374 0 13

(b) 7 C = 5

C = 2

−2 6

−1

1,171,875 arcsinx5 x2x2 2525 x252 625x25 x232 46,875x25 x2 82. (a) x 425 x232 dx C 128 16 64 128 5 1,171,875 x 425 x223 dx 14,381.0699 0 256

(b) 400

C = 100 C = 0 −4 4

−200

1 83. (a) dv 2x 3 dx ⇒ v 2x 312 dx 2x 332 3 u 2x ⇒ du 2 dx 2 2 2x2x 3 dx x2x 332 2x 332 dx 3 3 2 2 x2x 332 2x 352 C 3 15 2 2 2x 3323x 3 C 2x 332x 1 C 15 5 u 3 1 (b) u 2x 3 ⇒ x and dx du 2 2 u 3 1 1 1 2 2x2x 3 dx 2 u12 du u32 3u12 du u52 2u32 C 2 2 2 2 5 1 1 2 u32u 5 C 2x 3322x 3 5 C 2x 332x 1 C 5 5 5 Section 8.2 Integration by Parts 119

2 84. (a) dv 4 x dx ⇒ v 4 x12 dx 4 x32 3 u x ⇒ du dx 2 2 x4 x dx x4 x32 4 x32 dx 3 3 2 4 2 x4 x32 4 x52 C 4 x323x 8 C 3 15 15 (b) u 4 x ⇒ x u 4 and dx du

x4 x dx u 4u12 du u32 4u12 du

2 8 2 u52 u32 C u323u 20 C 5 3 15 2 2 4 x3234 x 20 C 4 x323x 8 C 15 15

x 85. (a) dv dx ⇒ v 4 x212x dx 4 x2 4 x2 u x2 ⇒ du 2x dx x3 dx x24 x2 2x4 x2 dx 4 x2 2 1 x24 x2 4 x232 C 4 x2 x2 8 C 3 3 1 (b) u 4 x2 ⇒ x2 u 4 and 2x dx du ⇒ x dx du 2 x3 x2 u 4 1 dx x dx du 4 x2 4 x2 u 2 1 1 2 u12 4u12 du u32 8u12 C 2 2 3 1 1 1 u12u 12 C 4 x2 4 x2 12 C 4 x2 x2 8 C 3 3 3

86. (a) dv 4 x dx ⇒ v 4 x12 dx (b) u 4 x ⇒ x 4 u and dx du

2 x4 x dx 4 uu du 4 x32 3 u x ⇒ du dx 4u12 u32 du 2 2 x4 x dx x4 x32 4 x32 dx 8 2 3 3 u32 u52 C 3 5 2 4 x4 x32 4 x52 C 2 3 15 u3220 3u C 15 2 4 x325x 24 x C 2 15 4 x3220 34 x C 15 2 4 x323x 8 C 2 15 4 x323x 8 C 15 120 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

87. n 0: ln x dx xln x 1 C

x2 n 1: x ln x dx 2 ln x 1 C 4 x3 n 2: x2 ln x dx 3 ln x 1 C 9 x 4 n 3: x3 ln x dx 4 ln x 1 C 16 x5 n 4: x 4 ln x dx 5 ln x 1 C 25 xn1 In general, xn ln x dx n 1ln x 1 C. n 12

88. n 0: ex dx ex C

n 1: xex dx xex ex C xex ex dx

n 2: x2ex dx x2ex 2xex 2ex C x2ex 2xex dx

n 3: x3ex dx x3ex 3x2ex 6xex 6ex C x3ex 3x2e x dx

n 4: x 4e x dx x 4ex 4x3ex 12x2ex 24xex 24ex C x 4ex 4x3ex dx

In general, x nex dx xnex nx n1ex dx.

89. dv sin x dx ⇒ v cos x 90. dv cos x dx ⇒ v sin x u xn ⇒ du nx n1 dx u x n ⇒ du nxn1 dx

x n sin x dx x n cos x nx n1 cos x dx x n cos x dx xn sin x nxn1 sin x dx

x n1 1 91. dv x n dx ⇒ v 92. dv eax dx ⇒ v eax n 1 a 1 u x n ⇒ du nx n1 dx u ln x ⇒ du dx x x neax n x neax dx x n1eax dx x n1 x n a a x n ln x dx ln x dx n 1 n 1 x n1 x n1 ln x C n 1 n 12 x n1 n 1ln x 1 C n 12 Section 8.2 Integration by Parts 121

93. Use integration by parts twice. 1 1 (1) dv eax dx ⇒ v eax (2) dv eax dx ⇒ v eax a a u sin bx ⇒ du b cos bx dx u cos bx ⇒ du b sin bx dx eax sin bx b eax sin bx dx eax cos bx dx a a eax sin bx b eax cos bx b eax sin bx b b2 eax sin bx dx eax cos bx eax sin bx dx a a a a a a2 a2 b2 eaxa sin bx b cos bx Therefore, 1 eax sin bx dx a2 a2 eaxa sin bx b cos bx eax sin bx dx C. a2 b2

94. Use integration by parts twice. 1 1 (1) dv eax dx ⇒ v eax (2) dv eax dx ⇒ v eax a a u cos bx ⇒ du b sin bx u sin bx ⇒ du b cos bx eax cos bx b eax cos bx b eax sin bx b eax cos bx dx eax sin bx dx eax cos bx dx a a a a a a eax cos bx beax sin bx b2 eax cos bx dx a a2 a2 b2 eaxa cos bx b sin bx Therefore, 1 eax cos bx dx a2 a2 eaxa cos bx b sin bx eax cos bx dx C. a2 b2

95. n 3, (Use formula in Exercise 91.) x 4 x3 ln x dx 4 ln x 1 C 16

96. n 2, (Use formula in Exercise 90.)

x2 cos x dx x2 sin x 2x sin x dx, (Use formula in Exercise 83.) n 1

x2 sin x 2x cos x cos x dx x2 sin x 2x cos x 2 sin x C

97. a 2, b 3, (Use formula in Exercise 94.) e2x2 cos 3x 3 sin 3x e2x cos 3x dx C 13 122 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

98. n 3, a 2, (Use formula in Exercise 92 three times.) x3e2x 3 x3e2x dx x2e2x dx, n 3, a 2 2 2 x3e2x 3 x2e2x xe2x dx, n 2, a 2 2 2 2 x3e2x 3x2e2x 3 xe2x 1 e2x dx 2 4 2 2 2 x3e2x 3x2e2x 3xe2x 3e2x C, n 1, a 2 2 4 4 8 e2x 4x3 6x2 6x 3 C 8

99. dv ex dx ⇒ v ex 100. dv ex3 dx ⇒ v 3ex3 u x ⇒ du dx u x ⇒ du dx 4 4 4 4 4 1 3 A xex dx xex ex dx ex x3 4 A xe dx 0 0 0 e 0 9 0 5 1 3 3 1 0.908 x3 x3 4 3xe 3 e dx e 9 0 0

3 3 1 9 9e x 3 0.4 9 e 0

1 1 −1 7 1 e e −1 4

− 1 2 1 0.264 −0.1 e

1 101. A ex sinxdx 102. A x sin x dx x cos x sin x 0 0 0 x 1 e sin x cos x See Exercise 89. 2 1 0 3 1 1 1 2 e 1 −1 4 1 2 1 e −1 0 1.5 0.395 See Exercise 93. 0

e e 103. (a) A ln x dx x x ln x 1 See Exercise 4. y 1 1 (b) Rx ln x, rx 0 2

e (,e 1) V ln x2 dx 1 1 e x xln x2 2x ln x 2x Use integration by parts twice, see Exercise 7. 123 1 e 2 2.257

—CONTINUED— Section 8.2 Integration by Parts 123

103. —CONTINUED— e x ln x dx e2 1 (c) px x, hx ln x (d) x 1 2.097 1 4 e x2 e 1e 2 V 2 x ln x dx 2 1 2 ln x 2 1 ln x dx e 2 4 1 y 0.359 1 1 2 e2 1 13.177 See Exercise 91. e2 1 e 2 2 x, y , 2.097, 0.359 4 2

104. y x sin x, 0 ≤ x ≤ y

3 (a) V x sin x2 dx x2 sin2 x dx 0 0 2 1 cos 2x 1 sin 2x 1 Let u x2, du 2x dx, dv sin2 x dx dx, v x . 2 2 4 x π π 2 1 sin 2x 1 sin 2x −1 x2 sin2 x dx x2 x x 2x dx 2 4 2 4 1 x2 sin 2x x sin 2x x3 x2 dx 2 4 2 1 x2 sin 2x x3 x sin 2x x3 dx 2 4 3 2 1 1 1 x3 x2 sin 2x sin 2x 2x cos 2x C (Integration by Parts) 6 4 8 1 1 1 1 1 V x2 sin2 x dx x3 x2 sin 2x sin 2x 2x cos 2x 4 2 0 6 4 8 0 6 4 (b) V 2xx sin x dx 22 cos x 2x sin x x2 cos x 2 2 4 23 8 0 0 (c) m x sinx dx sin x x cos x 0 0 1 2 Mx x sin x dx 0 2 1 1 1 3 (See part (a).) 2 6 4 1 1 3 12 8 2 My x x sin x dx 4 (See part (b).) 0

2 3 My 4 M 112 18 1 1 x 1.8684, y x 2 0.6975 m m 2 8

105. In Example 6, we showed that the centroid of an equivalent region was 1, 8. By y symmetry, the centroid of this region is 8, 1. You can also solve this problem directly.

1 π 1 2 A arcsin x dx x x arcsin x 1 x2 Example 3 0 2 2 0 0 1 1 x 2 2 1 1 1 My Mx 2 arcsin x x x arcsin x dx , y arcsin x dx 1 A 0 2 8 A 0 2 2 124 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

106. f x x2, gx 2x y (4, 16) f 2 g2 4, f 4 g4 16 16 f(x) = x2 4 x3 1 4 12 m x2 2x dx 2x 2 3 ln 2 2 8 g(x) = 2x 64 16 8 4 4 (2, 4) 3 ln 2 3 ln 2 x 1234 56 12 1.3543 3 ln 2 4 1 4 M x2 2xx2 2x dx 2 x x My x x 2 dx 2 2 2 4 1 56 12 x 4 22x dx 2 4.1855 2 2 ln 2 ln 2 1 x5 22x 4 M M x, y y, x 3.0905, 9.3318 2 5 2 ln 2 2 m m 1 1024 128 32 8 2 5 ln 2 5 ln 2 496 60 12.6383 5 ln 2

1 4t 107. Average value e cos 2t 5 sin 2t dt 0 1 4 cos 2t 2 sin 2t 4 sin 2t 2 cos 2t e4t 5e4t From Exercises 93 and 94 20 20 0 7 1 e4 0.223 10

2 2 108. (a) Average 1.6t ln t 1 dt 0.8t 2 ln t 0.4t 2 t 3.2ln 2 0.2 2.018 1 1

4 4 (b) Average 1.6t ln t 1 dt 0.8t 2 ln t 0.4t 2 t 12.8ln 4 7.2ln 3 1.8 8.035 3 3

109. c t 100,000 4000t, r 5%, t1 10 110. c t 30,000 500t, r 7%, t1 5 10 5 5 P 100,000 4000te0.05t dt P 30,000 500te0.07t dt 500 60 te0.07t dt 0 0 0 10 100 4000 25 te0.05t dt Let u 60 t, dv e0.07t dt, du dt, v e0.07t. 0 7 100 100 5 100 5 Let u 25 t, dv e0.05t dt, du dt, v e0.05t. P 50060 t e0.07t e0.07t dt 5 7 0 7 0 100 10 100 10 100 5 10,000 5 0.05t 0.05t 0.07t 0.07t P 4000 25 t e e dt 500 60 t e e 5 0 5 0 7 0 49 0 100 10 10,000 10 0.05t 0.05t $131,528.68 4000 25 t e e 5 0 25 0 $931,265 Section 8.2 Integration by Parts 125

x 1 x2 2x 2 111. x sin nx dx cos nx sin nx 2 2 112. x cos nx dx sin nx 2 cos nx 3 sin nx n n n n n 2 2 cos n cos n cos n cosn n n n2 n2 2 4 cos n cos n n n2 2n, if n is even 4n2, if n is even 2n, if n is odd 4n2, if n is odd 1n 4 n2

n 2 n 113. Let u x, dv sin x dx, du dx, v cos x. 2 n 2 1 1 1 n 2x n 2 n I1 x sin x dx cos x cos x dx 0 2 n 2 0 n 0 2 2 n 2 2 n 1 cos sin x n 2 n 2 0

2 n 2 2 n cos sin n 2 n 2 n 2 n Let u x 2, dv sin x dx, du dx, v cos x. 2 n 2 2 2 2 n 2 x 2 n 2 n I2 x 2 sin x dx cos x cos x dx 1 2 n 2 1 n 1 2 2 n 2 2 n 2 cos sin x n 2 n 2 1

2 n 2 2 n cos sin n 2 n 2

2 2 n 2 2 n 8h n hI I b h sin sin sin 1 2 n n 2 n 2 n2 2

114. For any integrable function, f x dx C f x dx, but this cannot be used to imply that C 0.

115. Shell Method: y Disk Method: b fb() yfx=() f a f b V 2 xfx dx V b2 a2 dy b2 f 1y2 dy a fa() 0 f a x2 f b dv x dx ⇒ v b2 a2 f a b2 f b f a f 1y2 dy 2 f a x u f x ⇒ du fx dx ab f b b2f b a2f a f 1y2 dy x2 x2 b f a V 2 f x fx dx 2 2 a

b b2f b a2f a x2 fx dx a

Since x f 1y, we have f x y and fx dx dy. When y f a, x a. When y f b, x b. Thus,

f b b f 1y2 dy x2fx dx f a a and the volumes are the same. 126 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

116. fx xex

(a) f x xe x dx xe x e x C (b) 1

Parts: u x, dv ex dx f 0 0 1 C ⇒ C 1 04 f x xex ex 1 0

n xn yn n xn yn 00 0 00 0 1 0.05 0 1 0.1 0 2 0.10 2.378 103 2 0.2 0.0090484 3 0.15 0.0069 3 0.3 0.025423 4 0.20 0.0134 4 0.4 0.047648 80 4.0 0.9064 40 4.0 0.9039

1 1

04 04 0 0

(e) The result in part (c) is better because h is smaller.

117. fx 3x sin2x, f 0 0 3 (a) f x 3x sin 2x dx 2x cos 2x sin 2x C (b) 3 4

(Parts: u 3x, dv sin 2x dx) 0 5 3 f 0 0 0 C ⇒ C 0 4 −5 3 f x 2x cos 2x sin 2x 4 (d) Using h 0.1, you obtain the points: 3 (c) Using h 0.05, you obtain the points: n xn yn

3 00 0 0 5 n xn yn 1 0.1 0 00 0 0 5 − 1 0.05 0.05 2 0.2 0.0060 5

2 0.10 7.4875 10 4 −5 3 0.3 0.0293 3 0.15 0.0037 4 0.4 0.0801 4 0.20 0.0104 40 4.0 1.0210 80 4.0 1.3181 Section 8.2 Integration by Parts 127

118. fx cos x, f 0 1

(a) Let w x, w2 x, 2w dw dx. (b) 3

cos x dx cos w 2w dw 09

Now use parts: u 2w, dv cos w dw. −3 cos x dx 2w sin w 2 cos w C (d) Using h 0.1, you obtain the points: 2x sin x 2 cos x C n xn yn f 0 1 2 C ⇒ C 1 00 1 f x 2x sin x 2 cos x 1 1 0.1 1.1 (c) Using h 0.05, you obtain the points: 2 0.2 1.1950

n xn yn 3 0.3 1.2852 00 1 4 0.4 1.3706 1 0.05 1.05 2 0.1 1.0988 80 4.0 1.8759 3 0.15 1.1463 4 0.2 1.1926 80 4.0 1.8404

2 2 119. On 0, , sin x ≤ 1 ⇒ x sin x ≤ x ⇒ x sin x dx ≤ x dx. 2 0 0

120. (a) A x sin x dx sin x x cos x 0 0

2 2 (b) x sin x dx sin x x cos x 2 3 A 3

3 3 (c) x sin x dx sin x x cos x 3 2 5 2 2 A 5 The area between y x sin x and y 0 on n, n 1 is 2n 1:

n1 n1 x sin x dx sin x x cos x ±n 1 ± n ±2n 1 n n A ±2n 1 2n 1 128 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

Section 8.3 Trigonometric Integrals

1. y sec x y sec x tan x sin x sec2 x

sin x sec2 x dx sec x C

Matches (c)

1 2. y cos x sec x 3. y x tan x tan3 x 3 y sin x sec x tan x y 1 sec2 x tan2 xsec2 x sin x sin x sec2 x tan2 x tan2 x1 tan2 x sin x1 sec2 x tan4 x sin x tan2 x 1 tan4 x dx x tan x tan3 x C 3 sin x tan2 x dx cos x sec x C Matches (d) Matches (a)

4. y 3x 2 sin x cos3 x 3 sin x cos x y 3 2 cos4 x 6 sin2 x cos2 x 3 cos2 x 3 sin2 x 3 2 cos4 x 6 cos2 x1 cos2 x 3 cos2 x 31 cos2 x 8 cos4 x

8 cos4 x dx 3x 2 sin x cos3 x 3 sin x cos x C

Matches (b)

5. Let u cos x, du sin x dx. 6. cos3 x sin4 x dx cos x1 sin2 x sin4 x dx

cos3 x sin x dx cos3 xsin x dx sin4 x sin6 x cos x dx 1 cos4 x C sin5 x sin7 x 4 C 5 7

7. Let u sin 2x, du 2 cos 2x dx. 8. Let u cos x, du sin x dx. 1 sin5 2x cos 2x dx sin5 2x2 cos 2x dx sin3 x dx sin x1 cos2 x dx 2 1 sin6 2x C cos2 xsin x dx sin x dx 12 1 cos3 x cos x C 3

9. Let u cos x, du sin x dx.

sin5 x cos2 x dx sin x1 cos2 x2 cos2 x dx

1 2 1 cos2 x 2 cos4 x cos6 xsin x dx cos3 x cos5 x cos7 x C 3 5 7 Section 8.3 Trigonometric Integrals 129

x 1 x 10. Let u sin , du cos dx. 11. cos3 sin d cos 1 sin2 sin 12 d 3 3 3 x x x cos3 dx cos 1 sin2 dx sin 12 sin 52 cos d 3 3 3 x 1 x 2 2 31 sin2 cos dx sin 32 sin 72 C 3 3 3 3 7 x 1 x 3sin sin3 C 3 3 3 x x 3 sin sin3 C 3 3

sin5 t 12. dt sin t1 cos2 t2cos t12 dt cos t

sin t1 2 cos2 t cos4 tcos t12 dt

4 2 cos t12 2cos t32 cos t72 sin t dt 2cos t12 cos t52 cos t92 C 5 9

1 cos 6x 1 cos 4x 1 1 13. cos2 3x dx dx 14. sin2 2x dx dx x sin 4x C 2 2 2 4 1 1 1 x sin 6x C 4x sin 4x C 2 6 8 1 6x sin 6x C 12

1 cos 2 1 cos 2 1 cos 4 1 cos 4 15. sin2 cos2 d d 16. sin4 2 d d 2 2 2 2 1 1 1 cos2 2 d 1 2 cos 4 cos2 4 d 4 4 1 1 cos 4 1 1 cos 8 1 d 1 2 cos 4 d 4 2 4 2 1 1 3 1 1 cos 4 d 2 cos 4 cos 8 d 8 4 2 2 1 1 1 3 1 1 sin 4 C sin 4 sin 8 C 8 4 4 2 2 16 1 3 1 1 4 sin 4 C sin 4 sin 8 C 32 8 8 64

17. Integration by parts: 1 cos 2x x sin 2x 1 dv sin2 x dx ⇒ v 2x sin 2x 2 2 4 4 u x ⇒ du dx 1 1 x sin2 x dx x2x sin 2x 2x sin 2x dx 4 4 1 1 1 1 x2x sin 2x x2 cos 2x C 2x2 2x sin 2x cos 2x C 4 4 2 8 130 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

18. Use integration by parts twice. 1 cos 2x x sin 2x 1 dv sin2 x dx ⇒ v 2x sin 2x 2 2 4 4 u x2 ⇒ du 2x dx 1 dv sin 2x dx ⇒ v cos 2x 2 u x ⇒ du dx 1 1 x2 sin2 x dx x22x sin 2x 2x2 x sin 2x dx 4 2 1 1 1 1 x3 x2 sin 2x x3 x sin 2x dx 2 4 3 2 1 1 1 1 1 x3 x2 sin 2x x cos 2x cos 2x dx 6 4 2 2 2 1 1 1 1 x3 x2 sin 2x x cos 2x sin 2x C 6 4 4 8 1 4x3 6x2 sin 2x 6x cos 2x 3 sin 2x C 24

2 2 2 2 4 8 19. cos3 x dx , n 3 20. cos5 x dx , n 5 0 3 0 3 5 15

2 2 4 6 16 2 1 21. cos7 x dx , n 7 22. sin2 x dx , n 2 0 3 5 7 35 0 2 2 4

2 1 3 5 5 2 2 4 6 16 23. sin6 x dx , n 6 24. sin7 x dx , n 7 0 2 4 6 2 32 0 3 5 7 35

1 1 25. sec3x dx lnsec 3x tan 3x C 26. sec22x 1 dx tan2x 1 C 3 2

27. sec4 5x dx 1 tan2 5x sec2 5x dx 28. sec6 3x dx 1 tan2 3x2 sec2 3x dx

1 tan3 5x tan 5x C 1 2 tan2 3x tan4 3x sec2 3x dx 5 3 tan 5x 1 2 1 3 tan2 5x C tan 3x tan3 3x tan5 3x C 15 3 9 15

1 2 ⇒ 29. dv sec x dx v tan x

u sec x ⇒ du sec x tan x dx 1 1 3 2 2 sec x dx sec x tan x sec x tan x dx sec x tan x sec x sec x 1 dx

1 3 2 sec x dx sec x tan x lnsec x tan x C1

1 sec3x dx sec x tan x lnsec x tan x C 2 Section 8.3 Trigonometric Integrals 131

x x x 30. tan2 x dx sec2 x 1 dx tan x x C 31. tan5 dx sec2 1 tan3 dx 4 4 4 x x x tan3 sec2 dx tan3 dx 4 4 4 x x x tan4 sec2 1 tan dx 4 4 4 x x x tan4 2 tan2 4 ln cos C 4 4 4

x x 1 x 32. tan3 sec2 dx tan4 C 33. u tan x, du sec2 x dx 2 2 2 2 1 sec2 x tan x dx tan2 x C 2

or, u sec x, du sec x tan x dx,

1 sec2 x tan x dx sec2 x C. 2

34. Let u sec 2t, du 2 sec 2t tan 2t. sec5 2t sec3 2t tan3 2t sec3 2t dt sec2 2t 1 sec3 2t tan 2t dt sec4 2t sec2 2tsec 2t tan 2t dt C 10 6

tan3 x 1 35. tan2 x sec2 x dx C 36. tan5 2x sec2 2x dx tan6 2x C 3 12

1 x x x 1 x x 37. sec6 4x tan 4x dx sec5 4x4 sec 4x tan 4x dx 38. sec2 tan dx 2sec sec tan dx 4 2 2 2 2 2 2 sec6 4x x C sec2 C or 24 2 x x x 1 x x sec2 tan dx 2tan sec2 dx tan2 C 2 2 2 2 2 2

39. Let u sec x, du sec x tan x dx. 40. tan3 3x dx sec2 3x 1 tan 3x dx

sec3 x tan x dx sec2 xsec x tan x dx 1 1 3 sin 3x tan 3x3 sec2 3x dx dx 3 3 cos 3x 1 sec3 x C 1 1 3 tan2 3x lncos 3x C 6 3

tan2 x sec2 x 1 tan2 x sin2 x 41. dx dx 42. cos5 x dx sec x sec x sec5 x cos2 x

sec x cos x dx sin2 x cos3 x dx

lnsec x tan x sin x C sin2 x1 sin2 x cos x dx

sin2 x sin4 x cos x dx

sin3 x sin5 x C 3 5 132 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 43. r sin4 d 1 cos22 d 44. s sin2 cos2 d 4 2 2 1 1 cos 1 cos 1 cos2 1 2 cos2 cos22 d d d 4 2 2 4 1 1 cos4 1 1 1 2 cos2 d sin2 d 1 cos 2 d 4 2 4 8 1 1 1 1 sin 2 sin2 sin4 C C 4 2 8 8 2 1 1 12 8 sin2 sin4 C 2 sin 2 C 32 16

45. y tan3 3x sec 3x dx 46. y tan x sec4 x dx

sec2 3x 1 sec 3x tan 3x dx tan12 xtan2 x 1 sec2 x dx

1 1 sec2 3x3 sec 3x tan 3x dx 3 sec 3x tan 3x dx tan52 x tan12 x sec2 x dx 3 3 1 1 2 2 sec3 3x sec 3x C tan72 x tan32 x C 9 3 7 3

47. (a) y 48. (a) y

4 1

x x − 4 1 1 1 (0, 4)

−1 −4

dy dy 1 (b) sin2 x, 0, 0 4 (b) sec2 x tan2 x, 0, dx dx 4

y sin2 x dx − 66y sec2 x tan2 x dx u tan x, du sec2 x dx

3 − tan x 1 cos 2x 4 y C dx 3 2 1 1 1 1 1 sin 2x 0, : C ⇒ y tan3 x x C 4 4 3 4 2 4 1 sin 2x 0, 0: 0 C, y x 2 4

dy 3 sin x dy 49. , y0 2 50. 3y tan2 x, y0 3 dx y dx

8 8 (0, 3)

−9 9 −1 1

−4 −2 Section 8.3 Trigonometric Integrals 133

1 51. sin 3x cos 2x dx sin 5x sin x dx 52. cos 4 cos3 d cos 4 cos 3 d 2 1 1 1 cos 5x cos x C cos 7 cos d 2 5 2 1 sin 7 sin cos 5x 5 cos x C C 10 14 2

1 53. sin sin 3 d cos 2 cos 4 d 54. sin4x cos 3x dx sin 4x cos 3x dx 2 1 1 1 1 sin 2 sin 4 C sin x sin 7x dx 2 2 4 2 1 1 1 2 sin 2 sin 4 C cos x cos 7x C 8 2 7 1 7 cos x cos 7x C 14

x 1 x 55. cot3 2x dx csc2 2x 1 cot 2x dx 56. Let u tan , du sec2 dx. 2 2 2 1 1 2 cos 2x x x x x x cot 2x2 csc2 2x dx dx tan4 sec4 dx tan4 tan2 1 sec2 dx 2 2 sin 2x 2 2 2 2 2 1 1 x x 1 x cot2 2x lnsin 2x C 2tan6 tan4 sec2 dx 4 2 2 2 2 2 1 2 x 2 x lncsc2 2x cot2 2x C tan7 tan5 C 4 7 2 5 2

57. Let u cot , du csc2 d. 58. u cot 3x, du 3 csc2 3x dx 1 csc4 d csc2 1 cot2 d csc2 3x cot 3x dx cot 3x3 csc2 3x dx 3 1 csc2 d csc2 cot2 d cot2 3x C 6 1 cot cot3 C 3

cot2 t csc2 t 1 cot3 t cos3 t 1 sin2 t cos t 59. dt dt 60. dt dt dt csc t csc t csc t sin2 t sin2 t cos t csc t sin t dt dt cos t dt sin2 t lncsc t cot t cos t C 1 sin t C csc t sin t C sin t

1 cos2 x 1 sin2 x sin2 x cos2 x 1 2 cos2 x 61. dx dx dx 62. dx dx sec x tan x sin x sin x cos x cos x

csc x sin x dx sec x 2 cos xdx

lncsc x cot x cos x C lnsec x tan x 2 sin x C 134 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

63. tan4 t sec4 t dt tan2 t sec2 ttan2 t sec2 t dt, tan2 t sec2 t 1

tan2 t sec2 t dt 2 sec2 t 1 dt 2 tan t t C

1 sec t cos t 1 1 cos 2x 64. dt dt 2 cos t 1 cos t 1 cos t 65. sin x dx 2 dx 0 2 1 sec t dt lnsec t tan t C x sin 2x 2 0

3 3 4 4 66. tan2 x dx sec2 x 1 dx 67. tan3 x dx sec2 x 1 tan x dx 0 0 0 0 3 4 4 sin x tan x x 3 sec2 x tan x dx dx 0 3 0 0 cos x 1 4 tan2 x lncos x 2 0 1 1 ln 2 2

68. Let u tan t, du sec2 t dt. 69. Let u 1 sin t, du cos t dt. 4 2 4 2 2 cos t 2 2 32 sec t tan t dt tan t dt ln1 sin t ln 2 0 3 0 3 0 1 sin t 0

1 70. sin 3 cos d sin 4 sin 2 d 71. Let u sin x, du cos x dx. 2 2 2 1 1 1 cos3 x dx 2 1 sin2 x cos x dx cos 4 cos 2 0 2 0 2 4 2 1 2 4 2sin x sin3 x 3 0 3

2 2 1 cos 2x 4 x 1 2 73. cos dx 6x 8 sin x sin 2x C 72. sin x 1 dx 1 dx 2 16 2 2 2 2 1 x x x x 3 1 4 sin cos3 6 sin cos 3x C cos 2x dx 8 2 2 2 2 2 2 2 2 6 3 1 3 x sin 2x C = 2 2 4 2 2 − 99 C = 0

−6

1 74. sin2 x cos2 x dx 4x sin 4x C 2 32

−66

−2 Section 8.3 Trigonometric Integrals 135

1 3 3 C = 10 75. sec5 x dx sec3 x tan x sec x tan x lnsec x tan x C 4 2 −33

C = −12 −3

tan21 x 1 76. tan31 x dx lncos1 x C 77. sec5 x tan x dx sec5 x C 2 5

2 5

C = 2 C = 0

−33 − 22

−2 −5

sec41 x 4 4 78. sec41 x tan1 x dx C 1 1 3 2 4 79. sin 2 sin 3 d sin sin 5 0 2 5 0 10

− 3.5 3.5

−2

2 2 3 1 2 2 2 1 3x 1 80. 1 cos d 2 sin sin 2 81. sin4 x dx sin 2x sin 4x 0 2 4 0 0 4 2 8 0 3 3 2 4 16

2 1 5x 3 1 2 5 82. sin6 x dx 2 sin 2x sin 4x sin3 2x 0 8 2 8 6 0 32

83. (a) Save one sine factor and convert the remaining sine 84. See guidelines on page 537. factors to cosine. Then expand and integrate. (b) Save one cosine factor and convert the remaining cosine factors to sine. Then expand and integrate. (c) Make repeated use of the power reducing formula to convert the integrand to odd powers of the cosine.

85. (a) Let u tan 3x, du 3 sec2 3x dx. 1 sec4 3x tan3 3x dx sec2 3x tan3 3x sec2 3x dx tan2 3x 1 tan3 3x3 sec2 3x dx 3 1 tan6 3x tan4 3x tan5 3x tan3 3x3 sec2 3x dx C 3 18 12 1 Or let u sec 3x, du 3 sec 3x tan 3x dx.

sec4 3x tan3 3x dx sec3 3x tan2 3x sec 3x tan 3x dx

1 sec6 3x sec4 3x sec3 3xsec2 3x 13 sec 3x tan 3x dx C 3 18 12

—CONTINUED— 136 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

85. —CONTINUED—

(b) 0.05

− 0.5 0.5

− 0.05

sec6 3x sec4 3x 1 tan2 3x3 1 tan2 3x2 (c) C C 18 12 18 12 1 1 1 1 1 1 1 tan6 3x tan4 3x tan2 3x tan4 3x tan2 3x C 18 6 6 18 12 6 12 tan6 3x tan4 3x 1 1 C 18 12 18 12 tan6 3x tan4 3x C 18 12 2

86. (a) Let u tan x, du sec2 x dx. (b) 8 1 sec2 x tan x dx tan2 x C 2 1

−44 Or let u sec x, du sec x tan x dx. −2 1 sec xsec x tan x dx sec2 x C 2 1 1 1 1 1 (c) sec2 x C tan2 x 1 C tan2 x C tan2 x C 2 2 2 2 2 2

2 1 87. A sin x sin3 x dx 88. A sin2x dx 0 0 2 2 1 1 cos2x sin x dx sin3 x dx dx 0 0 0 2 2 1 sin 2x 1 2 cos x Wallis's Formula x 0 3 2 4 0 2 1 1 1 3 3 2

4 4 89. A cos2 x sin2 x dx 90. A cos2 x sin x cos x dx 4 2 4 4 1 cos 2x cos 2x dx sin x cos x dx 4 2 2 sin 2x 4 1 sin 2x sin2 x 4 x 2 4 2 4 2 2 1 1 1 1 1 1 2 2 8 4 4 4 2 3 1 8 2 Section 8.3 Trigonometric Integrals 137

2 x x 91. Disks y 92. V cos2 sin2 dx 0 2 2 1 R x tan x, r x 0 2 1 4 2 cos x dx 2 0 V 2 tan x dx x 0 ππ 2 84 1 4 − sin x 2 2 0 2 sec x 1 dx −1 0 4 2tan x x 0 21 1.348 4

1 2 93. (a) V sin2 x dx 1 cos 2x dx x sin 2x 0 2 0 2 2 0 2 (b) A sin x dx cos x 1 1 2 y 0 0

1 ππ Let u x, dv sin x dx, du dx, v cos x. , ( 28( 1 1 1 1 x x sin x dx x cos x cos x dx x cos x sin x 2 A 0 2 0 0 2 0 2 1 1 1 1 x 2 ππ y sin x dx 1 cos 2x dx x sin 2x 2 2A 0 8 0 8 2 0 8 x, y , 2 8

2 2 1 2 2 94. (a) V cos2 x dx 1 cos 2x dx x sin 2x 0 2 0 2 2 0 4

2 2 (b) A cos x dx sin x 1 y 0 0

1 ππ− Let u x, dv cos x dx, du dx, v sin x. 2 , ( 28( 2 2 2 2 2 1 x x cos x dx x sin x sin x dx x sin x cos x 1 2 0 0 0 0 2 2 2 2 2 x 1 1 1 1 π π 2 y cos x dx 1 cos 2x dx x sin 2x 4 2 2 0 4 0 4 2 0 8 2 x, y , 2 8

95. dv sin x dx ⇒ v cos x u sinn1 x ⇒ du n 1 sinn2 x cos x dx

sinn x dx sinn1 x cos x n 1sinn2 x cos2 x dx sinn1 x cos x n 1sinn2 x1 sin2 x dx

sinn1 x cos x n 1sinn2 x dx n 1sinn x dx

Therefore, nsinn x dx sinn1 x cos x n 1sinn2 x dx

sinn1 x cos x n 1 sinn x dx sinn2 x dx. n n 138 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

96. dv cos x dx ⇒ v sin x u cosn1 x ⇒ du n 1 cosn2 x sin x dx

cosn x dx cosn1 x sin x n 1cosn2 x sin2 x dx

cosn1 x sin x n 1cosn2 x1 cos2 x dx

cosn1 x sin x n 1cosn2 x dx n 1cosn x dx

Therefore, ncosn x dx cosn1 x sin x n 1cosn2 x dx

cosn1 x sin x n 1 cosn x dx cosn2 x dx. n n

cosm1 x 97. Let u sinn1 x, du n 1 sinn2 x cos x dx, dv cosm x sin x dx, v . m 1 sinn1 x cosm1 x n 1 cosm x sinn x dx sinn2 x cosm2 x dx m 1 m 1 sinn1 x cosm1 x n 1 sinn2 x cosm x1 sin2 x dx m 1 m 1 sinn1 x cosm1 x n 1 n 1 sinn2 x cosm x dx sinn x cosm x dx m 1 m 1 m 1 m n sinn1 x cosm1 x n 1 cosm x sinn x dx sinn2 x cosm x dx m 1 m 1 m 1 cosm1 x sinn1 x n 1 cosm x sinn x dx cosm x sinn2 x dx m n m n

98. Let u secn2 x, du n 2 secn2 x tan x dx, dv sec2 x dx, v tan x.

secn x dx secn2 x tan x n 2 secn2 x tan2 x dx

secn2 x tan x n 2secn2 xsec2 x 1 dx

secn2 x tan x n 2secn x dx secn2 x dx

n 1secn x dx secn2 x tan x n 2secn2 x dx

1 n 2 secn x dx secn2 x tan x secn2 x dx n 1 n 1

sin4 x cos x 4 99. sin5 x dx sin3 x dx 5 5 sin4 x cos x 4 sin2 x cos x 2 sin x dx 5 5 3 3 1 4 8 sin4 x cos x sin2 x cos x cos x C 5 15 15 cos x 3 sin4 x 4 sin2 x 8 C 15 Section 8.3 Trigonometric Integrals 139

cos3 x sin x 3 100. cos4 x dx cos2 x dx 4 4 cos3 x sin x 3 cos x sin x 1 dx 4 4 2 2 1 3 3 cos3 x sin x cos x sin x x C 4 8 8 1 2 cos3 x sin x 3 cos x sin x 3x C 8

2x 5 2x 2 101. sec4 dx sec4 dx 5 2 5 5 5 1 2x 2x 2 2x 2 sec2 tan sec2 dx 2 3 5 5 3 5 5 5 2x 2x 2x sec2 tan 2 tan C 6 5 5 5 5 2x 2x tan sec2 2 C 6 5 5

cos3 x sin3 x 1 102. sin4 x cos2 x dx cos2 x sin2 x dx 6 2 cos3 x sin3 x 1 cos3 x sin x 1 cos2 x dx 6 2 4 4 1 1 1 cos x sin x x cos3 x sin3 x cos3 x sin x C 6 8 8 2 2 1 8 cos3 x sin3 x 6 cos3 x sin x 3 cos x sin x 3x C 48

t t 103. f t a a cos b sin 0 1 6 1 6 12 12 12 1 1 t 1 t a0 f t dt, a1 f t cos dt, b1 f t sin dt 12 0 6 0 6 6 0 6 1 12 0 (a) a 33.5 435.4 244.7 455.6 267.4 476.2 280.4 479.0 272.0 0 12 312 461.0 249.3 438.6 33.5 57.72 a1 23.36 b1 2.75 (Answers will vary.) t t Ht 57.72 23.36 cos 2.75 sin 6 6 t t (b)Lt 42.04 20.91 cos 4.33 sin (c) 90 6 6 H

L 0 14 10

Temperature difference is greatest in the summer t 4.9 or end of May. 140 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

104. (a) n is odd and n ≥ 3. 2 cosn1 x sin x 2 n 1 2 cosn x dx cosn2 x dx 0 n 0 n 0 n 1 cosn3 x sin x 2 n 3 2 n4 cos x dx n n 2 0 n 2 0 n 1 n 3 cosn5 x sin x 2 n 5 2 n6 cos x dx n n 2 n 4 0 n 4 0 n 1 n 3 n 5 2 n6 cos x dx n n 2 n 4 0 2 n 1 n 3 n 5 . . . cos x dx n n 2 n 4 0 n 1 n 3 n 5 2 . . . sin x n n 2 n 4 0 n 1 n 3 n 5 . . . 1 (Reverse the order.) n n 2 n 4 2 4 6 n 1 1 . . . 3 5 7 n 2 4 6 n 1 . . . 3 5 7 n (b) n is even and n ≥ 2. 2 n 1 n 3 n 5 2 n . . . 2 cos x dx cos x dx (From part (a)) 0 n n 2 n 4 0 n 1 n 3 n 5 x 1 2 . . . sin 2x n n 2 n 4 2 4 0 n 1 n 3 n 5 . . . (Reverse the order.) n n 2 n 4 4 1 3 5 n 1 . . . 2 2 4 6 n 1 3 5 n 1 . . . 2 4 6 n 2

1 sinm nx sinm nx 105. cosmx cosnx dx 0, m n 2 m n m n 1 sinmx sinnx dx cosm nx cosm nx dx 2 1 sinm nx sinm nx 0, m n 2 m n m n 1 sinmx cosnx dx sinm nx sinm nx dx 2 1 cosm nx cosm nx , m n 2 m n m n 1 cosm n cosm n cosm n cosm n 2 m n m n m n m n 0, since cos cos . 1 sin2mx sinmx cosmx dx 0 m 2 Section 8.4 Trigonometric Substitution 141

N 106. f x ai sin ix i1

N (a) f x sin nx ai sin ix sin nx (b) f x x i1 1 N a x sin x dx 2 1 f x sin nx dx ai sin ix sin nx dx i1 1 a x sin 2x dx 1 2 2 an sin nx dx by Exercise 106 1 2 1 cos2nx a x sin 3x dx 3 an dx 3 2 a sin2nx n x 2 2n a n a 2 n 1 Hence, an f x sin nx dx.

Section 8.4 Trigonometric Substitution

d x2 16 4 d 1. 4 ln x2 16 C 4 ln x2 16 4 4 lnx x2 16 C dx x dx xx2 16 4 x 4 x2 16 4 x x2 16 4x 4 x x2 16x2 16 4 x x2 16 4x2 4x2 16x2 16 4 x2x2 16 4 xx2 16x2 16 4 4x2 4x2 16 16x2 16 x2x2 16 4x2 xx2 16x2 16 4 x2 16x2 16 4x2 16 xx2 16x2 16 4 x2 16x2 16 4 xx2 16 x2 16 Indefinite integral: dx, matches (b). x

d 1 xx2 16 1 1 x 1 2. 8 lnx2 16 x xx2 16 C 8 x x2 16 dx 2 x2 16 x 2 x2 16 2 8x x2 16 x2 x2 16 x2 16x2 16 x 2x2 16 2 16 x2 x2 16 2x2 16 x2 x2 16 x2 Indefinite integral: , matches (d). x2 16 142 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

d x x16 x2 14 x1216 x2122x 16 x2 3. 8 arcsin C 8 dx 4 2 1 x42 2 8 x2 16 x2 16 x2 216 x2 2 16 x2 16 x2 x2 216 x2 216 x2 216 x2 16 x2 Matches (a)

d x 3 x 37 6x x2 1 1 1 3 x 1 4. 8 arcsin C 8 x 3 7 6x x2 dx 4 2 1 x 342 4 2 7 6x x2 2 8 x 32 16 x 32 16 x 32 216 x 32 2 16 x2 6x 9 16 x2 6x 9 216 x 32 216 x 32 216 x 32 16 x 32 7 6x x2

Indefinite integral:7 6x x2 dx, matches (c).

5. Let x 5 sin , dx 5 cos d, 25 x2 5 cos . 1 5 cos dx d 25 x232 5 cos 3 5 x 1 sec2 d 25 θ 25 − x2 1 tan C 25 x C 2525 x2

6. Same substitution as in Exercise 5 10 5 cos d 2 2 225 x2 dx 10 csc2 d cot C C x225 x2 25 sin2 5 cos 5 5 5x

7. Same substitution as in Exercise 5 25 x2 25 cos2 d 1 sin2 dx 5 d 5csc sin d x 5 sin sin 5 25 x2 5lncsc cot cos C 5 ln 25 x2 C x Section 8.4 Trigonometric Substitution 143

8. Same substitution as in Exercise 5 x2 25 sin2 25 dx 5 cos d 1 cos 2 d 25 x2 5 cos 2 25 1 25 sin 2 C sin cos C 2 2 2 25 x x 25 x2 1 x arcsin C 25 arcsin x25 x2 C 2 5 5 5 2 5

9. Let x 2 sec , dx 2 sec tan d, x2 4 2 tan .

x 1 2 sec tan d 2 dx sec d lnsec tan C x − 4 2 2 tan 1 x 4 θ x x2 4 2 ln C 2 2 1

2 2 lnx x 4 ln 2 C1 lnx x 4 C

10. Same substitution as in Exercise 9 x2 4 2 tan dx 2 sec tan d 2 tan2 d 2sec2 1 d x 2 sec x2 4 x x 2tan C 2 arcsec C x2 4 2 arcsec C 2 2 2

11. Same substitution as in Exercise 9

x3x2 4 dx 8 sec3 2 tan 2 sec tan d 32tan2 sec4 d

tan3 tan5 32tan2 1 tan2 sec2 d 32 C 3 5 32 32 x2 432 x2 4 tan3 5 3 tan2 C 5 3 C 15 15 8 4 1 1 x2 43220 3x2 4 C x2 4323x2 8 C 15 15

12. Same substitution as in Exercise 9 x3 8 sec3 dx 2 sec tan d 8sec4 d x2 4 2 tan tan3 8 81 tan2 sec2 d 8tan C tan 3 tan2 C 3 3 8 x2 4 x2 4 1 1 3 C x2 4 12 x2 4 C x2 4 x2 8 C 3 2 4 3 3

13. Let x tan , dx sec2 d, 1 x2 sec . 1 + x2 sec3 1 x1 x2 dx tan sec sec2 d C 1 x232 C x 3 3 θ Note: This integral could have been evaluated with the Power Rule. 1 144 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

14. Same substitution as in Exercise 13 9x3 tan3 sec3 dx 9 sec2 d 9sec2 1 sec tan d 9 sec C 1 x2 sec 3 3 sec sec2 3 C 31 x2 1 x2 3 C 31 x2x2 2 C

15. Same substitution as in Exercise 13 1 1 sec2 d dx dx 1 x22 1 x2 4 sec4 1 + x2 1 x cos2 d 1 cos 2 d 2 θ 1 1 sin 2 2 2 1 sin cos C 2 1 x 1 arctan x C 2 1 x2 1 x2 1 x arctan x C 2 1 x2

16. Same substitution as in Exercise 13 x2 x2 tan2 sec2 d dx dx sin2 d 1 x22 1 x2 4 sec4 1 1 sin 2 1 1 cos 2 d sin cos C 2 2 2 2 1 x 1 1 x arctan x C arctan x C 2 1 x2 1 x2 2 1 x2

17. Let u 3x, a 2, and du 3 dx. 1 4 9x2 dx 22 3x2 3 dx 3 1 1 3x4 9x2 4 ln 3x 4 9x2 C 3 2 1 2 x4 9x2 ln 3x 4 9x2 C 2 3

18. Let u x, a 1, and du dx. 1 1 x2 dx x1 x2 ln x 1 x2 C 2

25 5 19. 25 4x2 dx 2 x2 dx, a 4 2 1 25 2x 25 2 arcsin x x2 C 2 4 5 4 25 2x x arcsin 25 4x2 C 4 5 2 Section 8.4 Trigonometric Substitution 145

20. 2x2 1 dx 2x2 1 dx, u 2x, du 2 dx

1 1 2x2x2 1 ln2x 2x2 1 C 2 2 x 2 2x2 1 ln 2x 2x2 1 C 2 4

x 1 x 1 21. dx x2 9122x dx 22. dx 9 x2 1 22x dx x2 9 2 9 x2 2 x2 9 C 9 x212 C (Power Rule) (Power Rule)

1 x 1 x 23. dx arcsin C 24. dx arcsin C 16 x2 4 25 x2 5

25. Let x 2 sin , dx 2 cos d, 4 x2 2 cos .

16 4x2 dx 24 x2 dx 2 x

22 cos 2 cos d θ

4 − x 2 8cos2 d

41 cos 2 d

1 4 sin 2 C 2 4 4 sin cos C x 4 arcsin x4 x2 C 2

26. Let u 16 4x2, du 8x dx. 1 1 2 x16 4x2 dx 16 4x2128x dx 16 4x232 C 4 x232 C 8 12 3

27. Let x 3 sec , dx 3 sec tan d, x2 9 3 tan . 1 3 sec tan d dx 2 3 tan x 9 x x2 − 9 sec d θ 3 lnsec tan C1 x x2 9 ln C 3 3 1 lnx x2 9 C 146 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

28. Let u 1 t2, du 2t dt. t 1 1 232 dt 1 t 2t dt C 1 t23 2 2 1 t2

3 29. Let x sin , dx cos d, 1 x2 cos . 30. Let 2x 3 tan , dx sec2 d, 4x2 9 3 sec . 2 1 x2 cos cos d dx 4x2 9 3 sec 32 sec2 d 4 4 dx x sin 4 4 4 1 x 3 2 tan x cot2 csc2 d 8 cos θ d 4 9 sin 49x2 + − 2 1 1 x 2x cot3 C 8 3 C 27 sin3 θ 1 x232 3 C 8 3x3 csc3 C 27 4x2 932 C 27x3

31. Same substitution as in Exercise 30 32. Let 2x 4 tan , dx 2 sec2 d, 4x2 16 4 sec . 3 3 1 2 sec2 d x tan , dx sec2 d dx 2 2 x4x2 16 2 tan 4 sec 1 32 sec2 d 1 sec 1 dx d csc d x4x2 9 32 tan 3 sec 4 tan 4 1 1 csc d lncsc cot C 3 4 1 1 x2 4 2 lncsc cot C ln C 3 4 x 1 4x2 9 3 ln C 2 3 2x x + 4 x

θ 4x2 + 9 2 2x

33. Let x 5 tan , dx 5 sec2 d, x2 5 5 sec2 . 5x 55 tan dx 5 sec2 d 2 32 2 32 x 5 5 sec x2 + 5 x tan 5 d sec θ 5 5sin d

5 cos C 5 5 C x2 5 5 C x2 5 Section 8.4 Trigonometric Substitution 147

34. Let x 3 tan , dx 3 sec2 d, x2 3 3 sec2 . 1 3 sec2 d dx 2 32 3 x 3 3 3 sec x2 + 3 x 1 1 x cos d sin C C θ 3 3 3x2 3 3

35. Let u 1 e2x, du 2e2x dx. 1 1 e2x1 e2x dx 1 e2x122e2x dx 1 e2x32 C 2 3

36. Let u x2 2x 2, du 2x 2 dx. 1 1 x 1x2 2x 2 dx x2 2x 2122x 2 dx x2 2x 232 C 2 3

37. Let ex sin , ex dx cos d, 1 e2x cos . 38. Let x sin , x sin2 , dx 2 sin cos d, 1 x cos . ex1 e2x dx cos2 d 1 x cos 2 sin cos d dx sin 1 x 1 cos 2 d 2 2cos2 d 1 sin 2 2 2 1 cos 2 d 1 sin cos C 2 sin cos C 1 arcsin ex ex1 e2x C arcsinx x1 x C 2

1 1 ex x

θ θ 1 − x 1 − e2x

39. Let x 2 tan , dx 2 sec2 d, x2 2 2 sec2 . 1 1 2 sec2 d dx dx 4 4x2 x4 x2 22 4 sec4 2 cos2 d 4 x2 + 2 x 2 1 1 cos 2 d θ 4 2 2 2 1 sin 2 C 8 2 2 sin cos C 8 2 x x 2 arctan 8 2 x2 2 x2 2

1 x 1 x 2 arctan C 4 x 2 2 2 148 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

40. Let x tan , dx sec2 d, x2 1 sec2 . x3 x 1 1 4x3 4x 1 dx dx dx 4 2 4 2 2 2 x 2x 1 4 x 2x 1 x 1 x2 + 1 x 1 sec2 d 4 2 ln x 2x 1 θ 4 sec4 1 1 1 lnx2 1 1 cos 2 d 2 2 1 1 lnx2 1 sin cos C 2 2 1 x lnx2 1 arctan x C 2 x2 1

1 41. Use integration by parts. Since x > 2, 1 u arcsec 2x ⇒ du dx, dv dx ⇒ v x 2 x 4x 1 2x 4x2 − 1 1 arcsec 2x dx x arcsec 2x dx θ 4x2 1 1 1 2x sec , dx sec tan d, 4x2 1 tan 2 12 sec tan d arcsec 2x dx x arcsec 2x tan 1 x arcsec 2x sec d 2 1 x arcsec 2x lnsec tan C 2 1 x arcsec 2x ln 2x 4x2 1 C. 2

1 x2 42. u arcsin x ⇒ du dx, dv x dx ⇒ v 1 x2 2 x2 1 x2 x arcsin x dx arcsin x dx 2 2 1 x2 x sin , dx cos d, 1 x2 cos x2 1 sin2 x2 1 x arcsin x dx arcsin x cos d arcsin x 1 cos 2d 2 2 cos 2 4 x2 1 1 x2 1 arcsin x sin 2 C arcsin x sin cos C 2 4 2 2 4 x2 1 1 arcsin x arcsin x x1 x2 C 2x2 1 arcsin x x1 x2 C 2 4 4

1 1 x 2 43. dx dx arcsin C 4x x2 4 x 22 2 Section 8.4 Trigonometric Substitution 149

44. Let x 1 sin , dx cos d, 1 x 12 2x x2 cos . x2 x2 dx dx 2x x2 1 x 12 1 − 1 sin 2cos d x 1 cos θ

11− (x − )2 1 2 sin sin2 d

3 1 2 sin cos 2 d 2 2 3 1 2 cos sin 2 C 2 4 3 1 2 cos sin cos C 2 2 3 1 arcsinx 1 22x x2 x 12x x2 C 2 2 3 1 arcsinx 1 2x x2x 3 C 2 2

45. Let x 2 2 tan , dx 2 sec2 d, x 22 4 2 sec .

2 2 (x + 2) + 4 x x 2 tan 22 sec d x + 2 dx dx 2 2 2 sec x 4x 8 x 2 4 θ 2 2tan 1sec d

2 sec lnsec tan C1 x 22 4 x 22 4 x 2 2 ln C 2 2 2 1 2 2 x 4x 8 2 ln x 4x 8 x 2 ln 2 C1 x2 4x 8 2 lnx2 4x 8 x 2 C

46. Let x 3 2 sec , dx 2 sec tan d, x 32 4 2 tan . x x dx dx x2 6x 5 x 32 4 2 sec 3 2 sec tan d 2 tan

2 sec2 3 sec d

2 tan 3 lnsec tan C1 x 33 4 x 3 x 32 4 2 3 ln C 2 2 2 1 x2 6x 5 3 lnx 3 x2 6x 5 C 150 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

47. Let t sin , dt cos d, 1 t2 cos2 . t2 sin2 cos d (a) dt 1 t232 cos3 1 t tan2 d sec2 1 d θ

tan C 1 − t2 t arcsin t C 1 t2 3 2 t2 t 32 32 3 Thus, dt arcsin t arcsin 3 0.685. 1 t232 2 2 3 0 1 t 0 1 4 (b) When t 0, 0. When t 32, 3. Thus, 3 2 t2 3 dt tan 3 0.685. 232 0 1 t 0 3

48. Same substitution as in Exercise 47 1 cos d (a) dt sec4 d tan2 1 sec2 d 1 t252 cos5 1 1 t 3 t tan3 tan C C 3 3 1 t 2 1 t 2 3 2 1 t3 t 32 Thus, dt 1 t252 31 t232 2 0 1 t 0 3 3 8 3 2 3 3 2 3 3.464. 3143 2 14 (b) When t 0, 0. When t 32, 3. Thus, 3 2 1 1 3 1 dt tan3 tan 3 3 3 23 3.464. 252 0 1 t 3 0 3

49. (a) Let x 3 tan , dx 3 sec2 d, x2 9 3 sec . x3 27 tan3 3 sec2 d dx x2 9 3 sec

27sec2 1 sec tan d

1 27 sec3 sec C 9sec3 3 sec C 3 x2 9 3 x2 9 1 9 3 C x2 932 9x2 9 C 3 3 3 3 x3 1 3 Thus, dx x2 932 9x2 9 2 0 x 9 3 0 1 542 272 9 27 3 18 92 92 2 5.272. (b) When x 0, 0. When x 3, 4. Thus, 3 x3 4 dx 9sec3 3 sec 922 32 91 3 92 2 5.272. 2 0 x 9 0 Section 8.4 Trigonometric Substitution 151

3 50. (a) Let 5x 3 sin , dx cos d, 9 25x2 3 cos . 5 3 9 25x2 dx 3 cos cos d 5 9 1 cos 2 d 5 2 9 1 sin 2 C 10 2 9 sin cos C 10 9 5x 5x 9 25x2 arcsin C 10 3 3 3 35 3 5 9 5x 5x9 25x2 9 9 Thus, 9 25x2 dx arcsin . 10 3 9 10 2 20 0 0 3 (b) When x 0, 0. When x , . 5 2 3 5 9 2 9 9 Thus, 9 25x2 dx sin cos . 0 10 0 10 2 20

51. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan . x2 9 sec2 dx 3 sec tan d 2 x 9 3 tan x x2 − 9 9sec3 d θ 3 1 1 9 sec tan sec d (8.3 Exercise 98 or Example 5, Section 8.2) 2 2 9 sec tan lnsec tan 2 9 x x2 9 x x2 9 ln 23 3 3 3 Hence, 6 6 x2 9 xx2 9 x x2 9 dx ln 2 4 x 9 2 9 3 3 4 9 627 27 47 4 7 ln 2 ln 2 9 3 9 3 3 9 6 27 4 7 93 27 ln ln 2 3 3 9 6 33 93 27 ln 2 4 7 9 4 72 3 93 27 ln 12.644. 2 3

—CONTINUED— 152 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

51. —CONTINUED— 4 (b) When x 4, arcsec . When x 6, arcsec2 . 3 3 6 x2 9 3 dx sec tan lnsec tan 2 4 x 9 2 arcsec43 9 9 4 7 4 7 2 3 ln2 3 ln 2 23 3 3 3 9 6 33 93 27 ln 12.644 2 4 7

52. (a) Let x 3 sec , dx 3 sec tan d, (b) When x 3, 0; when x 6, . Hence, 3 x2 9 3 tan . 6x2 9 3 dx lnsec tan sin x2 9 3 tan x2 dx 3 sec tan d 3 0 x2 9 sec2 3 ln 2 3 . tan2 sin2 2 d d sec cos 1 cos2 d cos

sec cos d

lnsec tan sin C x x2 9 x2 9 ln C 3 3 x Hence, 6x2 9 x x2 9 x2 9 6 dx ln x2 3 3 x 3 3 3 ln 2 3 . 2

dy dy 53. x x2 9, x ≥ 3, y3 1 54. x2 4 1, x ≥ 2, y0 4 dx dx x2 9 dy 1 y dx x dx x2 4 Let x 3 sec , dx 3 sec tan d, x2 9 3 tan . 1 y dx x2 4 3 tan y 3 sec tan d 3tan2 d 3 sec Let x 2 tan , x2 4 4 sec2 , dx 2 sec2 d. 1 3sec2 1 d 3tan C y 2 sec2 d sec d 2 sec x2 9 x2 9 lnsec tan C 3 arctan C 3 3 x2 4 x ln C x2 9 2 2 x2 9 3 arctan C 3 2 ln x 4 x C1 ⇒ y 3 1: 1 0 3 0 C C 1 ⇒ ⇒ y 0 4 4 ln2 C1 C1 4 ln 2 x2 9 y x2 9 3 arctan 1 y lnx2 4 x 4 ln 2 3 Section 8.4 Trigonometric Substitution 153

x2 1 55. dx x2 10x 9 x 15 33 lnx 5 x2 10x 9 C x2 10x 9 2

1 75 56. x2 2x 1132 dx x 1x2 2x 26x2 2x 11 ln x2 2x 11 x 1 C 4 2

x2 1 57. dx xx2 1 lnx x2 1 C x2 1 2

1 1 58. x2x2 4 dx x3x2 4 xx2 4 2 ln x x2 4 C 4 2

59. (a) u a sin 60. (a) Substitution: u x2 1, du 2x dx (b) u a tan (b) Trigonometric substitution: x sec (c) u a sec

61. (a) u x2 9, du 2x dx x 1 du 1 1 dx lnu C lnx2 9 C x2 9 2 u 2 2 (b) Let x 3 tan , x2 9 9 sec2 , dx 3 sec2 d. x 3 tan dx 3 sec2 d tan d 2 2 x 9 9 sec x2 + 9 x lncos C1 θ 3 ln C 3 x2 9 1 2 ln 3 ln x 9 C1 1 lnx2 9 C 2 2 The answers are equivalent.

x2 x2 9 9 9 x 62. (a) dx dx 1 dx x 3 arctan C x2 9 x2 9 x2 9 3 (b) Let x 3 tan , x2 9 9 sec2 , dx 3 sec2 d. x2 9 tan2 dx 3 sec2 d x2 9 9 sec2

3tan2 d 3sec2 1 d

3 tan 3 C1 x x 3 arctan C 3 1 The answers are equivalent.

63. True 64. False dx cos d x2 1 tan d dx sec tan d tan2 d 1 x2 cos x sec 154 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

65. False 3 dx 3 sec2 d 3 cos d 2 3 3 0 1 x 0 sec 0

66. True

1 1 2 2 x21 x2 dx 2 x21 x2 dx 2 sin2 cos cos d 2 sin2 cos2 d 1 0 0 0

a b 67. A 4 a2 x2 dx 68. x2 y2 a2 0 a x ±a2 y2 4b a a2 x2 dx a a 0 A 2 a2 y2 dy 4b 1 x a h 2 2 2 a arcsin x a x a a 2 a 0 y a2 arcsin ya2 y2 (Theorem 8.2) a 2b h a2 ab a 2 h a2 a2 arcsin ha2 h2 2 a Note: See Theorem 8.2 for a2 x2 dx. a2 h y 2 2 2 b 22− a arcsin h a h yax= a 2 a

x −aa

−b

1 69. (a) x2 y k2 25 y (b) Area square circle 4 Radius of circle 5 1 25 52 251 k2 52 52 50 (0,k ) 4 4 k 52 5 5 1 (c) Area r 2 r 2 r 21 4 4 5 x

70. (a) Place the center of the circle at 0, 1; x2 y 12 1. The depth d satisfies 0 ≤ d ≤ 2. The volume is d 1 d V 3 2 1 y 12 dy 6 arcsin y 1 y 11 y 12 (Theorem 8.2 (1)) 0 2 0 3arcsind 1 d 11 d 12 arcsin1 3 3 arcsind 1 3d 12d d 2. 2

(b) 10 (c) The full tank holds 3 9.4248 cubic meters. The horizontal lines 3 3 9 y , y , y 4 2 4 02 0 intersect the curve at d 0.596, 1. 0, 1.404. The dipstick would have these markings on it. —CONTINUED— Section 8.4 Trigonometric Substitution 155

70. —CONTINUED—

d 2 (d) V 6 1 y 1 dy (e)0.3 The minimum occurs at d 1, 0 which is the widest part of the dV dV dd tank. 61 d 12 dt dt dd dt

1 1 02 ⇒ dt 0 4 241 d 12

71. Let x 3 sin , dx cos d, 1 x 32 cos . y Shell Method: 2

4 1 2 V 4 x 1 x 3 dx x 2 13 2 −1 2 4 3 sin cos d −2 2 3 2 2 4 1 cos 2 d cos2 sin d 2 2 2 3 1 1 2 4 sin 2 cos3 6 2 2 2 3 2

72. Let x h r sin , dx r cos d, r 2 x h2 r cos . y Shell Method:

hr 2 2 V 4 x r x h dx x hr hr− hhr+ 2 2 4 h r sin r cos r cos d 4r 2 h r sin cos2 d 2 2 h 2 2 4r 2 1 cos 2 d r sin cos2 d 2 2 2 r xh− 2 3 2 2 1 3 cos 2 2 2r h sin 2 4r 2 r h θ 2 2 3 2 rxh22− () −

1 1 x2 1 73. y ln x, y , 1 y2 1 x x2 x2 Let x tan , dx sec2 d, x2 1 sec . x2 + 1 5 2 5 2 x 1 x 1 x s 2 dx dx 1 x 1 x θ b sec b sec 2 2 1 sec d 1 tan d a tan a tan b b csc sec tan d lncsc cot sec a a 5 x2 1 1 ln x2 1 x x 1 26 1 ln 26 ln2 1 2 5 52 1 26 1 ln 26 2 4.367 or ln 26 2 26 1 52 1 156 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 74. y x2, y x, 1 y2 1 x2 2 4 1 4 s 1 x2 dx xx2 1 lnx x2 1 (Theorem 8.2) 0 2 0 1 417 ln4 17 9.2936 2

75. Length of one arch of sine curve: y sin x, y cos x 2 L1 1 cos x dx 0 Length of one arch of cosine curve: y cos x, y sin x

2 2 L2 1 sin x dx 2 2 1 cos2x dx, u x , du dx 2 2 2 0 1 cos2 u du 2 1 cos u du L1 0

2 4 2 y 76. (a) Along line: d1 a a a 1 a

Along parabola: y x2, y 2x ()a, a2

a 2 d2 1 4x dx 0 2 1 a y = x 2x4x2 1 ln2x 4x2 1 (Theorem 8.2) x 4 0 (0, 0) 1 2a4a2 1 ln2a 4a2 1 4 5 1 (b) For a 1, d 2 and d ln2 5 1.4789. 1 2 2 4 For a 10, d1 10 101 100.4988, d2 101.0473. → (c) As a increases, d2 d1 0.

77. (a) 60 (c) y x 0.005x 2, y 1 0.01x, 1 y2 1 1 0.01x2 Let u 1 0.01x, du 0.01 dx, a 1. (See Theorem 8.2.)

200 200 2 2 − 25 250 s 1 1 0.01x dx 100 1 0.01x 1 0.01 dx 0 0 − 10 200 501 0.01x1 0.01x2 1 ln1 0.01x 1 0.01x2 1 (b)y 0 for x 200 (range) 0 502 ln1 2 2 ln1 2 2 1 1002 50 ln 229.559 2 1 Section 8.4 Trigonometric Substitution 157

78. (a) 25

− 10 80

−5

(b)y 0 for x 72

x2 x x 2 (c) y x , y 1 , 1 y 2 1 1 72 36 36

72 72 x 2 x 2 1 s 1 1 dx 36 1 1 dx 0 36 0 36 36 72 36 x x 2 x x 2 1 1 1 ln1 1 1 2 36 36 36 36 0 2 1 182 ln1 2 2 ln1 2 362 18 ln 82.641 2 1

79. Let x 3 tan , dx 3 sec2 d, x2 9 3 sec . y 4 3 4 dx b 3 sec2 d A 2 dx 6 6 2 2 3 0 x 9 0 x 9 a 3 sec 4 b 4 1 b x2 9 x 2 6 sec d 6 lnsec tan 6 ln 6 ln 3 (0, 0.422) 1 a a 3 0 4 x x 0 (by symmetry) −4 −224 1 1 4 3 2 9 4 1 3 1 x 4 2 4 y dx dx arctan arctan 0.422 2 2 2 A 4 x 9 12 ln 3 4 x 9 4 ln 3 3 3 4 4 ln 3 3 1 4 x, y 0, arctan 0, 0.422 2 ln 3 3

80. First find where the curves intersect. y

6 1 (4, 4) y 2 16 x 42 x4 4 16 2 2 2 4 x 16 16 x 4 x −224610 −2 2 2 2 4 16 16x 128x 16 x −4 −6 yx= 16−− ( 4)2 x4 16x2 128x 0 xx 4x2 4x 32 ⇒ x 0, 4 4 1 1 1 4 16 A x2 dx 42 x3 4 4 0 4 4 12 0 3 4 1 8 2 2 My x x dx x 16 x 4 dx 0 4 4 x4 4 8 8 x 416 x 42 dx 416 x 42 dx 16 0 4 4 1 8 x 4 8 16 16 x 4232 216 arcsin x 416 x 42 3 4 4 4 1 64 112 16 1632 216 16 16 16 3 2 3 3

—CONTINUED— 158 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

80. —CONTINUED— 4 1 1 2 8 1 2 2 Mx x dx 16 x 4 dx 0 2 4 4 2 1 x5 4 x 43 8 8x 32 5 0 6 4 32 64 416 64 32 5 6 15

My 1123 16 112 48 28 12 x 4.89 A 163 4 16 12 4 3 M 41615 104 y x 1.55 A 163 4 54 3 x, y 4.89, 1.55

81. y x2, y 2x, 1 y 1 4x2 1 2x tan , dx sec2 d, 1 4x2 sec 2 (For sec5 d and sec3 d, see Exercise 98 in Section 8.3.)

2 b tan 2 1 S 2 x21 4x2 dx 2 sec sec2 d 0 a 2 2 b b b sec3 tan2 d sec5 d sec3 d 4 a 4 a a 1 3 1 b sec3 tan sec tan lnsec tan sec tan lnsec tan 4 4 2 2 a 1 1 2 1 4x2322x 1 4x2122x ln1 4x2 2x 4 4 8 0 542 62 1 ln3 22 4 4 8 8 512 ln3 22 1022 ln3 22 13.989 4 4 8 32

82. Let r L tan , dr L sec2 d, r 2 L2 L2 sec2 .

2 2 1 R 2mL 2mL b L sec2 d rL+ dr r 2 232 3 3 R 0 r L R a L sec θ 2m b cos d L RL a 2m b sin RL a

2m r R 2 2 RL r L 0 2m LR 2 L2 Section 8.4 Trigonometric Substitution 159

83. (a) Area of representative rectangle: 21 y2 y y Force: 262.43 y1 y 2 y xy= 1 − 2 1 2 F 124.8 3 y1 y2 dy 1 1 1 x −2 2 124.83 1 y2 dy y1 y 2 dy 1 1 3 1 2 1 124.8 arcsin y y1 y2 1 y232 2 2 3 1 62.43arcsin 1 arcsin1 187.2 lb

1 1 1 (b) F 124.8 d y1 y 2 dy 124.8d 1 y 2 dy 124.8 y1 y 2 dy 1 1 1 d 1 124.8 arcsin y y1 y2 124.80 62.4d lb 2 1

0.8 2 84. (a) Finside 48 0.8 y 2 1 y dy 1 0.8 0.8 960.8 1 y2 dy y1 y 2 dy 1 1 0.8 1 0.8 96 arcsin y y1 y2 1 y 232 961.263 121.3 lbs 2 3 1

0.4 2 (b) Foutside 64 0.4 y 2 1 y dy 1 0.4 0.4 0.4 1 0.4 1280.4 1 y 2 dy y1 y 2 dy 128 arcsin y y1 y2 1 y232 92.98 1 1 2 3 1

85. Let u a sin , du a cos d, a2 u2 a cos . 1 cos 2 a2 u2 du a2 cos2 d a2 d 2 a2 1 a2 sin 2 C sin cos C 2 2 2 a 2 u u a2 u2 1 u arcsin C a2 arcsin ua2 u2 C 2 a a a 2 a Let u a sec , du a sec tan d, u2 a2 a tan .

u2 a2 du a tan a sec tan d a2tan2 sec d

a 2sec2 1 sec d a2sec3 sec d

1 1 1 1 a2 sec tan sec d a2sec d a2 sec tan lnsec tan 2 2 2 2 a2 u u2 a2 u u2 a2 ln C 2 a a a a 1 1 uu2 a2 a2 lnu u2 a2 C 2

—CONTINUED— 160 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

85. —CONTINUED— Let u a tan , du a sec2 d, u2 a2 a sec .

u2 a2 du a sec a sec2 d

1 1 a2sec3 d a2 sec tan lnsec tan C 2 2 1 a2 u2 a2 u u2 a2 u 1 ln C uu2 a2 a2 lnu u2 a2 C 2 a a a a 1 2

86. y sin x on 0, 2 y cos x 2 s1 2 1 cos x dx 3.820197789 0 Ellipse: x2 2y2 2 1 Upper half: y 1 x2, 2 ≤ x ≤ 2 2 x y 21 12x2 2 x2 2 x2 s 2 1 dx 2 1 dx 2 2 2 2 4 1 1 2 x 2 4 2x Let x 2 sin , dx 2 cos d, x2 2 sin2 , 4 2x2 4 4 sin2 4 cos2 . 2 2 sin2 s 2 1 2 cos d 2 2 2 4 cos 2 2 2 4 cos 2 sin 2 2 cos d 2 2 cos 2 2 2 cos2 2 d 2 2 2 2 1 cos2 d 2 2 2 1 cos d s1 0

87. Large circle: x2 y2 25 y

y 25 x2, upper half 6 4 (3, 4) From the right triangle, the center of the small circle is 0, 4 . 2 2 2 x x y 4 9 -6 −4 −2 246 −2 2 y 4 9 x , upper half −4 3 2 2 A 2 4 9 x 25 x dx 0 1 x 1 x 3 24x 9 arcsin x9 x2 25 arcsin x25 x2 2 3 2 5 0 9 25 3 212 arcsin1 arcsin 6 2 2 5 9 3 12 25 arcsin 10.050 2 5 Section 8.5 Partial Fractions 161

Section 8.5 Partial Fractions

5 5 A B 4x 2 3 A B C 1. 2. x 2 10x xx 10 x x 10 x 53 x 5 x 52 x 53

2x 3 2x 3 A Bx C x 2 x 2 A B 3. 4. x3 10x xx 2 10 x x 2 10 x 2 4x 3 x 1x 3 x 1 x 3

16 A B 2x 1 A Bx C Dx E 5. 6. xx 10 x x 10 xx 2 12 x x 2 1 x 2 12

1 1 A B 1 1 A B 7. 8. x 2 1 x 1x 1 x 1 x 1 4x 2 9 2x 32x 3 2x 3 2x 3 1 Ax 1 Bx 1 1 A2x 3 B2x 3 1 3 1 When x 1, 1 2A, A 2. When x 2, 1 6A, A 6. 1 3 1 When x 1, 1 2B, B 2. When x 2, 1 6B, B 6. 1 1 1 1 1 1 1 1 1 dx dx dx dx dx dx x 2 1 2 x 1 2 x 1 4x2 9 6 2x 3 2x 3 1 1 1 lnx 1 lnx 1 C ln2x 3 ln2x 3 C 2 2 12 1 x 1 1 2x 3 ln C ln C 2 x 1 12 2x 3

3 3 A B x 1 x 1 9. 10. dx dx x 2 x 2 x 1x 2 x 1 x 2 x2 4x 3 x 1x 3 3 Ax 2 Bx 1 1 dx lnx 3 C x 3 When x 1, 3 3A, A 1. When x 2, 3 3B, B 1. 3 1 1 dx dx dx x 2 x 2 x 1 x 2 lnx 1 lnx 2 C x 1 ln C x 2

5 x 5 x A B 5x2 12x 12 A B C 11. 12. 2x 2 x 1 2x 1x 1 2x 1 x 1 xx 2x 2 x x 2 x 2 5 x Ax 1 B2x 1 5x2 12x 12 Ax2 4 Bxx 2 Cxx 2 1 9 3 When x 2, 2 2A, A 3. When x 0, 12 4A ⇒ A 3. When x 2, 16 8B ⇒ B 2. When x 2, When x 1, 6 3B, B 2. 32 8C ⇒ C 4. 5 x 1 1 dx 3 dx 2 dx 5x2 12x 12 2x 2 x 1 2x 1 x 1 dx x3 4x 3 ln2x 1 2 lnx 1 C 3 2 4 2 dx dx dx x x 2 x 2 3 lnx 2 lnx 2 4 lnx 2 C 162 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

x 2 12x 12 A B C 13. xx 2x 2 x x 2 x 2 x 2 12x 12 Ax 2x 2 Bxx 2 Cxx 2 When x 0, 12 4A, A 3. When x 2, 8 8B, B 1. When x 2, 40 8C, C 5. x 2 12x 12 1 1 1 dx 5 dx dx 3 dx x3 4x x 2 x 2 x 5 lnx 2 lnx 2 3 lnx C

x3 x 3 2x 1 A B 14. x 1 x 1 x 2 x 2 x 2x 1 x 2 x 1 2 x 1 Ax 1 Bx 2 When x 2, 3 3A, A 1. When x 1, 3 3B, B 1. x3 x 3 1 1 dx x 1 dx x 2 x 2 x 2 x 1 x 2 x2 x lnx 2 lnx 1 C x lnx 2 x 2 C 2 2

2x3 4x 2 15x 5 x 5 A B 15. 2x 2x x 2 2x 8 x 4x 2 x 4 x 2 x 5 Ax 2 Bx 4 3 1 When x 4, 9 6A, A 2. When x 2, 3 6B, B 2. 2x3 4x 2 15x 5 32 12 dx 2x dx x2 2x 8 x 4 x 2 3 1 x 2 lnx 4 lnx 2 C 2 2

x 2 A B 4x 2 2x 1 A B C 16. 17. xx 4 x 4 x x 2x 1 x x 2 x 1 x 2 Ax Bx 4 4 x 2 2x 1 Axx 1 Bx 1 Cx 2 3 When x 4, 6 4A, A 2. When x 0, B 1. When x 1, C 1. When 1 x 1, A 3. When x 0, 2 4B, B 2. 4x 2 2x 1 3 1 1 x 2 32 12 dx dx dx dx x3 x 2 x x 2 x 1 x 2 4x x 4 x 1 3 1 3 lnx lnx 1 C lnx 4 lnx C x 2 2 1 lnx 4 x3 C x

2x 3 A B 18. x 12 x 1 x 12 2 x 3 Ax 1 B When x 1, B 1. When x 0, A 2. 2x 3 2 1 1 dx dx 2 lnx 1 C x 12 x 1 x 12 x 1 Section 8.5 Partial Fractions 163

x2 3x 4 x2 3x 4 A B C 19. x3 4x2 4x xx 22 x x 2 x 22 x2 3x 4 Ax 22 Bxx 2 Cx When x 0, 4 4A ⇒ A 1. When x 2, 6 2C ⇒ C 3. When x 1, 0 1 B 3 ⇒ B 2. x2 3x 4 1 2 3 dx dx dx dx x3 4x2 4x x x 2 x 22 3 lnx 2 lnx 2 C x 2

4x2 4x2 4x2 A B C 20. x3 x2 x 1 x2x 1 x 1 x2 1x 1 x 1 x 1 x 12 4 x2 Ax 12 Bx 1x 1 Cx 1 When x 1, 4 2C ⇒ C 2. When x 1, 4 4A ⇒ A 1. When x 0, 0 1 B 2 ⇒ B 3. 4x2 1 3 2 dx dx dx dx x3 x2 x 1 x 1 x 1 x 12 2 lnx 1 3 lnx 1 C x 1

x 2 1 A Bx C 21. xx 2 1 x x 2 1 x 2 1 Ax 2 1 Bx Cx When x 0, A 1. When x 1, 0 2 B C. When x 1, 0 2 B C. Solving these equations we have A 1, B 2, C 0. x 2 1 1 2x dx dx dx x3 x x x 2 1 lnx lnx2 1 C x 2 1 ln C x

6x 6x A Bx C 22. x3 8 x 2x2 2x 4 x 2 x2 2x 4 6 x Ax2 2x 4 Bx Cx 2 When x 2, 12 12A ⇒ A 1. When x 0, 0 4 2C ⇒ C 2. When x 1, 6 7 B 21 ⇒ B 1. 6x 1 x 2 dx dx dx x3 8 x 2 x2 2x 4 1 x 1 3 dx dx dx x 2 x2 2x 4 x2 2x 1 3 1 3 x 1 lnx 2 lnx2 2x 4 arctan C 2 3 3 1 3x 1 lnx 2 lnx2 2x 4 3 arctan C 2 3 164 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

x 2 A B Cx D 23. x 4 2x 2 8 x 2 x 2 x 2 2 x 2 Ax 2x 2 2 Bx 2x 2 2 Cx Dx 2x 2 When x 2, 4 24A. When x 2, 4 24B. When x 0, 0 4A 4B 4D, and when x 1, 1 1 1 1 9A 3B 3C 3D. Solving these equations we have A 6, B 6, C 0, D 3. x 2 1 1 1 1 dx dx dx 2 dx 4 2 2 x 2x 8 6 x 2 x 2 x 2

1 x 2 x ln 2 arctan C 6 x 2 2

x2 x 9 Ax B Cx D 24. x2 92 x2 9 x2 92 x2 x 9 Ax Bx2 9 Cx D Ax3 Bx2 9A Cx 9B D By equating coefficients of like terms, we have A 0, B 1, D 0, and C 1. x2 x 9 1 x 1 x 1 dx dx arctan C x2 92 x2 9 x2 92 3 3 2x2 9

x A B Cx D 25. 2x 12x 14x 2 1 2x 1 2x 1 4x 2 1 x A2x 14x 2 1 B2x 14x 2 1 Cx D2x 12x 1 1 1 1 1 When x 2, 2 4A. When x 2, 2 4B. When x 0, 0 A B D, and when x 1, 1 1 1 1 15A 5B 3C 3D. Solving these equations we have A 8, B 8, C 2, D 0. x 1 1 1 x 1 4x2 1 dx dx dx 4 dx ln C 16x 4 1 8 2x 1 2x 1 4x 2 1 16 4x2 1

x 2 4x 7 A Bx C 26. x 1x 2 2x 3 x 1 x 2 2x 3 x 2 4x 7 Ax 2 2x 3 Bx Cx 1 When x 1, 12 6A. When x 0, 7 3A C. When x 1, 4 2A 2B 2C. Solving these equations we have A 2, B 1, C 1. x 2 4x 7 1 x 1 dx 2 dx dx x3 x 2 x 3 x 1 x 2 2x 3 1 2 lnx 1 lnx 2 2x 3 C 2

x 2 5 A Bx C 27. x 1x 2 2x 3 x 1 x 2 2x 3 x 2 5 Ax 2 2x 3 Bx Cx 1 A Bx 2 2A B Cx 3A C When x 1, A 1. By equating coefficients of like terms, we have A B 1, 2A B C 0, 3A C 5. Solving these equations we have A 1, B 0, C 2. x 2 5 1 1 dx dx 2 dx x3 x 2 x 3 x 1 x 12 2 x 1 lnx 1 2 arctan C 2 Section 8.5 Partial Fractions 165

x 2 x 3 Ax B Cx D 3 A B 28. 29. x 2 32 x 2 3 x 2 32 2x 1x 2 2x 1 x 2 x 2 x 3 Ax Bx 2 3 Cx D 3 Ax 2 B2x 1

3 2 1 Ax Bx 3A Cx 3B D When x 2, A 2. When x 2, B 1. By equating coefficients of like terms, we have A 0, 1 3 1 2 1 1 dx dx dx 2 B 1, 3A C 1, 3B D 3. Solving these equa- 0 2x 5x 2 0 2x 1 0 x 2 tions we have A 0, B 1, C 1, D 0. 1 ln2x 1 lnx 2 x 2 x 3 1 x 0 dx dx x4 6x 2 9 x 2 3 x 2 32 ln 2 1 x 1 arctan C 3 3 2x 2 3

x 1 A B C 30. x 2x 1 x x 2 x 1 x 1 Axx 1 Bx 1 Cx 2 When x 0, B 1. When x 1, C 2. When x 1, 0 2A 2B C. Solving these equations we have A 2, B 1, C 2. 5 x 1 5 1 5 1 5 1 dx 2 dx dx 2 dx 2 2 1 x x 1 1 x 1 x 1 x 1 1 5 2 lnx 2 lnx 1 x 1

5 x 1 2 ln x 1 x 1 5 4 2 ln 3 5

x 1 A Bx C 1 x 2 x 1 1 2x 1 31. 32. dx dx dx 2 2 2 2 x x 1 x x 1 0 x x 1 0 0 x x 1 x 1 Ax 2 1 x Cx 1 x lnx 2 x 1 0 When x 0, A 1. When x 1, 2 2A B C. When x 1, 0 2A B C. Solving these equations 1 ln 3 we have A 1, B 1, C 1. 2 x 1 2 1 2 x 2 1 dx dx dx dx 2 2 2 1 x x 1 1 x 1 x 1 1 x 1 1 2 lnx lnx 2 1 arctan x 2 1 1 8 ln arctan 2 2 5 4 0.557

3x dx 9 33. 3 lnx 3 C 30 x 2 6x 9 x 3

9 ⇒ 4, 0 : 3 ln4 3 C 0 C 9 4 3 −610 (4, 0)

−10 166 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

6x 2 1 x 1 1 2 7 34. dx 3 ln C 4 x 2x 13 x x x 1 2x 12

1 1 2 7 1 (2, 1) 2, 1: 3 ln C 1⇒ C 2 3 ln − 4.7 4.7 2 2 1 2 2

−4

x 2 x 2 2 x 1 x 3 1 2 35. dx arctan C 36. dx lnx 2 4 C x 2 22 2 2 2x 2 2 x 2 42 2 x 2 4 1 5 1 2 22 1 0, 1: 0 C 1 ⇒ C 3, 4: ln 5 C 4 ⇒ C ln 5 4 4 2 5 5 2

3 8

(0, 1) (3, 4)

− 33 − 88

−1 −2

2x 2 2x 3 1 2x 1 20 37. dx lnx 2 lnx 2 x 1 3 arctan C 3 2 x x x 2 2 3 (3, 10) 1 7 1 7 3, 10: 0 ln 13 3 arctan C 10 ⇒ C 10 ln 13 3 arctan − 2 3 2 3 2 6 −5

x2x 9 1 5 10 38. dx 2 lnx 2 C x 3 6x 2 12x 8 x 2 x 22

3, 2: 0 1 5 C 2 ⇒ C 4 (3, 2) − 10 10

−3

1 1 x 2 x 2 x 2 39. dx ln C 40. dx arctan x lnx 1 C x 2 4 4 x 2 x 3 x 2 x 1 1 4 1 1 1 2, 6: arctan 2 0 C 6 ⇒ C 6 arctan 2 6, 4: ln C 4 ⇒ C 4 ln 4 ln 2 4 8 4 2 4 10

(2, 6)

(6, 4) −2 5 − 10 10 −2

−3

41. Let u cos x du sin x dx. 1 A B uu 1 u u 1 1 Au 1 Bu When u 0, A 1. When u 1, B 1, u cos x, du sin x dx. sin x 1 dx du cos xcos x 1 uu 1 1 1 u cos x du du lnu lnu 1 C ln C ln C u u 1 u 1 cos x 1 Section 8.5 Partial Fractions 167

42. Let u cos x, du sin x dx. 3 cos x 1 43. dx 3 du 2 2 1 A B sin x sin x 2 u u 2 uu 1 u u 1 u 1 ln C 1 Au 1 Bu u 2 When u 0, A 1. When u 1, B 1, u cos x, 1 sin x ln C du sin dx. 2 sin x sin x 1 (From Exercise 9 with u sin x, du cos x dx ) dx du cos x cos2 x uu 1 1 1 du du u 1 u lnu 1 lnu C u 1 ln C u cos x 1 ln C cos x ln1 sec x C

1 A B 44. , u tan x, du sec2 x dx 45. Let u ex, du ex dx. uu 1 u u 1 1 A B 1 Au 1 Bu u 1u 4 u 1 u 4 When u 0, A 1. 1 Au 4 Bu 1 When u 1, 1 B ⇒ B 1. When u 1, A 1. When u 4, B 1, u ex, sec2 x dx 1 5 5 du du ex dx. tan xtan x 1 uu 1 ex 1 1 1 dx du du ex 1ex 4 u 1u 4 u u 1 1 1 1 lnu lnu 1 C du du 5 u 1 u 4 u ln C 1 u 1 u 1 ln C 5 u 4 tan x ln C 1 ex 1 tan x 1 ln C 5 ex 4

46. Let u ex, du ex dx. 1 A Bu C u2 1u 1 u 1 u2 1 1 Au2 1 Bu Cu 1 1 When u 1, A 2. When u 0, 1 A C. When u 1, 1 2A 2B 2C. 1 1 1 x x Solving these equations we have A 2, B 2, C 2, u e , du e dx. e x 1 dx du e 2x 1e x 1 u2 1u 1 1 1 u 1 du du 2 u 1 u2 1 1 1 lnu 1 lnu2 1 arctan u C 2 2 1 2 lne x 1 lne 2x 1 2 arctan e x C 4 168 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 A B 1 A B 47. 48. xa bx x a bx a2 x2 a x a x 1 Aa bx Bx 1 Aa x Ba x When x 0, 1 aA ⇒ A 1a. When x a, A 12a. When x ab, 1 abB ⇒ B ba. When x a, B 12a. 1 1 1 b 1 1 1 1 dx dx dx dx xa bx a x a bx a2 x2 2a a x a x 1 1 lnx lna bx C lna x lna x C a 2a 1 x 1 a x ln C ln C a a bx 2a a x

x A B 1 A B C 49. 50. a bx2 a bx a bx2 x2a bx x x2 a bx x Aa bx B 1 Axa bx Ba bx Cx2 When x ab, B ab. When x 0, 1 Ba ⇒ B 1a. When x ab, When x 0, 0 aA B ⇒ A 1b. 1 Ca2b2 ⇒ C b2a2. When x 1, 1 a bA a bB C ⇒ A ba 2. x 1b ab dx dx a bx2 a bx a bx2 1 ba2 1a b2a2 dx dx x2a bx x x2 a bx 1 1 a 1 dx dx b a bx b a bx2 b 1 b lnx lna bx C a2 ax a2 1 a 1 lna bx C b2 b2 a bx 1 b a bx ln C ax a2 x 1 a lna bx C b2 a bx 1 b x ln C ax a2 a bx

dy 6 dy 4 51. , y0 3 52. , y0 5 53. Dividing x3 by x 5 dx 4 x2 dx x2 2x 3

10 8

−2 2 −1 3

−4 −2

Nx A A A 54. (a) 1 2 . . . m 55. (a) Substitution: u x2 2x 8 Dx px q px q2 px qm Nx A B x A B x (b) Partial fractions (b) 1 1 . . . n n Dx ax2 bx c ax2 bx cn (c) Trigonometric substitution (tan) or inverse tangent rule Section 8.5 Partial Fractions 169

3 10 1 12 56. A dx 3, matches (c). 57. A dx 2 2 1 x x 1 0 x 5x 6

y 12 12 A B 2 5 x 5x 6 x 2x 3 x 2 x 3

4 12 Ax 3 Bx 2 3

2 Let x 3: 12 B1 ⇒ B 12 1 Let x 2: 12 A1 ⇒ A 12 x 12345 1 12 12 A dx 0 x 2 x 3 1 12 lnx 2 12 lnx 3 0 12ln 3 ln 4 ln 2 ln 3 9 12 ln 1.4134 8

3 7 3 3 1 1 80 124p 58. A 2 1 dx 2 dx 14 dx 59. Average cost dp 2 2 0 16 x 0 0 16 x 80 75 75 10 p 100 p 3 80 14 4 x 1 124 1240 2x ln (From Exercise 48) dp 8 4 x 0 5 75 10 p 11 100 p 11 y 7 1 124 1240 80 6 ln 7 2.595 5 ln10 p ln100 p 2 4 5 11 11 75 2 3 2 1 24.51 4.9 5

x Approximately $490,000 −3 −2 −1 1 2 3

dy A B 60. (a) y (d) yL y y L y 6 5 1 1 1 AL y By ⇒ A , B 4 L L 3 2 dy k dt 1 yL y t 1 2 3 4 5 6 1 1 1 dy dy k dt (b) The slope is negative because the function is decreasing. L y L y 1 (c) For y > 0, lim yt 3. lny lnL y kt C t → L 1 (e) k 1, L 3 y ln kLt LC 15 L y 1 i y0 5: y 5 5 2e3t y kLt 1 C2e ii y0 : L y 2 y y y 0 ⇒ 0 kLt 32 05When t 0, C2 e . y 0 L y0 L y L y0 12 52e3t y L Solving for y,you obtain y 0 . 3 kLt y0 L y0 e 1 5e3t

—CONTINUED— 170 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

60. —CONTINUED— dy (f) kyL y dt d 2y dy dy ky L y 0 dt2 dt dt dy dy ⇒ y L y dt dt L ⇒ y 2 From the first derivative test, this is a maximum.

3 2x 2 3 x2 61. V dx 4 dx 2 2 2 0 x 1 0 x 1 3 1 1 4 dx (partial fractions) 2 2 2 0 x 1 x 1

3 1 x 4 arctan x arctan x 2 (trigonometric substitution) 2 x 1 0

3 x 3 2 arctan x 2 2 arctan 3 5.963 x 1 0 10 3 2x 3 A dx lnx2 1 ln 10 y 2 0 x 1 0 2 1 3 2x2 1 3 2 x dx 2 dx (1.521, 0.412) 2 2 A 0 x 1 ln 10 0 x 1 1

3 1 2 x 2x 2 arctan x 3 arctan 3 1.521 123 ln 10 0 ln 10 −1 1 1 3 2x 2 2 3 x2 y dx dx 2 2 2 A 2 0 x 1 ln 10 0 x 1 2 3 1 1 dx (partial fractions) 2 2 2 ln 10 0 x 1 x 1

3 2 1 x arctan x arctan x 2 (trigonometric substitution) ln 10 2 x 1 0

3 3 2 1 x 1 x 1 3 arctan x 2 arctan x 2 arctan 3 0.412 ln 10 2 2x 1 0 ln 10 x 1 0 ln 10 10 x, y 1.521, 0.412

2 x2 62. y2 , 0, 1 y 1 x2 2 − x 2 y = 1 + x 1 2 x2 V dx 2 0 1 x x − 1 4 1 4x 1 x2 2 2 dx dx dx 2 2 2 0 1 x 0 1 x 0 1 x −2 3 2 4 ln 2 2 2 ln 2 2 11 6 ln 2 11 12 ln 2 2 2 Section 8.5 Partial Fractions 171

1 A B 1 63. , A B x 1n x x 1 n x n 1 1 1 1 dx kt C n 1 x 1 n x 1 x 1 ln kt C n 1 n x 1 1 When t 0, x 0, C ln . n 1 n 1 x 1 1 1 ln kt ln n 1 n x n 1 n 1 x 1 1 ln ln kt n 1 n x n nx n ln n 1kt n x nx n en1kt n x nen1kt 1 x Note: lim x n n e n 1 kt t →

1 A B 64. (a) , (b) (1) If y0 < z0, lim x y0. t → y0 x z0 x y0 x z0 x (2) If y0 > z0, lim x z0. t → 1 1 A , B , (Assume y0 z0.) z0 y0 z0 y0 (3) If y0 z0, then the original equation is: 1 1 1 1 dx k dt dx kt C 2 z0 y0 y0 x z0 x y0 x 1 z x y x1 kt C ln 0 kt C, when t 0, x 0 0 1 z y y x 0 0 0 1 x 0 when t 0 ⇒ C 1 z y 1 C ln 0 0 z y y 0 0 0 1 1 kty 1 kt 0 z x z y x y y 1 0 0 0 0 0 ln ln kt z0 y0 y0 x y0 y y x 0 y z x 0 kty 1 0 0 0 ln z0 y0 kt z0 y0 x y 0 x y0 y0 z0 x kty0 1 z0 y0 kt e z0 y0 x → → As t , x y0 x0. z0 y0 kt y0z0 e 1 x z0 y0 kt z0e y0 172 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

x Ax B Cx D 65. 1 x4 x2 2 x 1 x2 2 x 1 x Ax Bx2 2 x 1 Cx Dx2 2 x 1 A Cx3 B D 2 A 2 Cx2 A C 2 B 2 Dx B D 0 A C ⇒ C A 0 B D 2 A 2 C 22 A 0 ⇒ A 0 and C 0 2 2 1 A C 2 B 2 D 22 B 1 ⇒ B and D 4 4 0 B D ⇒ D B Thus, 1 1 x 2 4 2 4 4 dx dx 2 2 0 1 x 0 x 2 x 1 x 2 x 1 1 2 1 1 2 2 dx 4 0 x 22 12 x 22 12

2 1 x 22 x 22 1 arctan arctan 4 1 2 1 2 1 2 0 1 1 arctan2 x 1 arctan2 x 1 2 0 1 arctan2 1 arctan2 1 arctan 1 arctan1 2 1 arctan2 1 arctan2 1 . 2 4 4 Since arctan x arctan y arctanx y1 xy, we have: 1 x 1 2 1 2 1 dx arctan 1 2 1 4 arctan 0 1 x 2 1 2 12 1 2 2 2 2 2 4 2 8

66. The partial fraction decomposition is: x 41 x4 4 x 6 4x 5 5x 4 4x2 4 1 x2 1 x2 1 x 41 x4 x7 2x6 4 1 dx x5 x3 4x 4 arctan x 2 0 1 x 7 3 3 0 1 2 4 1 4 4 7 3 3 4 22 7 Note: You can easily verify this calculation with a graphing utility. Section 8.6 Integration by Tables and Other Integration Techniques 173

Section 8.6 Integration by Tables and Other Integration Techniques

x 2 x 1. By Formula 6: dx 2 x ln1 x C 1 x 2

2. By Formula 13: b 2, a 5 2 1 2 1 5 4x 4 x dx ln C 3 x22x 52 3 25 x5 2x 5 2x 5 8 x 2 4x 5 ln C 375 2x 5 75 x2x 5

1 3. By Formula 26: e x1 e 2x dx e xe 2x 1 lne x e 2x 1 C 2 u ex, du ex dx

1 1 x 2 4. By Formula 29: a 3 5. By Formula 44: dx C x 21 x 2 x 1 x2 9 1 x dx x2 9 arcsec C 3 x 3 3

x 1 2x 6. By Formula 41: dx dx 9 x 4 2 32 x 22 1 x2 arcsin C 2 3

1 7. By Formulas 50 and 48: sin42x dx sin42x2 dx 2 1 sin32x cos2x 3 sin22x2 dx 2 4 4 1 sin32x cos2x 3 2x sin 2x cos 2x C 2 4 8 1 6x 3 sin 2x cos 2x 2 sin3 2x cos 2x C 16

cos3 x 1 8. By Formulas 51 and 47: dx 2 cos3 x dx x 2x cos2 x sin x 2 1 2 2 cos x dx sinxcos2 x 2 C 3 3 2x 3 1 u x, du dx 2x

1 1 1 9. By Formula 57: dx 2 dx x1 cos x 1 cos x 2x 2cot x csc x C 1 u x, du dx 2x 174 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

10. By Formula 71: 11. By Formula 84: 1 1 1 1 1 dx 5 dx dx x ln1 e 2x C 1 tan 5x 5 1 tan 5x 1 e 2x 2 1 1 u lncos u sin u C 5 2 1 5x lncos 5x sin 5x C 10 u 5x, du 5 dx

1 12. By Formula 85: a , b 2 2 ex2 1 ex2 sin 2x dx sin 2x 2 cos 2x C 14 4 2 4 1 ex2 sin 2x 2 cos 2x C 17 2

13. By Formula 89: x 4 x 3 ln x dx 4 lnx 1 C 16

14. By Formulas 90 and 91: ln x3 dx xln x3 3ln x2 dx

xln x3 3x2 2 ln x ln x2 C xln x3 3ln x2 6 ln x 6 C

15. (a) By Formulas 83 and 82: x 2ex dx x 2ex 2xex dx

2 x x x e 2 x 1 e C1 x 2ex 2xex 2ex C (b) Integration by parts: u x 2, du 2x dx, dv ex dx, v ex

x 2ex dx x 2ex 2xex dx

Parts again: u 2x, du 2 dx, dv ex dx, v ex

x 2ex dx x 2ex 2xex 2ex dx x 2ex 2xex 2ex C

x5 x5 1 16. (a) By Formula 89: x 4 ln x dx 1 4 1 ln x C x5 ln x C 52 25 5 1 x5 (b) Integration by parts: u ln x, du dx, dv x 4 dx, v x 5 x5 x5 1 x5 x5 x4 ln x dx ln x dx ln x C 5 5 x 5 25 Section 8.6 Integration by Tables and Other Integration Techniques 175

17. (a) By Formula: 12,a b 1, u x, and (b) Partial fractions: 1 1 1 1 x 1 A B C dx ln C x 2x 1 1 x 1 1 x x 2x 1 x x 2 x 1 1 Axx 1 Bx 1 Cx 2 1 x ln C x 1 x x 0: 1 B x 1: 1 C 1 x 1 ln C x x x 1: 1 2A 2 1 ⇒ A 1 1 1 1 1 dx dx x 2x 1 x x 2 x 1 1 lnx lnx 1 C x 1 x ln C x x 1

18. (a) By Formula 24:a 75, x u, and (b) Partial fractions: 1 1 x 75 1 A B dx ln C x 2 75 275 x 75 x 2 75 x 75 x 75 3 x 75 1 Ax 75 Bx 75 ln C 30 x 75 1 1 3 x 75: 1 2A75 ⇒ A 275 103 30 3 x 75: 1 2B75 ⇒ B 30 1 330 330 dx dx x 2 75 x 75 x 75 3 x 75 ln C 30 x 75

1 19. By Formula 79: x arcsecx 2 1 dx arcsecx 2 1(2x dx 2 1 x 2 1 arcsecx 2 1 lnx 2 1 x 4 2x 2 C 2 u x 2 1, du 2x dx

1 20. By Formula 79: arcsec 2x dx 2x arcsec 2x ln2x 4x 2 1 C 2 u 2x, du 2 dx

1 x 2 4 21. By Formula 35: dx C x 2x 2 4 4x

1 2 2x 2 22. By Formula 14: dx arctan C arctanx 1 C x 2 2x 2 4 2

2x x 2 1 23. By Formula 4: dx 2 dx ln1 3x C 1 3x2 1 3x2 9 1 3x 176 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

24. By Formula 56: 25. By Formula 76: 2 1 1 d 32 d ex arccos ex dx ex arccos ex 1 e2x C 1 sin 3 3 1 sin 3 1 u ex, du ex dx tan 3 sec 3 C 3

26. By Formula 71: 27. By Formula 73: ex 1 x 1 2x dx ex lncos ex sin ex C dx dx 1 tan ex 2 1 sec x 2 2 1 sec x 2 u ex, du ex dx 1 x 2 cot x 2 csc x 2 C 2

1 1 1 28. By Formula 23: dt dt arctanln t C t1 ln t2 1 ln t2 t 1 u ln t, du dt t

cos 2 1 sin 29. By Formula 14: d arctan C b2 4 < 12 4ac 3 2 sin sin2 2 2 u sin , du cos d

1 30. By Formula 27: x 22 3x2 dx 3x222 3x2 3 dx 27 1 3x18x 2 22 9x 2 4 ln3x 2 9x 2 C 827

1 3 31. By Formula 35: dx 3 dx 2 2 2 x 2 9x 3x2 2 3x2 32 9x 2 C 6x 2 9x 2 C 2x

2 3 32. By Formula 77: x arctanx32 dx arctanx32 x dx 3 2 2 x32 arctanx32 ln 1 x3 C 3

ln x 1 33. By Formula 3: dx 2 lnx 3 ln3 2 lnx C x3 2 ln x 4 1 u ln x, du dx x Section 8.6 Integration by Tables and Other Integration Techniques 177

e x e x 34. By Formula 45: dx C 1 e 2x32 1 e 2x u ex, du ex dx

x 1 2x 6 6 35. By Formulas 1, 25, and 33: dx dx x 2 6x 102 2 x 2 6x 102 1 1 x 2 6x 1022x 6 dx 3 dx 2 x 32 12 1 3 x 3 arctanx 3 C 2x 2 6x 10 2 x 2 6x 10 3x 10 3 arctanx 3 C 2x 2 6x 10 2

36. By Formula 27: 1 2x 322x 32 4 dx 2x 322x 3)2 42 dx 2 1 2x 32x 32 22x 32 4 ln2x 3 2x 32 4 C 8 u 2x 3, du 2 dx

x 1 2x 37. By Formula 31: dx dx x 4 6x 2 5 2 x 2 32 4 1 lnx 2 3 x 4 6x 2 5 C 2 u x 2 3, du 2x dx

cos x 38. By Formula 31: dx lnsin x sin2 x 1 C sin2 x 1

u sin x, du cos x dx

x3 8 sin3 2 cos d 39. dx 2 2 cos 4 x 2 x 2 8 1 cos sin d θ

4 − x 2 8sin cos2 sin d

8 cos3 8 cos C 3 4 x2 8 4 x2 3 8 C 2 3 2 1 4 x2 4 4 x2 C 3 4 x 2 x 2 8 C 3 x 2 sin , dx 2 cos d, 4 x 2 2 cos 178 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

3 x 3 x 40. dx dx 41. By Formula 8: 3 x 9 x 2 e3x ex2 dx ex dx 1 x 1 ex3 1 ex3 3 dx dx 2 2 9 x 9 x 2 1 ln1 ex C x 1 ex 21 ex2 3 arcsin 9 x 2 C 3 u ex, du ex dx

1 42. By Formula 67: 43. xex2 dx 0 tan2 tan3 d tan d By Formula 81: 2 1 1 2 1 2 1 tan2 xex dx ex e 1 lncos x C 2 2 2 0 0

44. By Formula 21: 45. By Formula 89:

3 3 3 3 3 x x 2 2 dx 2 x1 x x ln x dx 1 3 lnx 1 9 1 0 1 x 3 0 1 26 2 2 8 31 3 ln 3 9 ln 3 12 2 9 9 3 3 3

46. By Formula 52: 47. By Formula 23, and letting u sin x: 2 cos x 2 dx arctansin x x sin x dx sin x x cos x 2 1 sin x 0 0 2 2 arctan1 arctan1 2

48. By Formula 7: 4 x2 1 25 4 dx 3x 10 ln 3x 5 2 2 3x 5 27 3x 5 2 1 25 64 10 12 10 ln 7 6 25 ln 7 27 7 63 27

49. By Formulas 54 and 55:

t3 cos t dt t3 sin t 3t2 sin t dt

t3 sin t 3t2 cos t 2t cos t dt

t3 sin t 3t2 cos t 6t sin t sin t dt

t3 sin t 3t2 cos t 6t sin t 6 cos t C Thus,

2 2 t3 cos t dt t3 sin t 3t2 cos t 6t sin t 6 cos t 0 0 3 3 3 6 6 3 8 8 Section 8.6 Integration by Tables and Other Integration Techniques 179

50. By Formula 26:

1 1 1 1 3 3 x 2 dx xx 2 3 3 lnx x 2 3 2 3 ln 3 3 ln 3 1 ln 3 0 2 0 2 4

u2 1 2abu a2b2 1 A B 51. a bu2 b2 a bu2 b2 a bu a bu2 2a a2 u Aa bu B aA B bAu b b2 Equating the coefficients of like terms we have aA B a2b2 and bA 2ab. Solving these equations we have A 2ab2 and B a2b2. u2 1 2a 1 1 a2 1 1 1 2a a2 1 du du b du b du u lna bu C a bu2 b2 b2 b a bu b2 b a bu2 b2 b3 b3 a bu 1 a2 bu 2a lna bu C b3 a bu

du 2 52. Integration by parts: w un, dw nun1 du, dv , v a bu a bu b un 2un 2n du a bu un1a bu du a bu b b 2un 2n a bu a bu un1a bu du b b a bu 2un 2n aun1 bun a bu du b b a bu 2un 2na un1 un a bu du 2n du b b a bu a bu un 2 un1 Therefore,2n 1 du una bu na du and a bu b a bu

un 2 un1 una bu na du. a bu 2n 1b a bu

53. When we have u2 a2: When we have u2 a2: u a tan u a sec du a sec2 d du a sec tan d u2 a2 a2 sec2 u2 a2 a2 tan2 1 a sec2 d 1 a sec tan d du du u2 a232 a3 sec3 u2 a23 2 a3 tan3 1 1 cos 1 cos d d csc cot d a2 a2 sin2 a2 1 1 sin C csc C a2 a2 u u C C a2u2 a2 a2u2 a2

22 ua+ u u ua22−

θ θ a a 180 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 2u 54. uncos u du un sin u nun1sin u du 55. arctan u du u arctan u du 2 1 u2 w un, dv cos u du, dw nun1 du, v sin u 1 u arctan u ln1 u2 C 2 u arctan u ln1 u2 C du w arctan u, dv du, dw , v u 1 u2

1 56. ln un du uln un nln un1 u du uln un nln un1 du u

1 w ln un, dv du, dw nln un1 du, v u u

1 21 x 57. dx C 8 x321 x x

1 212 ()1, 5 , 5: C 5 ⇒ C 7 2 2 1 2 −0.5 1.5

21 x − y 7 2 x

1 58. xx 2 2x dx 2x 2 2x32 3x 1x 2 2x 3 lnx 1 x 2 2x C 6 1 15 0, 0: 3 ln1 C 0 ⇒ C 0 6

−66 (0, 0)

−15

1 1 x 3 59. dx tan 1x 3 C 2 x2 6x 102 2 x2 6x 10 1 0 3, 0: 0 C 0 ⇒ C 0 −88 2 10 (3, 0) 1 x 3 y tan1x 3 −2 2 x2 6x 10

3 2 2 2 2x x 3 2 2x x (0,2 ) 60. dx 2 2x x 2 3 ln C x 1 x 1 −2.5 1

0, 2: 2 3 ln3 2 C 2 ⇒ C 3 ln3 2

−6

1 61. d csc C 10 sin tan

π 2 (), 2 , 2: C 2 ⇒ C 2 2 2 4 2 − 2 2 − y csc 2 2 2 Section 8.6 Integration by Tables and Other Integration Techniques 181

sin 1 sin 1 sin 10 62. d ln C cos 1 sin 21 sin cos

1 sin 1 sin −88 0, 1: C 1 ⇒ y ln 1 (0, 1) − 2 1 sin cos 2

2 du 1 1 u2 sin sin 63. d , u tan 64. d d 2 3 sin 2u 2 1 cos2 1 cos 2 2 3 1 u2 arctancos C 2 du 21 u2 6u 1 du u2 3u 1 1 du 3 2 5 u 2 4 3 5 u 1 2 2 ln C 5 3 5 u 2 2 1 2u 3 5 ln C 5 2u 3 5

2 tan 3 5 1 2 ln C 5 2 tan 3 5 2

2 du 2u 2 1 1 1 u2 2 1 1 1 u2 65. d 66. d 2 2 0 1 sin cos 0 2u 1 u 0 3 2 cos 0 2 1 u 1 3 1 u2 1 u2 1 u2 1 1 du 1 1 1 u 2 du 0 2 0 5u 1 1 ln1 u 2 1 0 arctan5 u 5 0 ln 2 2 arctan 5 u tan 5 2

sin 1 2 sin cos cos 1 cos 67. d d 68. d d 3 2 cos 2 3 2 cos 1 cos 1 cos 1 cos 1 cos cos2 lnu C d 2 sin2 1 ln3 2 cos C csc cot cot2 d 2 u 3 2 cos , du 2 sin d csc cot csc2 1 d

csc cot C 182 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

cos 1 1 1 69. d 2cos d 70. d d 2 sec tan 1cos sin cos 2 sin C cos d 1 1 sin u , du d 2 ln1 sin C u 1 sin , du cos d

2 8 y x y x 71. A dx 72. A 2 dx 1 ex 0 x 1 4 0 1 2 8 22 x 3 2 x 1 1 2x dx 2 3 0 x 1 2 2 0 1 e 4

4 1 2 1 2 12 2 x x 3 x x ln 1 e 1 2 2468 2 0 40 13.333 square units 1 1 3 4 ln1 e4 ln 2 2 2

0.337 square units

1 73. Arctangent Formula, Formula 23, 74. Log Rule: du, u ex 1 75. Substitution: u x2, du 2x dx u 1 Then Formula 81. du, u ex u2 1

76. Integration by parts 77. Cannot be integrated. 78. Formula 16 with u e2x

1 x2 79. (a) n 1: u ln x, du dx, dv x dx, v x 2 x2 x2 1 x2 x2 x ln x dx ln x dx ln x C 2 2 x 2 4 1 x3 n 2: u ln x, du dx, dv x2 dx, v x 3 x3 x3 1 x3 x3 x2 ln x dx ln x dx ln x C 3 3 x 3 9 1 x 4 n 3: u ln x, du dx, dv x3 dx, v x 4 x 4 x 4 1 x 4 x 4 x3 ln x dx ln x dx ln x C 4 4 x 4 16 xn1 xn1 (b) xn ln x dx ln x C n 1 n 12

80. A reduction formula reduces an 81. False. You might need to convert 82. True integral to the sum of a function your integral using substitution or and a simpler integral. For exam- algebra. ple, see Formula 50, 54. Section 8.6 Integration by Tables and Other Integration Techniques 183

5 5 500x 83. W 2000xex dx 84. W dx 2 0 0 26 x 5 x 5 2000 xe dx 212 0 250 26 x 2x dx 0 5 2000 xex1 dx 5 2 0 500 26 x 0 5 2000xe x e x 0 500 26 1 6 2000 1 2049.51 ft lbs e5 1919.145 ft lbs

3 2 85. V 202 dy 2 0 1 y W 148 80 ln 3 10 3 80 lny 1 y 2 11,840 ln 3 10 0 21,530.4 lb 80 ln3 10 145.5 cubic feet By symmetry, x 0. 3 2 3 M 2 dy 4 lny 1 y 2 4 ln3 10 2 0 1 y 0 3 2y 3 M 2 dy 41 y 2 410 1 x 2 0 1 y 0 M 410 1 y x 1.19 M 4 ln3 10

Centroid: x, y 0, 1.19

1 2 5000 2500 2 1.9 dt 86. dt 4.81.9t 4.81.9t 2 0 0 1 e 1.9 0 1 e 2500 2 4.8 1.9t ln1 e4.81.9t 1.9 0

2500 1 ln1 e 4.8 ln1 e4.8 1.9

2500 1 e 3.8 ln 401.4 1.9 1 e4.8

4 4 8 k 15.417 87. (a) dx 10 (b) dx 0 2 3x 0 2 3x

10 10 10 k 4 1 1 0.6486 0 4 dx ln 7 − 1 0 2 3x 3 30 15.417 ln 7 184 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

20 k 88. (a) 6x 2ex2 dx 50 0 By trial and error, k 5.51897.

0 5.52 5.51897 −1 (b) 6x 2ex2 dx 0

2 dx 89. Let I . 2 0 1 tan x For x u, dx du, and 2

0 du 2 du 2 tan u2 I 2 2 2 du. 2 1 tan 2 u 0 1 cot u 0 tan u 1 2 dx 2 tan x2 2 2I dx dx 2 2 0 1 tan x 0 tan x 1 0 2 Thus, I . 4

Section 8.7 Indeterminate Forms and L’Hôpital’s Rule

sin 5x 5 1. lim 2.5 exact: 3 x →0 sin 2x 2

x 0.1 0.01 0.001 0.001 0.01 0.1 −1 1 f x 2.4132 2.4991 2.500 2.500 2.4991 2.4132 −1

1 ex 2. lim 1 1 x →0 x

−1 1 x 0.1 0.01 0.001 0.001 0.01 0.1 f x 0.9516 0.9950 0.9995 1.00005 1.005 1.0517 −2

5 x100 3. lim x e 0 3 1011 x →

x 110 102 103 104 105 f x 0.9900 90,4843.7 109 4.5 1010 0 0 0 1500 0

6x 6 4. lim 3.4641 exact: 5 x → 3x2 2x 3

x 110 102 103 104 105

f x 6 3.5857 3.4757 3.4653 3.4642 3.4641 0 100 0 Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 185

2x 3 2x 3 2 1 5. (a) lim lim lim x →3 x2 9 x →3 x 3x 3 x →3 x 3 3 2x 3 ddx2x 3 2 2 1 (b) lim lim lim x →3 x2 9 x →3 ddxx2 9 x →3 2x 6 3

2x2 x 3 2x 3x 1 6. (a) lim lim lim 2x 3 5 x →1 x 1 x →1 x 1 x →1 2x2 x 3 ddx2x2 x 3 4x 1 (b) lim lim lim 5 x →1 x 1 x →1 ddxx 1 x →1 1

x 1 2 x 1 2 x 1 2 x 1 4 1 7. (a) lim lim lim lim 1 x →3 x 3 x →3 x 3 x 1 2 x →3 x 3x 1 2 x →3 x 1 2 4 x 1 2 ddxx 1 2 12x 1 1 (b) lim lim lim x →3 x 3 x →3 ddxx 3 x →3 1 4

sin 4x sin 4x sin 4x ddxsin 4x 4 cos 4x 8. (a) lim lim 2 21 2 (b) lim lim lim 2 x →0 2x x →0 4x x →0 2x x →0 ddx2x x →0 2

5x2 3x 1 5 3x 1x2 5 9. (a) lim lim x → 3x2 5 x → 3 5x2 3

5x2 3x 1 ddx5x2 3x 1 10x 3 ddx10x 3 10 5 (b) lim lim lim lim lim x → 3x2 5 x → ddx3x2 5 x → 6x x → ddx6x x → 6 3

2x 1 2x 1x2 0 10. (a) lim lim 0 x → 4x2 x x → 4 1x 4 2x 1 ddx2x 1 2 (b) lim lim lim 0 x → 4x2 x x → ddx4x2 x x → 8x 1

x2 x 2 2x 1 x2 x 2 2x 1 11. lim lim 3 12. lim lim 3 x →2 x 2 x →2 1 x →1 x 1 x →1 1

4 x2 2 x4 x2 4 x2 x4 x2 13. lim lim 0 14. lim lim x →0 x x →0 1 x →2 x 2 x →2 1 x lim x →2 4 x2

ex 1 x ex 1 ln x2 2 ln x 15. lim lim 2 16. → → lim lim x 0 x x 0 1 x→1 x2 1 x→1 x2 1 2x lim x→1 2x 1 lim 1 x→1 x2

ex 1 x ex 1 17. lim lim x →0 x3 x →0 3x2 ex lim x →0 6x 186 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

18. Case 1: n 1 ex 1 x ex 1 lim lim 0 x →0 x x →0 1 Case 2: n 2 ex 1 x ex 1 ex 1 lim lim lim x →0 x2 x →0 2x x →0 2 2 Case 3: n ≥ 3

x x x e 1 x e 1 e lim lim lim x →0 xn x →0 nxn 1 x →0 nn 1xn 2

sin 2x 2 cos 2x 2 sin ax a cos ax a arcsin x 11 x2 19. lim lim 20. lim lim 21. lim lim 1 x→0 sin 3x x→0 3 cos 3x 3 x→0 sin bx x→0 b cos bx b x →0 x x →0 1

2 3x2 2x 1 6x 2 arctan x 4 1 1 x 1 22. lim lim 23. lim 2 lim x →1 x 1 x →1 1 2 x→ 2x 3 x→ 4x 6 3 lim x→ 4 2

x 1 1 x2 2x 3 2x 2 24. lim lim 0 25. lim lim x → x2 2x 3 x → 2x 2 x→ x 1 x → 1

x3 3x2 3 2 x 3x 26. lim lim 27. lim lim x→ x 1 x→ 1 x→ e x 2 x→ 12e x 2 6x 6 lim lim 0 x→ 14e x 2 x→ 18e x 2

x2 2x 2 x 1 28. lim lim lim 0 29. lim lim 1 x→ ex x→ ex x→ ex x→ x2 1 x→ 1 1x2 Note: L’Hôpital’s Rule does not work on this limit. See Exercise 79.

x2 x cos x 30. lim lim 31. lim 0 by Squeeze Theorem x→ x2 1 x→ 1 1x2 x→ x cos x 1 ≤ , for x > 0 x x

sin x ln x 1x 1 32. lim 0 33. lim lim lim 0 x→ x x→ x2 x→ 2x x→ 2x2 Note: Use the Squeeze Theorem for x > . 1 sin x 1 ≤ ≤ x x x

4 e x e x ex x2 x2 ln x 4 ln x 4 x 35. lim lim lim e 1 2 e 34. lim 3 lim 3 lim 2 → 2 → → 36. lim lim x→ x x→ x x→ 3x x x x 2x x 2 x→ x x→ 1 4 lim 0 x→ 3x3 Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 187

37. (a)lim x ln x, not indeterminate 38. (a) lim x3 cot x 0 x→ x→0 (b) lim x ln x x3 3x2 x→ (b) lim x3 cot x lim lim 0 x→0 x→0 tan x x→0 sec2 x (c) 3

0 4 0 3

−1

1 1 39. (a) lim x sin 0 40. (a) lim x tan 0 x→ x x→ x 1 sin1x 1 tan1x (b) lim x sin lim (b) lim x tan lim x→ x x→ 1x x→ x x→ 1x 1x2cos1x 2 2 1 x sec 1 x lim 2 lim x→ 1x x→ 1x2 1 1 lim cos 1 lim sec2 1 x→ x x→ x

−11 110

− 0.5 −1

41. (a)lim x1x 0 0, not indeterminate 42. (a) lim e x x2x 1 x →0 x→0 (See Exercise 106). (b) Let y lim e x x2x. x→0 1x (b) Let y x 2 lne x x ln y lim 1 x→0 x ln y ln x1x ln x. x 2e x 1e x x 1 lim 4 Since x → 0 , ln x → . Hence, x→0 1 x Thus, ln y 4 ⇒ y e4 54.598. ln y → ⇒ y → 0. (c) 60 Therefore, lim x1x 0. x→0

0 2 0

− 0.5 2

− 0.5

1x 0 43. (a) lim x (c) 2 x → (b) Let y lim x1x. x→ ln x 1x ln y lim lim 0 − 5 20 x→ x x→ 1 −0.5 Thus,ln y 0 ⇒ y e0 1. Therefore, lim x1x 1. x → 188 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 x 44. (a) lim 1 1 45. (a) lim 1 x1x 1 x→ x x →0 (b) Let y lim 1 x1x. 1 x x →0 (b) Let y lim 1 . x→ x ln1 x ln y lim x →0 x 1 ln1 1x ln y lim x ln 1 lim → → 11 x x x x 1 x lim 1 x →0 1 1x2 ⇒ 1 1 1x 1 Thus, ln y 1 y e e. lim lim 1 x→ 1x2 x→ 1 1x Therefore, lim 1 x1x e. x →0 1 Thus,ln y 1 ⇒ y e e. Therefore, (c) 6 1 x lim 1 e. x→ x

0 10

−1

46. (a) lim 1 x1x 0 47. (a) lim 3xx2 00 x → x→0 (b) Let y lim 1 x1x. (b) L et y lim 3xx2. x→ x→0 x ln 1 x ln y lim ln 3 ln x ln y lim → x→ x x 0 2 ln x 1 1 x lim ln 3 lim 0 x→0 x→ 1 2 x 1x Thus, ln y 0 ⇒ y e0 1. lim ln 3 lim → → 2 Therefore, lim 1 x1x 1. x 0 x 0 2 x x → x (c) 5 lim ln 3 lim x→0 x→0 2 ln 3

x2 0 10 Hence, lim 3x 3. x→0 −1 (c) 7

− 6 6

−1

x4 0 48. (a) lim 3x 4 0 (c) 2 x→4 (b) Let y lim 3x 4x4. x→4 ln y lim x 4ln3x 4 x→4 ln3x 4 4 7 lim 0 x→4 1x 4 1x 4 lim x→4 1x 42 lim x 4 0 x→4 Hence, lim 3x 4x4 1. x→4 Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 189

x x1 0 0 49. (a) lim ln x 0 50. (a) lim cos x 0 x→1 x→0 2 (b) Let y lim ln xx1 x→1 x (b) Let y lim cos x . lim x 1ln x 0. x→0 2 x→1 Hence, lim ln xx1 1. x→1 ln y lim x lncos x x→0 2 (c) 6 0 0 0 x − 4 8 Hence, lim cos x 1. x→0 2

−2 (c) 2

0 3 0

8 x 1 x 1 51. (a) lim 52. (a) lim x →2 x2 4 x 2 x →2 x2 4 x2 4 8 x 8 xx 2 1 x 1 1 x 1 (b) lim lim (b) lim lim x →2 x2 4 x 2 x →2 x2 4 x →2 x2 4 x2 4 x →2 x2 4 2 x4 x 12x 1 lim lim x →2 x 2x 2 x →2 2x 1 1 x 4 3 lim lim x →2 x 2 2 x →2 4xx 1 8

−7 5 2 4

−4 −0.25

3 2 10 3 53. (a) lim 54. (a) lim x →1 ln x x 1 x→0 x x2 3 2 3x 3 2 ln x 10 3 10x 3 (b) lim lim (b) lim lim 2 2 x →1 ln x x 1 x →1 x 1ln x x→0 x x x→0 x

−20

−1 4

−4

x 3 1 55. (a) 3 (b) lim lim x →3 ln2x 5 x →3 22x 5 2x 5 1 lim x →3 2 2 −1 7

−1 190 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

x 56. (a) 2 (b) Let y sin x , then ln y x lnsin x. lnsin x cos xsin x x2 2x lim lim lim lim 0 x →0 x →0 2 x →0 x →0 2 1 x 1 x tan x sec x − 2 Therefore, since ln y 0, y 1 and lim sin xx 1. x →0 −1

x2 5x 2 x 57. (a) 10 (b) lim x2 5x 2 x lim x2 5x 2 x x→ x→ x2 5x 2 x x2 5x 2 x2 lim x→ x2 5x 2 x −810

− 5x 2 2 lim x→ x2 5x 2 x 5 2x 5 lim x→ 1 5x 2x2 1 2

x3 3x2 6x 6 58. (a) 1 (b) lim lim lim lim 0 x→ e2x x→ 2e2x x→ 4e2x x→ 8e2x − 15

−3

0 59. , , 0 , 1, 00, , 0 60. See Theorem 8.4. 0

61. (a) Let f x x2 25 and gx x 5. 62. Let f x x 25 and gx x. (b) Let f x x 52 and gx x2 25. (c) Let f x x2 25 and gx x 53.

63. x 10 102 104 106 108 1010 ln x4 2.811 4.498 0.720 0.036 0.001 0.000 x

64. x 151020 30 40 50 100 ex 2.718 0.047 0.220 151.614 4.40 105 2.30 109 1.66 1013 2.69 1033 x5

x2 2x 2 x3 3x2 6x 6 65. lim lim lim 0 66. lim lim lim lim 0 x → e5x x → 5e5x x → 25e5x x→ e2x x→ 2e2x x→ 4e2x x→ 8e2x Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 191

ln x3 3ln x21x ln x2 2 ln xx 67. lim lim 68. lim lim x→ x x→ 1 x→ x3 x→ 3x2 3ln x2 2 ln x lim lim x→ x x→ 3x3 6ln x1x 2x 2 lim lim lim 0 x→ 1 x→ 9x2 x→ 9x3 6ln x 6 lim lim 0 x→ x x→ x

ln xn nln xn1x m m1 x mx 69. lim m lim m1 70. lim lim x→ x x→ mx x→ enx x→ nenx nln xn1 m2 m m 1 x lim m lim x→ mx x→ n2enx nn 1ln xn2 . . . m! lim 2 m lim 0 x→ m x x→ nmenx n! . . . lim 0 x→ mnxm

71. y x1x, x > 0 72. y xx, x > 0 Horizontal asymptote:y 1 (See Exercise 43.) lim xx and lim xx 1 x→ x→0 1 4 ln y ln x No horizontal asymptotes 4 x (e , e1/e) ln y x ln x 1 dy 1 1 1 1 ln x 1, 1 e y dx x x x2 1 dy 1 (e (e ( ( − 0 6 x ln x 14 0 y dx x − dy 1 1 1x 1x2 dy x 2 1 ln x x 1 ln x 0 xx1 ln x 0 dx x dx Critical number: x e Critical number: x e 1 Intervals: 0, e e, Intervals: 0, e 1 e 1, 0 Sign of dydx: Sign of dydx: y f x: Increasing Decreasing y f x: Decreasing Increasing 1e Relative maximum: e, e 1e 1 1 1 Relative minimum: e1, e1e , e e

ln x 73. y 2xex 74. y x 2x 2 Horizontal asymptote:y 0 (See Exercise 29.) lim lim 0 3 → x → x x e x e 2 (1, e ) dy x1x ln x1 1 ln x − 0 Horizontal asymptote: y 0 2 10 dx x2 x2 dy 2xex 2ex Critical number: x e dx −5 Intervals: 0, e e, 2ex 1 x 0 Sign of dydx: Critical number: x 1 y f x: Increasing Decreasing Intervals: , 1 1, 1 1 Relative maximum: e, Sign of dydx: e − 14 e, 1 y f x : Increasing Decreasing ( e ( 2 Relative maximum: 1, e −4 192 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

e2x 1 0 sin x 1 75. lim 0 76. lim 0 (Numerator is bounded) x →0 ex 1 x→ x Limit is not of the form 00 or . Limit is not of the form 00 or . L’Hôpital’s Rule does not apply. L’Hôpital’s Rule does not apply.

x 1 e 0 77. lim x cos 1 78. lim 0 x → x x → 1 e x 1 0 Limit is not of the form 00 or . Limit is not of the form 00 or . L’Hôpital’s Rule does not apply. L’Hôpital’s Rule does not apply.

tan x 79. (a) Applying L’Hôpital’s Rule twice results in the 80. (a)lim is indeterminant: x→2 sec x original limit, so L’Hôpital’s Rule fails: tan x sec2 x x 1 lim lim lim lim x→2 sec x x→2 sec x tan x x→ x2 1 x→ xx2 1 sec x x2 1 lim lim x→2 tan x x→ x sec x tan x xx2 1 lim lim x→2 sec2 x x→ 1 tan x x lim , the original problem! lim x→2 sec x x→ x2 1 x xx tan x sin x (b) lim lim cos x (b) lim lim x→2 sec x x→2 cos x x→ x2 1 x→ x2 1x lim sin x 1 1 → lim x 2 → 2 x 1 1x (c) 2 1 1 1 0 0 π (c) 1.5 −1

−66

−1.5

81. fx sin3x, gx sin4x 82. fx e3x 1, gx x fx 3 cos3x, gx 4 cos4x fx 3e3x , gx 1 fx sin 3x fx 3 cos 3x fx e3x 1 fx y , y y , y 3e3x 1 gx sin 4x 2 gx 4 cos 4x 1 gx x 2 gx

sin 3x 3 cos 3x 3x y = y = y = 3e sin 4x 4 cos 4x 5 1.5 e3x−1 y = x

− 0.5 0.5 − 0.5 0.5 0 0.5 → → → → → → As x 0, y1 3 and y2 3 As x 0, y1 0.75 and y2 0.75 By L’Hôpital’s Rule, By L’Hôpital’s Rule, e3x 1 3e3x sin 3x 3 cos 3x 3 lim lim 3 lim lim x→0 x x→0 1 x→0 sin 4x x→0 4 cos 4x 4 Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 193

v kekt 321 ekt 0 32 321 ekt kt 83. lim lim lim v0e k→0 k k→0 k k→0 kt v 32 0 te 0 lim lim 32t v0 k→0 1 k→0 ekt

r nt 84. A P1 n r ln1 r n ln A ln P nt ln1 ln P n 1 nt r r 1 ln1 n n2 1 rn 1 lim lim lim rt rt n → 1 n → 1 n → r 1 nt n2t n Since lim ln A ln P rt, we have lim A eln Prt eln Pert Pert. Alternatively, n→ n→ r nt r nr rt lim A lim P1 lim P1 Pert. n→ n→ n n→ n

85. Let N be a fixed value for n. Then x N1 N 1x N2 N 1N 2x N3 N 1! lim lim lim . . . lim 0. (See Exercise 70.) x→ ex x→ ex x→ ex x→ ex

2 dy y y 144 x 86. (a) m y (0,y + 144 − x2 ( dx x 0 12 144 x2 x 144 − x2 12

(,x y ) y x x 2 4 6 8 10 12

144 x2 (b) y dx x 12 Let x 12 sin , dx 12 cos d, 144 x2 12 cos . x 12 cos 1 sin2 θ y 12 cos d 12 d 12 sin sin 144 − x2

12csc sin d 12 lncsc cot 12 cos C 30

12 144 x2 144 x2 12 ln 12 C x x 12 12 144 x2 0 12 12 ln 144 x2 C 0 x 12 144 x2 When x 12, y 0 ⇒ C 0. Thus, y 12 ln 144 x2. x 12 144 x2 Note: > 0 for 0 < x ≤ 12 x

—CONTINUED— 194 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

86. —CONTINUED— (c) Vertical asymptote: x 0 (d) y 144 x2 12 ⇒ y 12 144 x2 Thus, 12 144 x2 12 144 x2 12 ln 144 x2 x 12 144 x2 1 ln x xe1 12 144 x2

2 xe1 122 144 x2 x2e2 24xe1 144 144 x2 x2e2 1 24xe1 0 xxe2 1 24e1 0 24e1 x 0 or x 7.77665. e2 1 Therefore,

12 2 2 12 2 2 144 x x 144 x s 1 dx 2 dx 7.77665 x 7.77665 x 12 12 12 dx 12 lnx 12ln 12 ln 7.77665 5.2 meters. 7.77665 x 7.77665

1 87. f x x3, gx x2 1, 0, 1 88. f x , gx x2 4, 1, 2 x f b f a f c f 2 f 1 fc g b g a g c g2 g1 gc 2 f 1 f 0 3c 12 1c2 g 1 g 0 2c 3 2c 1 3c 1 1 1 2 6 2c3 2 c 2 c3 6 3 c 3 3

89. f x sin x, gx cos x, 0, 90. f x ln x, gx x3, 1, 4 2 f 4 f 1 fc f 2 f 0 fc g4 g1 gc g2 g0 gc ln 4 1c 1 1 cos c 63 3c2 3c3 1 sin c 3 c3 ln 4 63 1 cot c 21 c3 c ln 4 4 21 c 3 2.474 ln 4 Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 195

91. False. L’Hôpital’s Rule does not apply since 92. False. If y exx2, then lim x2 x 1 0. x2ex 2xex xex x 2 ex x 2 x →0 y 4 4 3 . x2 x 1 1 x x x lim lim x 1 1 x →0 x x →0 x

93. True 94. False. Let f x x and gx x 1. Then x lim 1, but lim x x 1 1. x → x 1 x →

1 95. Area of triangle: 2x1 cos x x x cos x 2 Shaded area: Area of rectangle Area under curve

x x 2x1 cos x 2 1 cos t dt 2x1 cos x 2t sin t 0 0 2x1 cos x 2x sin x 2 sin x 2x cos x x x cos x 1 x sin x cos x Ratio: lim lim x →0 2 sin x 2x cos x x →0 2 cos x 2x sin x 2 cos x 1 x sin x cos x lim x →0 2x sin x x cos x sin x sin x lim x →0 2x cos x 2 sin x x cos x 2 sin x 1cos x lim x →0 2x cos x 2 sin x 1cos x x 2 tan x lim x →0 2x 2 tan x 1 2 sec2 x 3 lim x →0 2 2 sec2 x 4

96. (a) sin BD cos DO ⇒ AD 1 cos 1 1 1 1 Area ABD bh 1 cos sin sin sin cos 2 2 2 2 1 (b) Area of sector: 2 1 1 1 1 1 Shaded area: Area OBD cos sin sin cos 2 2 2 2 2 12 sin 12 sin cos sin sin cos (c) R 12 12 sin cos sin cos sin 12 sin 2 (d) lim R lim →0 →0 12 sin 2 cos cos 2 sin 2 sin 2 cos 4 cos 2 3 lim lim lim →0 1 cos 2 →0 2 sin 2 →0 4 cos 2 4 196 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

4x 2 sin 2x 4 4 cos 2x 97. lim lim 98. Let y ex x1x. x→0 2x3 x→0 6x2 1 lnex x 8 sin 2x ln y lnex x lim x x x→0 12x lnex x ex 1 2 16 cos 2x 16 4 lim lim 2 lim x→0 x x→0 ex x 1 x→0 12 12 3 Hence, limex x1x e2. 4 x→0 Let c . 3 Let c e2 7.389.

a cos bx 99. lim 2 100. We use mathematical induction. x→0 x2 x1 1 Near x 0, cos bx 1 and x2 0 ⇒ a 1. For n 1, lim lim 0. x→ ex x→ ex Using L’Hôpital’s Rule, xk Assume that lim 0. 1 cos bx b sin bx x→ ex lim lim x→0 x2 x→0 2x xk1 k 1xk Then, lim lim b2 cos bx x→ ex x→ e x lim 2. x→0 2 xk k 1lim Hence,b2 4 and b ±2. x→0 ex Answer: a 1, b ±2 k 10 0.

fx h fx h fx h1 fx h1 101. (a) lim lim h→0 2h h→0 2 f x h f x h lim h→0 2 fx fx fx 2

(b) y

x x − hx x + h

Graphically, the slope of the line joining x h, fx h and x h, fx h is approximately fx. And, as h → 0, fx h fx h lim fx. h→0 2h

fx h 2fx fx h fx h1 fx h1 102. lim lim h→0 h2 h→0 2h f x h f x h lim h→0 2h f x h 1 f x h 1 lim h→0 2 f x h f x h lim h→0 2 fx fx fx 2 Section 8.7 Indeterminate Forms and L’Hôpital’s Rule 197

e1x2, x 0 103. gx y 0, x 0 1.5 gx g0 e1x2 g0 lim lim x →0 x 0 x →0 x 0.5 1x2 1x2 2 e e 1 1 x ln x x Let y , then ln y ln ln x . Since − − x x x2 x2 2 1 12 −0.5 ln x 1x x2 lim x2 ln x lim lim lim 0 x →0 x →0 1x2 x →0 2x3 x →0 2 1 x2 ln x we have lim . Thus, lim y e 0 ⇒ g0 0. x →0 x2 x →0 Note: The graph appears to support this conclusion—the tangent line is horizontal at 0, 0.

x k 1 104. f x 6 k k = 1 k 1, f x x 1 k = 0.1 k = 0.01 0.1 x 1 − 210 k 0.1, f x 10x 0.1 1 0.1 −2 x 0.01 1 k 0.01, f x 100x 0.01 1 0.01 x k 1 xkln x lim lim ln x k →0 k k →0 1

105. (a)lim x ln x is the form 0 . (c) 1 x→0 ln x 1x −1 5 (b) lim lim lim x 0 x→0 1x x→0 1x2 x→0

−3

106. lim f xgx 107. lim f xgx x →a x →a y f xgx y f xgx ln y gx ln f x ln y gx ln f x lim gx ln f x lim gx ln f x x →a x →a As x → a, ln y ⇒ , and hence y 0. Thus, As x → a, ln y ⇒ , and hence y . Thus, lim f xgx 0. lim f xgx . x →a x →a

b b b 108. fab a f tt b dt fab a ftt b ft dt a a a b fab a faa b f t f b f a a

dv ft dt ⇒ v ft u t b ⇒ du dt 198 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

2a3x x4 a 3 a2x ln 21ln x 0 110. lim 109. (a)lim x is of form 0 . 4 x →0 x→a a ax3 Let y xln 21ln x 1 a 2a3x x4122a3 4x3 a2x23a2 ln 2 2 3 ln y ln x lim 1 ln x x→a 1 ax334 4 ln 21x lim ln y ln 2. x →0 1x 1 a3 a4122a3 a323 2 3 Thus, lim xln 21ln x 2. x→0 1 ax3343ax2 (b)lim xln 21ln x is of form 0. 4 x→ Let y xln 2)1ln x a a 3 ln 2 ln y ln x 1 1 ln x a33a3 4 ln 21x lim ln y ln 2. 4 x→ 1x a 3 16 a Thus, lim x ln 2 1 ln x 2. 3 9 x→ (c)lim x 1ln 2x is of form 1. 4 x→0 Let y x 1ln 2)x ln 2 ln y lnx 1 x ln 21x 1 lim ln y lim ln 2. x→0 x→0 1 Thus, lim x 1ln 2x 2. x→0

x sin x x sin x x sin x sin x 111. (a) hx (b) hx 1 , x > 0 x x x x x

3 sin x Hence, lim hx lim 1 1 0 1. x→ x→ x lim hx 1 x→ (c) No. hx is not an indeterminate form.

−2 20 0

1 ax 1 1x 112. Let fx . x a 1

For a > 1 and x > 0, 1 1 ln x lnax 1 lna 1 ln fx ln lnax 1 lna 1 . x x x x x ln x lna 1 lnax 1 ln1 axax ln1 ax As x → , → 0, → 0, and ln a → ln a. x x x x x Hence, ln fx → ln a.

For 0 < a < 1 and x > 0, ln x ln1 ax ln1 a ln fx → 0 as x → . x x x

a if a > 1 Combining these results, lim f x . x→ 1 if 0 < a < 1 Section 8.8 Improper Integrals 199

Section 8.8 Improper Integrals

1 dx 2 2dx 1 1. is improper because 3x 2 0 when x . 2. is not improper because is continuous on 1, 2. 2 2 0 3x 2 3 1 x x

1 2x 5 1 2x 5 3. dx dx 4. lnx2dx 2 0 x 5x 6 0 x 2 (x 3 1 is not improper because is improper because the upper limit of integration is . 2x 5 is continuous on 0, 1. x 2x 3

5. Infinite discontinuity at x 0. 6. Infinite discontinuity at x 3. 4 1 4 1 dx lim dx 4 1 4 b →0 32 0 x b x dx lim x 3 dx 32 → 3 x 3 b 3 b 4 lim 2x 4 b →0 b lim 2x 3 1 2 b →3 b lim 4 2b 4 b →0 2 lim 2 Converges b →3 b 3 Diverges

7. Infinite discontinuity at x 1. 2 1 1 1 2 1 dx dx dx 2 2 2 0 x 1 0 x 1 1 x 1 b 1 2 1 lim dx lim dx → 2 → 2 b 1 0 x 1 c 1 c x 1 1 b 1 2 lim lim 1 1 b→1 c→1 x 1 0 x 1 c Diverges

8. Infinite discontinuity at x 1.

2 1 1 1 2 1 dx dx dx 23 23 23 0 x 1 0 x 1 1 x 1

b 1 2 1 lim dx lim dx → 23 → 23 b 1 0 x 1 c 1 c x 1

b 2 lim 33 x 1 lim 33 x 1 0 3 3 0 6 b→1 0 c→1 c Converges

9. Infinite limit of integration. 10. Infinite limit of integration.

b 0 0 ex dx lim ex dx e2x dx lim e2x dx b→ → 0 0 b b b 1 0 1 1 lim ex 0 1 1 lim e2x 0 b→ 0 b→ 2 b 2 2 Converges Converges 200 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 1 2 2 8 11. dx 2 12. dx because the integral is not defined 2 3 1 x 2 x 1 9 because the integrand is not defined at x 0. at x 1. The integral diverges. Diverges

13. ex dx 0. You need to evaluate the limit. 14. sec x dx 0 because sec x is not defined at x 2. 0 0 b b The integral diverges. lim ex dx lim ex → → b 0 b 0

lim eb 1 1 b→

1 b 1 5 b 5 15. dx lim dx 16. 2 2 dx lim dx x b→ x 3 b→ 3 1 1 1 x 1 x 1 b 5 b 5 lim 1 lim x2 b→ → x 1 b 2 1 2

b 3 4 b 17. dx lim 3x13 dx 18. dx lim 4x14 dx 3 → 4 → 1 x b 1 1 x b 1 9 b 16 b lim x23 lim x34 Diverges → → b 2 1 b 3 1 Diverges

0 0 1 0 1 19. xe2x dx lim xe2x dx lim 2x 1e2x lim 1 2b 1e2b (Integration by parts) → → → b b b 4 b b 4 Diverges

b b 20. xex2 dx lim xex2 dx lim ex22x 4 lim eb22b 4 4 4 → → → 0 b 0 b 0 b

b b b2 2b 2 21. x2ex dx lim x2ex dx lim exx2 2x 2 lim 2 2 → → → b 0 b 0 b 0 b e b2 2b 2 Since lim 0 by L’Hôpital’s Rule. b→ eb

b b b 22. x 1ex dx lim x 1ex dx lim xex lim 0 0 by L’Hôpital’s Rule. → → → b 0 b 0 b 0 b e

b ax c 1 e a sin bx b cos bx 23. e x cos x dx lim e x cos x sin x 24. eax sin bx dx lim b→ 0 → 2 2 0 2 0 c a b 0 1 1 b b 0 1 0 2 2 a2 b2 a2 b2 Section 8.8 Improper Integrals 201

1 b 1 ln x b ln x 25. dx lim ln x3 dx 26. dx lim dx 3 → → 4 x ln x b 4 x 1 x b 1 x 1 b ln x2 b lim ln x2 lim → → b 2 4 b 2 1 1 1 Diverges ln b2 ln 42 2 2 1 1 1 2 2 ln 22 8ln 22

2 0 2 2 x3 b x b x 27. dx dx dx 28. dx lim dx lim dx 2 2 2 2 2 b→ 2 b→ 2 2 4 x 4 x 0 4 x 0 x 1 0 x 1 0 x 1 0 2 c 2 1 1 b lim dx lim dx lim lnx2 1 → 2 → 2 → 2 b b 4 x c 0 4 x b 2 2 x 1 0 x 0 x c 1 lim arctan lim arctan → → b 2 b c 2 0 2 Diverges 0 0 2 2

1 b ex ex b 29. dx lim dx x 30. dx lim ln 1 e ln 2 x x b→ 2x x → 0 e e 0 1 e 0 1 e b 0 b lim arctanex Diverges b→ 0 2 4 4

1 b x x b 31. cos x dx lim sin x 32. sin dx lim 2 cos → → 0 b 0 0 2 b 2 0 x Diverges since sin b does not approach a limit as b → . Diverges since cos does not approach a limit as x → . 2

1 1 1 1 1 48 48 4 33. dx lim lim 1 1 34. dx lim dx lim 8 ln x 2 b→ → b→0 b→0 0 x 0 x b b 0 b 0 x b x b Diverges Diverges

8 1 b 1 6 4 b 35. dx lim dx 36. dx lim 46 x12 dx 3 → 3 → 0 8 x b 8 0 8 x 0 6 x b 6 0 b b 12 3 23 lim 86 x lim 8 x 6 → b 6 b→8 2 0 0 80 86 86

1 x2 x2 1 e e x ln x dx lim lnx 2 37. 38. ln x dx lim 2 ln x dx b→0 2 4 b → 0 0 b 0 0 1 b2 ln b b2 1 e lim 2x ln x 2x lim → b→0 4 2 4 4 b 0 b

2 lim 2e 2e 2b ln b 2b since lim b ln b 0 by L’Hôpital’s Rule. b→0 b→0 0 202 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

2 b 2 b 39. tan d lim lnsec 40. sec d lim lnsec tan b→2 b→2 0 0 0 0 Diverges Diverges

4 2 4 2 2 1 x b 41. dx lim dx 42. dx lim arcsin 2 → 2 2 b→2 2 x x 4 b 2 b x x 4 0 4 x 2 0 2 x 4 lim arcsec b→2 2 b b lim arcsec 2 arcsec b→2 2 0 3 3

4 1 4 43. lim ln x x2 4 ln4 23 ln 2 ln2 3 1.317 2 → 2 x 4 b 2 b

2 1 b 1 1 1 1 2 x b 44. dx lim dx lim ln 0 2 b→2 b→2 0 4 x 0 4 2 x 2 x 4 2 x 0 Diverges

2 1 1 1 2 1 45. dx dx dx 3 3 3 0 x 1 0 x 1 1 x 1 3 b 3 2 3 3 lim x 123 lim x 123 0 b→1 2 0 c→1 2 c 2 2

3 2 2 3 46. dx 2x 283 dx 2x 283 dx 83 1 x 2 1 2 b 3 lim 2x 283 dx lim 2x 283 dx → → b 2 1 c 2 c 6 b 6 3 lim x 253 lim x 253 → → b 2 5 1 c 2 5 c Diverges

4 1 4 4 47. dx dx dx 0 xx 6 0 xx 6 1 xx 6 Let u x, u2 x, 2u du dx. 4 42u du du 8 u 8 x dx 8 arctan C arctan C xx 6 uu2 6 u2 6 6 6 6 6

1 c 4 8 x 8 x Thus, dx lim arctan lim arctan b→0 c→ 0 x x 6 6 6 b 6 6 1 8 1 8 8 8 1 arctan 0 arctan 6 6 6 6 2 6 6

8 26 . 26 3 Section 8.8 Improper Integrals 203

1 1 b 1 b 48. dx lnln x C 49. If p 1, dx lim dx lim ln x → → x ln x 1 x b 1 x b 1 Thus, lim ln b . b→ 1 e 1 1 dx dx dx Diverges. For p 1, 1 x ln x 1 x ln x e x ln x e 1 x1p b b1p 1 lim lnln x lim lnln x . dx lim lim . → → p b→ b→ b 1 1 c e 1 x 1 p 1 1 p 1 p Diverges 1 This converges to if 1 p < 0 or p > 1. p 1

11 1 50. If p 1, dx lim ln x lim ln a . → → 0 x a 0 a a 0 Diverges. If p 1, 1 1 x1p 1 1 a1p dx lim lim . p → → 0 x a 0 1 p a a 0 1 p 1 p 1 This converges to if 1 p > 0 or p < 1. 1 p

51. For n 1 we have b xex dx lim xex dx → 0 b 0 b lim exx ex Parts: u x, dv ex dx → b 0 lim ebb eb 1 b→ b 1 lim 1 1 (L’Hôpital’s Rule). b→ eb eb Assume that xnex dx converges. Then for n 1 we have 0

xn1ex dx xn1ex n 1xnex dx

by parts u xn1, du n 1xn dx, dv ex dx, v ex. Thus,

b xn1ex dx lim xn1ex n 1 xnex dx 0 n 1 xnex dx, which converges. → 0 b 0 0 0

52. (a) Assume gx dx L (converges). a Since 0 ≤ f x ≤ gx on a, , 0 ≤ f x dx ≤ gx dx L and f x dx converges. a a a (b) gx dx diverges, because otherwise, by part (a), if gx dx converges, then so does f x dx. a a a

1 1 1 1 1 3 53. dx diverges. 54. dx converges. 3 3 0 x 0 x 1 1 3 2 1 (See Exercise 50, p 3 1. See Exercise 50, p 3 . 204 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 1 1 55. dx converges. 56. x 4ex dx converges. 3 1 x 3 1 2 0 (See Exercise 49, p 3. (See Exercise 51.)

1 1 1 1 57. Since ≤ on 1, and dx converges by Exercise 49, dx converges. 2 2 2 2 x 5 x 1 x 1 x 5

1 1 1 1 58. Since ≥ on 2, and dx diverges by Exercise 49, dx diverges. x 1 x 2 x 2 x 1

1 1 1 1 59. Since ≥ on 2, and dx diverges by Exercise 49, dx diverges. 3 3 2 3 2 3 xx 1 x 2 x 2 xx 1

1 ≤ 1 1 1 60. Since 32 on 1, and 32 dx converges by Exercise 49, dx converges. x1 x x 1 x 1 x1 x

61. Since ex2 ≤ ex on 1, and ex dx converges (see Exercise 9), ex2 dx converges. 0 0

1 1 1 1 62. ≥ since x ln x < x on 2, . Since dx diverges by Exercise 49, dx diverges. x ln x x 2 x 2 x ln x

1 1 0 1 1 1 63. Answers will vary. 64. See the definitions, 65. dx dx dx x3 x3 x3 pages 578, 581. 1 1 0 These two integrals diverge by Exercise 50.

10 10 66. ⇒ x 0, 2. x2 2x xx 2 You must analyze three improper integrals, and each must converge in order for the original integral to converge. 3 1 2 3 f x dx f x dx f x dx f x dx 0 0 1 2

1 1 67. A e x dx 68. A ln x dx 0 1 1 lim e x dx lim ln x dx b→ → b b 0 b 1 1 lim ex lim x ln x x b→ b b→0 b lim 0 1 b ln b b lim e eb e b→0 b→ 1 ln b 1b Note: lim b ln b lim lim 0 b→0 b→0 1b b→0 1b2 Section 8.8 Improper Integrals 205

1 8 69. A dx 70. A dx 2 2 x 1 x 4 0 1 b 1 0 8 b 8 lim dx lim dx lim dx lim dx → 2 → 2 → 2 → 2 b b x 1 b 0 x 1 b b x 4 b 0 x 4 0 b x 0 x b lim arctanx lim arctanx → → lim 4 arctan lim 4 arctan b b b 0 b→ 2 b b→ 2 0 lim 0 arctan b lim arctan b 0 b b b→ b→ lim 0 4 arctan lim 4 arctan 0 b→ 2 b→ 2 2 2 4 4 4 2 2

1 1 71. (a) A ex dx 72. (a) A 2 dx 1 0 1 x x 1 b (b) Disk: lim ex 0 1 1 b→ 0 1 b V dx lim (b) Disk: 4 → 3 1 x b 3x 1 3 V ex 2 dx (c) Shell: 0 b 1 b V 2 x 2 dx lim 2 ln x 1 x b→ 1 lim e 2x 1 b→ 2 0 2 Diverges (c) Shell: V 2 xex dx 0 b lim 2exx 1 2 b→ 0

73. x23 y23 4 y 8 (0, 8) 2 2 x13 y13y 0 3 3 (−8, 0) 2 (8, 0) 13 x y −8 −2 28 y x13

y 23 x23 y23 4 2 −8 (0, −8) 1 y2 1 , x > 0 x 23 x23 x23 x13 8 2 3 8 s 4 dx lim 8 x23 48 13 → 0 x b 0 2 b

74. y 16 x2, 0 ≤ x ≤ 4 x y 16 x2 4 x2 4 4 s 1 dx dx 2 2 0 16 x 0 16 x t 4 lim dx → 2 t 4 0 16 x x t lim 4 arcsin t→4 4 0 t lim 4 arcsin 2 t→4 4 206 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

75. x 22 y2 1 2x 2 2yy 0 x 2 y y 1 1 y2 1 x 2 2y 2 (Assume y > 0. y 3x 3 x 3 x 2 2 S 4 dx 4 dx 4 dx 2 2 2 1 y 1 1 x 2 1 1 x 2 1 x 2

b lim 41 x 22 2 arcsinx 2 40 2 arcsin1 2 arcsin1 8 2 a→1 a b→3

x 76. y 2e y

x 2 y 2e − y = 2e x x 2x S 2 2e 1 4e dx x 0 3 4 Let u ex, du ex dx. −2 ex1 4e2x dx 1 4u2 du

1 2u4u2 1 ln 2u 4u2 1 C 4 1 2ex4e2x1 ln 2ex4e2x1 C 4

b S 4 lim ex1 4e2x dx → b 0 b lim 2ex4e2x1 ln 2ex4e2x1 b→ 0 25 ln2 5 18.5849

K K 77. (a) Fx , 5 , K 80,000,000 x2 40002 80,000,000 80,000,000 b W dx lim 20,000 mi-ton 2 → 4000 x b x 4000 W 80,000,000 b 80,000,000 (b) 10,000 20,000 2 x 4000 b 80,000,000 10,000 b b 8000 Therefore, 4000 miles above the earth’s surface.

k k 2 b 2 2 W 10 4000 10 4000 78. (a) F x 2, 10 2, k 10 4000 (b) 20,000 40,000 x 4000 2 x 4000 b 1040002 1040002 b 1040002 W dx lim 2 → 20,000 4000 x b x 4000 b 1040002 b 8000 40,000 mi-ton 4000 Therefore, 4000 miles above the earth’s surface. Section 8.8 Improper Integrals 207

1 1 b 2 2 b 79. (a) et7 dt et7 dt lim et7 1 80. (a) e2t5 dt e2t5 dt lim e2t5 1 b→ → 7 0 7 0 5 0 5 b 0 41 4 4 2 4 (b) et7 dt et7 e47 1 (b) e2t5 dt e2t5 e85 1 0 7 0 0 5 0 0.4353 43.53% 0.7981 79.81% b 1 b 2 5 5 (c) t et7 dt lim tet7 7et7 (c) t e 2t 5 dt lim te2t 5 e 2t 5 → 5 b→ 2 0 2 0 7 b 0 0 0 7 7

5 25,000 5 81. (a) C 650,000 25,000 e0.06t dt 650,000 e0.06t $757,992.41 0 0.06 0 10 (b) C 650,000 25,000e0.06t dt $837,995.15 0 25,000 b (c) C 650,000 25,000e0.06t dt 650,000 lim e0.06t $1,066,666.67 → 0 b 0.06 0

5 82. (a) C 650,000 25,0001 0.08te0.06t dt 0 1 t 1 5 0.06t 0.06t 0.06t 650,000 25,000 e 0.08 e 2 e $778,512.58 0.06 0.06 0.06 0

10 (b) C 650,000 25,0001 0.08te0.06t dt 0 1 t 1 10 0.06t 0.06t 0.06t 650,000 25,000 e 0.08 e 2 e $905,718.14 0.06 0.06 0.06 0 (c) C 650,000 25,0001 0.08te0.06t dt 0 t t 1 b 0.06t 0.06t 0.06t 650,000 25,000 lim e 0.08 e 2 e $1,622,222.22 b→ 0.06 0.06 0.06 0

2 NI r 83. Let K . Then k rx22+ 1 x P K dx. 2 232 θ c r x r Let x r tan , dx r sec2 d, r2 x2 r sec . 1 r sec2 d 1 dx cos d r2 x232 r3 sec3 r2 1 1 x sin C C r2 r2 r2 x2 Hence, b 1 x P K 2 lim r b→ r2 x2 c K c 1 r 2 r2 c2 2 2 K r c c r2r2 c2 2 NI r2 c2 c . krr2 c2 208 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

GM 84. F dx 85. False. f x 1x 1 is continuous on 2 0 a x 0, , lim 1x 1 0, but b x → GM lim b → 1 b a x 0 dx lim lnx 1 . x 1 b→ 0 0 GM a Diverges

86. False. This is equivalent to 87. True 88. True Exercise 85.

1 b 89. (a) dx lim lnx → (b) It would appear to converge. 1 x b 1 y 1 1 b dx lim 1 2 → 1.00 1 x b x 1 0.75 1 ≤ 0.50 n dx will converge if n > 1 and will diverge if n 1. 1 x 0.25 ⇒ x (c) Let dv sin x dx v cos x −52015 − 1 1 0.25 u ⇒ du dx. x x2 sin x cos x b cos x dx lim dx → 2 1 x b 0 x 1 1 x cos x cos 1 2 dx 1 x Converges

90. (a) Yes, the integrand is not defined at x 2. (b) 5 (c) As n → , the integral approaches 44 . 2 4 (d) I dx 0 n n 2 0 1 tan x −2 I2 3.14159 I4 3.14159 I8 3.14159 I12 3.14159

91. n xn1ex dx 0

b (a) 1 ex dx lim ex 1 → 0 b 0 b 2 xex dx lim ex x 1 1 → 0 b 0 b 3 x2ex dx lim x2ex 2xex 2ex 2 → 0 b 0 b b (b) n 1 xnex dx lim xnex lim n xn1ex dx 0 nn u xn, dv ex dx → → 0 b 0 b 0 (c) n n 1! Section 8.8 Improper Integrals 209

92. For n 1, b x 1 b 2 4 1 1 1 I1 dx lim x 1 2x dx 2 4 b→ lim 2 3 . 0 x 1 2 0 b→ 6 x 1 0 6 For n > 1, x2n1 x2n2 b n 1 x2n3 n 1 I dx lim dx 0 I n 2 n3 → 2 n2 2 n2 n 1 0 x 1 b 2 n 2 x 1 0 n 2 0 x 1 n 2 x 1 Parts: u x 2n2, du 2n 2x 2n3 dx, dv dx, v x2 1n3 2n 2x2 1n2 x 1 b 1 (a) dx lim 2 4 → 2 3 0 x 1 b 6 x 1 0 6 x3 1 x 1 1 1 (b) dx dx 2 5 2 4 0 x 1 4 0 x 1 4 6 24 x5 2 x3 2 1 1 (c) dx 2 6 2 5 0 x 1 5 0 x 1 5 24 60

93. f t 1 94. f t t 1 b 1 1 b Fs est dt lim est , s > 0 Fs test dt lim st 1est b→ → 2 0 s 0 s 0 b s 0 1 , s > 0 s2

95. ft t2 96. f t eat 1 b Fs t 2est dt lim s2 t 2 2st 2est Fs eat est dt etas dt → 3 0 b s 0 0 0 1 b 2 lim et as , s > 0 → s3 b a s 0 1 1 0 , s > a a s s a

97. ft cos at 98. f t sin at Fs est cos at dt Fs est sin at dt 0 0

st b st b e e lim 2 2 s cos at a sin at lim 2 2 s sin at a cos at b→ s a 0 b→ s a 0 s s a a 0 , s > 0 0 , s > 0 s2 a2 s2 a2 s2 a2 s2 a2

99. f t cosh at eat eat 1 Fs est cosh at dt est dt etsa etsa dt 0 0 2 2 0 1 1 1 b 1 1 1 lim et sa et sa 0 b→ 2 s a s a 0 2 s a s a 1 1 1 s , s > a 2 s a s a s2 a2 210 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

100. f t sinh at eat eat 1 Fs est sinh at dt est dt et sa et sa dt 0 0 2 2 0 1 1 1 b 1 1 1 lim et sa et sa 0 b→ 2 s a s a 0 2 s a s a 1 1 1 a , s > a 2 s a s a s2 a2

1 2 101. (a) f x e x 70 18 0.4 102. (a) y 32 3 90 f x dx 1.0 50 90 50 1

−0.2 (b) P72 ≤ x < 0.2525 x -2-1 1 2 (c) 0.5 P70 ≤ x ≤ 72 0.5 0.2475 0.2525 -1 These are the same answers because by symmetry, 1 (b) Area 22 2 P70 ≤ x < 0.5 2 and 1 Arc length is also 22 2. 2 0.5 P70 ≤ x < P70 ≤ x ≤ 72 P72 ≤ x < . Hence, the corresponding integrals are equal. x Let y 4 x2, y 4 x2 4 2 1 y2 ⇒ 1 y2 . 4 x2 4 x2 2 2 2 Thus, 4 x2 dx dx. 2 2 2 4 x (area) (arc length)

1 c b 1 c 103. dx lim dx 2 → 2 0 x 1 x 1 b 0 x 1 x 1 b lim lnx x2 1 c lnx 1 b→ 0 b b2 1 2 c lim ln b b 1 ln b 1 lim ln c b→ b→ b 1 This limit exists for c 1, and you have b b2 1 lim ln ln 2. b→ b 1

cx 1 b cx 1 104. dx lim dx 2 → 2 1 x 2 3x b 1 x 2 3x c 1 b x2 2c2 b b2 2c2 2 c2 lim ln x 2 lnx lim ln 13 lim ln 13 ln 3 b→ 2 3 1 b→ x 1 b→ b This limit exists if c 13, and you have

2 16 b 2 16 16 ln 3 lim ln ln 3 ln 3 . b→ b1 3 6 Section 8.8 Improper Integrals 211

y x ln x, 0 < x ≤ 2 105. fx 0, x 0 2 2 y = x ln x 2 V x ln x dx 1 0

u u x Let u ln x, e x, e du dx. 1 2 2 2 ln ln −1 V e2u u2e4 du e3u u2 du −2 u2 2u 2 ln 2 ln 22 2 ln 2 2 lim e3u 8 2.0155 b→ 3 9 27 b 3 9 27

1 y 106. V ln x2 dx 3 0 1 2 − lim ln x dx y = ln x → b 0 b

1 x 2 lim xln x 2 ln x 2 1 2 b→0 b

lim 2 bln b2 2b ln b 2b b→0 −3 2

107. u x, u2 x, 2u du dx 108. u 1 x, 1 x u2, 2u du dx 1sin x 1sinu2 1 1 cos x 0 cos1 u2 dx 2u du 2 sinu2 du dx 2u du 0 x 0 u 0 0 1 x 1 u Trapezoidal Rule n 5: 0.6278 1 2 cos1 u2 du 0 Trapezoidal Rule n 5: 1.4997

x2 109. (a) 3 (b) Let y e , 0 ≤ x < . ln y x2 ≤ −3 3 x ln y for 0 < y 1 x2 −1 The area bounded by y e , x 0 andy 0 is 1 ex2 dx ln y dy, . 0 0 2

110. Assume a < b. The proof is similar if a > b. y

a a d 3 fx dx fx dx lim f x dx lim f x dx → → 2 a c c d a 1 a b d lim fx dx lim fx dx fx dx x c→ d→ ab c a b -1 a b d -2 lim fx dx fx dx lim fx dx → → c c a d b a b d b d lim fx dx fx dx lim fx dx lim fx dx lim fx dx → → → → c c a d b c c d b b fx dx fx dx b 212 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

Review Exercises for Chapter 8

1 1 1. xx2 1 dx x2 1122x dx 2. xex21 dx ex212x dx 2 2

2 32 1 x 1 1 2 C ex 1 C 2 32 2 1 x2 132 C 3

x 1 2x x 1 3. dx dx 4. dx 1 x2122x dx x2 1 2 x2 1 1 x2 2 1 1 1 x212 lnx2 1 C C 2 2 12 1 x2 C

1 1 5. Let u ln2x, du dx. 6. Let u 2x 3, du 2 dx, x u 3. x 2 e ln2x 1 ln 2 2 1 1 12 dx u du 2x 2x 3 dx u 3 u du 1 x ln 2 32 0 2 u2 1ln 2 1 1 u32 3u12 du 2 ln 2 2 0 1 1 2 1 1 2 ln 2 ln 22 ln 22 u 52 2u32 2 2 5 0 1 1 2 ln 2 1.1931 2 2 2 5 6 5

16 x x4 2x2 x 1 x 7. dx 16 arcsin C 8. 1 16 x2 4 x4 2x2 1 x2 12 x4 2x2 x 1 1 2x dx dx dx x2 12 2 x2 12 1 x C 2x2 1

1 2 9. e2x sin 3x dx e2x cos 3x e2x cos 3x dx 3 3 1 2 1 2 e2x cos 3x e2x sin 3x e2x sin 3x dx 3 3 3 3 13 1 2 e2x sin 3x dx e2x cos 3x e2x sin 3x 9 3 9 e2x e2x sin 3x dx 2 sin 3x 3 cos 3x C 13 1 1 (1)dv sin 3x dx ⇒ v cos 3x (2) dv cos 3x dx ⇒ v sin 3x 3 3 u e2x ⇒ du 2e2x dx u e2x ⇒ du 2e2x dx Review Exercises for Chapter 8 213

10. x2 1ex dx x2 1ex 2xex dx x2 1ex 2xex 2ex dx exx2 2x 1 1

(1)dv ex dx ⇒ v ex (2) dv ex dx ⇒ v ex u x2 1 ⇒ du 2x dx u x ⇒ du dx

2 2 11. u x, du dx, dv x 512 dx, v x 532 12. u arctan 2x, du dx, dv dx,v x 3 1 4x2 2 2 2x xx 5 dx xx 532 x 532 dx arctan 2x dx x arctan 2x dx 3 3 1 4x2 2 4 1 xx 532 x 552 C x arctan 2x ln1 4x2 C 3 15 4 2 4 x 532 x x 5 C 3 15 6 4 x 532 x C 15 3 2 x 5323x 10 C 15

1 1 13. x2 sin 2x dx x2 cos 2x x cos 2x dx 14. lnx2 1 dx lnx2 1 dx 2 2 1 1 1 1 x2 x2 cos 2x x sin 2x sin 2x dx x lnx2 1 dx 2 2 2 2 x2 1 1 x 1 1 1 x2 cos 2x sin 2x cos 2x C x lnx2 1 dx dx 2 2 4 2 x2 1 1 1 1 x 1 (1) dv sin 2x dx ⇒ v cos 2x x lnx2 1 x ln C 2 2 2 x 1 u x2 ⇒ du 2x dx dv dx ⇒ v x 1 (2) dv cos 2x dx ⇒ v sin 2x 2x 2 u lnx2 1 ⇒ du dx x2 1 u x ⇒ du dx

x2 x2 15. x arcsin 2x dx arcsin 2x dx 2 1 4x2 x2 1 22x2 arcsin 2x dx 2 8 1 2x2 x2 1 1 arcsin 2x 2x1 4x2 arcsin 2x C (by Formula 43 of Integration Tables) 2 8 2 1 8x2 1arcsin 2x 2x1 4x2 C 16 x2 dv x dx ⇒ v 2 2 u arcsin 2x ⇒ du dx 1 4x2 214 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

e2x 16. ex arctanexdx ex arctanex dx 1 e2x 1 ex arctanex ln1 e2x C 2

dv ex dx ⇒ v ex ex u arctan ex ⇒ du dx 1 e2x

17. cos3x 1 dx 1 sin2x 1 cosx 1 dx

1 1 sinx 1 sin3x 1 C 3 1 sinx 13 sin2x 1 C 3 1 sinx 13 1 cos2x 1 C 3 1 sinx 12 cos2x 1 C 3

x 1 1 1 1 18. sin2 dx 1 cos x dx x sin x C x sin x C 2 2 2 2

x x x 19. sec4 dx tan2 1 sec2 dx 2 2 2 x x x tan2 sec2 dx sec2 dx 2 2 2 2 x x 2 x x tan3 2 tan C tan3 3 tan C 3 2 2 3 2 2

1 1 20. tan sec4 d tan3 tan sec2 d tan4 tan2 C 4 2 1 or

1 tan sec4 d sec3 sec tan d sec4 C 4 2

1 1 1 sin 1 sin 21. d d d sec2 sec tan d tan sec C 1 sin 1 sin 1 sin cos2

22. cos 2sin cos 2 d cos2 sin2 sin cos 2 d

1 sin cos 3cos sin d sin cos4 C 4 Review Exercises for Chapter 8 215

34 6 23. A sin4x dx. Using the Table of Integrals, 24. A cos3xcos x dx 4 0 sin3x cos x 3 6 1 sin4x dx sin2x dx cos 2x cos 4x dx 4 4 0 2 sin3x cos x 3 1 sin 2x sin 4x 6 xsin x cos x C 4 4 2 4 8 0 3 4 sin3x cos x 3 3 34 3 3 sin4x dx x sin x cos x 0 4 4 8 8 4 8 16 1 9 3 1 3 3 33 16 32 16 16 32 16 16 3 1 1.0890 16 2

12 24 cos d x2 9 3 tan 25. dx 26. dx 3 sec tan d x24 x2 4 sin2 2 cos x 3 sec

3csc2 d 3tan2 d

3 cot C 3sec2 1 d 34 x2 C x 3tan C 2 x 2 sin , dx 2 cos d , 4 x 2 cos x x2 9 3 arcsec C 3 2 x x 3 sec , dx 3 sec tan d, x2 9 3 tan

4 − x 2 x x2 − 9 θ 3

27. x 2 tan x2 + 4 dx 2 sec2 d x 4 x2 4 sec2 θ 2 x3 8 tan3 dx 2 sec2 d 4 x2 2 sec

8tan3 sec d

8sec2 1tan sec d

sec3 8 sec C 3 x2 432 x2 4 8 C 24 2 1 x2 4 x2 4 4 C 3 1 8 x2x2 4 x2 4 C 3 3 1 x2 412x2 8 C 3 216 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 0 1 x 0 28. 9 4x2 dx 9 2x2 2 dx 29. 4 x2 dx 4 arcsin x4 x2 2 2 2 2 2 1 1 2x 1 9 arcsin 2x9 4x2 C 0 4 arcsin1 2 2 3 2 9 2x x 1 arcsin 9 4x2 C 4 4 3 2 2 2 Note: The integral represents the area of a quarter circle 1 2 of radius 2: A 4 2 .

30. Let u cos , du sin d. 2 sin 0 1 d du 2 2 0 1 2 cos 1 1 2u 1 1 du 2 0 1 2u 1 1 1 1 du, a 2 2 0 1 2 u 2 1 1 2 arctan2u 2 0 2 arctan2 2

x3 x2 31. (a) Let x 2 tan , dx 2 sec2 d. (b) dx x dx 4 x2 4 x2 x3 8 tan3 2 dx 2 sec d 2 2 u 4u du 4 x 2 sec u 3 + 2 8 tan sec d 4 x x u2 4 du sin3 θ 8 d cos 4 2 1 u3 4u C 3 81 cos2 cos4 sin d u u2 12 C 3 8cos4 cos2 sin d 4 x2 x2 8 C cos3 cos1 3 8 C 3 1 u2 4 x2, 2u du 2x dx 8 sec sec2 3 C 3 8 4 x2 4 x2 3 C 3 2 4 1 4 x2 x2 8 C 3 x3 (c) dx x24 x2 2x4 x2 dx 4 x2 2 4 x2 x24 x2 4 x232 C x2 8 C 3 3 x dv dx ⇒ v 4 x2 4 x2 u x2 ⇒ du 2x dx Review Exercises for Chapter 8 217

32. (a) x4 x dx 64tan3 sec3 d (b) x4 x dx 2u4 4u2 du

2u3 64sec4 sec2 sec tan d 3u2 20 C 15 64 sec3 24 x32 3 sec3 5 C 3x 8 C 15 15 24 x32 u2 4 x, dx 2u du 3x 8 C 15 x 4 tan2 , dx 8 tan sec2 d, 4 x 2 sec

2x 2 (c) x4 x dx u32 4u12 du (d) x4 x dx 4 x32 4 x32 dx 3 3 2u32 2x 4 3u 20 C 4 x32 4 x52 C 15 3 15 24 x32 24 x32 3x 8 C 3x 8 C 15 15 2 u 4 x, du dx dv 4 x dx ⇒ v 4 x32 3 u x ⇒ du dx

x 28 A B 33. x2 x 6 x 3 x 2 x 28 Ax 2 Bx 3 x 2 ⇒ 30 B5 ⇒ B 6 x 3 ⇒ 25 A5 ⇒ A 5 x 28 5 6 dx dx 5 lnx 3 6 lnx 2 C x2 x 6 x 3 x 2

2x3 5x2 4x 4 4 3 34. 2x 3 x2 x x x 1 2x3 5x2 4x 4 4 3 dx 2x 3 dx x2 3x 4 lnx 3 lnx 1 C x2 x x x 1

x2 2x A Bx C 35. x 1x2 1 x 1 x2 1 x2 2x Ax2 1 Bx Cx 1 3 3 1 Let x 1: 3 2A ⇒ A Let x 0: 0 A C ⇒ C Let x 2: 8 5A 2B C ⇒ B 2 2 2 x2 2x 3 1 1 x 3 dx dx dx x3 x2 x 1 2 x 1 2 x2 1 3 1 1 2x 3 1 dx dx dx 2 x 1 4 x2 1 2 x2 1 3 1 3 lnx 1 lnx2 1 arctan x C 2 4 2 1 6 lnx 1 lnx2 1 6 arctan x C 4 218 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

4x 2 A B 36. 3x 12 x 1 x 12 4 x 2 3Ax 1 3B 2 Let x 1: 2 3B ⇒ B 3 4 Let x 2: 6 3A 3B ⇒ A 3 4x 2 4 1 2 1 4 2 2 1 dx dx dx lnx 1 C 2 lnx 1 C 3x 12 3 x 1 3 x 12 3 3x 1 3 x 1

x2 15 2x 37. 1 x2 2x 15 x2 2x 15 15 2x A B x 3x 5 x 3 x 5 15 2x Ax 5 Bx 3 9 Let x 3: 9 8A ⇒ A 8 25 Let x 5: 25 8B ⇒ B 8

x2 9 1 25 1 9 25 dx dx dx dx x lnx 3 lnx 5 C x2 2x 15 8 x 3 8 x 5 8 8

sec2 1 1 1 38. d du du du tan tan 1 uu 1 u 1 u tan 1 lnu 1 lnu C ln C ln1 cot C tan u tan , du sec2 d

1 A B uu 1 u u 1

1 Au 1 Bu Let u 0: 1 A ⇒ A 1 Let u 1: 1 B

x 1 2 x 24 3x 39. dx ln2 3x C 40. dx 2 3x C (Formula 21) 2 3x2 9 2 3x 2 3x 27 (Formula 4) 6x 8 2 3x C 27

41. Let u x2, du 2x dx. 42. Let u x2, du 2x dx. 2 4 1 x 1 1 1 x 1 1 dx du 2 dx du 2 x u 0 1 sin x 2 0 1 sin u 0 1 e 2 0 1 e 1 4 1 1 tan u sec u u ln1 eu 2 0 2 0 1 1 1 2 0 1 1 ln1 e ln 2 2 2 2 1 2 1 1 ln 2 2 1 e Review Exercises for Chapter 8 219

x 1 1 43. dx lnx2 4x 8 4 dx x2 4x 8 2 x2 4x 8 (Formula 15) 1 2 2x 4 lnx2 4x 8 2 arctan C (Formula 14) 2 32 16 32 16 1 x lnx2 4x 8 arctan1 C 2 2

3 3 1 1 1 1 44. dx 3 dx u 3x 45. dx dx u x 2x9x2 1 2 3x3x2 1 sin x cos x sin x cos x 3 1 arcsec3x C (Formula 33) lntan x C (Formula 58) 2

1 1 1 46. dx dx u x 1 tan x 1 tan x 1 1 x lncos x sin x C (Formula 71) 2

47. dv dx ⇒ v x 48. tann x dx tann2xsec2 x 1 dx 1 u ln xn ⇒ du nln xn1 dx x tann2 x sec2 x dx tann2 x dx

ln xn dx xln xn nln xn1 dx 1 tann1 x tann2 x dx n 1

1 49. sin cos d sin 2 d 2 1 1 1 1 1 cos 2 cos 2 d cos 2 sin 2 C sin 2 2 cos 2 C 4 4 4 8 8 1 dv sin 2 d ⇒ v cos 2 2 u ⇒ du d

csc2x 1 x14 uu3 50. dx 2 csc 2x dx 51. dx 4 du x 2x 1 x1 2 1 u2 2 lncsc2x cot2x C 1 4u2 1 du u2 1 1 u 2x, du dx 1 2x 4 u3 u arctan u C 3 4 x34 3x14 3 arctanx14 C 3 u 4 x, x u 4, dx 4u3 du

5 3 4u 4u 4 52. 1 x dx u4u3 4u du 4u4 4u2 du C 1 x3 23x 2 C 5 3 15 u 1 x, x u4 2u2 1, dx 4u3 4u du 220 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

1 cos x 1 cos x 3x3 4x Ax B Cx D 53. 1 cos x dx dx 54. 1 1 cos x x2 12 x2 1 x2 12 sin x 3 x3 4x Ax Bx2 1 Cx D dx 1 cos x Ax3 Bx2 A Cx B D 1 cos x12sin x dx A 3, B 0, A C 4 ⇒ C 1, B D 0 ⇒ D 0 21 cos x C 3x3 4x x x u 1 cos x, du sin x dx dx 3 dx dx x2 12 x2 1 x2 12 3 1 lnx2 1 C 2 2x2 1

55. cos x lnsin x dx sin x lnsin x cos x dx sin x lnsin x sin x C

dv cos x dx ⇒ v sin x cos x u lnsin x ⇒ du dx sin x

56. sin cos 2 d sin2 2 sin cos cos2 d

1 1 1 sin 2 d cos 2 C 2 cos 2 C 2 2

9 3 x 3 57. y dx ln C (by Formula 24 of Integration Tables) x2 9 2 x 3

4 x2 2 cos 2 cos d 2x2 x 58. y dx 59. y lnx2 xdx x lnx2 x dx 2x 4 sin x2 x 2x 1 csc sin d x lnx2 x dx x 1 lncsc cos cos C 1 2 x lnx x 2dx dx 2 4 x2 4 x2 x 1 ln C x 2 x lnx2 x 2x lnx 1 C x 2 sin , dx 2 cos d, 4 x2 2 cos dv dx ⇒ v x 2x 1 u lnx2 x ⇒ du dx x2 x

sin 60. y 1 cos d d 1 cos 12sin d 21 cos C 1 cos u 1 cos, du sin d

1 5 1 5 1 x 1 61. xx2 432 dx x2 452 62. dx 2 lnx 4 lnx 2 0 2 5 2 5 0 x 2 x 4 2 ln 3 2 ln 4 ln 2 9 ln 0.118 8 Review Exercises for Chapter 8 221

4 ln x 1 4 1 2 e3x 2 1 63. dx ln x2 ln 42 2ln 22 0.961 64. xe3x dx 3x 1 5e6 1 224.238 1 x 2 1 2 0 9 0 9

3 x 22 x 3 4 4 8 65. x sin x dx x cos x sin x 66. dx 1 x 0 0 0 1 x 3 0 3 3 3

4 0 4 1 2 68. A dx 67. A x 4 x dx 4 u u 2u du 2 0 2 0 25 x 0 1 x 5 4 1 1 1 4 2 2 u 4u du ln ln ln 9 0.220 2 10 x 5 0 10 9 10 u5 4u3 0 128 2 5 3 2 15 u 4 x, x 4 u2, dx 2u du

1 69. By symmetry, x 0, A . 70. By symmetry, y 0. y 2 3 1 A 4 5 2 1 1 1 1 4 2 y 1 x22 dx x x3 1 2 1 3 1 3 1 4 4 (3.4, 0) x x 4 1 345 4 −1 x, y 0, − 3 17 2 3.4 − 5 3 x, y 3.4, 0

71. s 1 cos2 x dx 3.82 72. s 1 sin2 2x dx 3.82 0 0

ln x2 21xln x sin x cos x 1 73. lim lim 0 74. lim lim x →1 x 1 x →1 1 x →0 sin 2x x →0 2 cos 2x 2 2

2x 2x 2x 2 x 1 e 2e 4e x 75. lim lim lim 76. lim xe lim x2 lim x2 0 x → x2 x → 2x x → 2 x → x → e x → 2xe

77. y lim ln x2x x → 2 lnln x 2x ln x ln y lim lim 0 x → x x → 1 Since ln y 0, y 1.

78. y lim x 1ln x x →1

ln y lim ln x lnx 1 x →1 1 1 2 ln x lnx 1 x 1 ln2 x x lim lim lim lim x →1 1 x →1 1 1 x →1 x 1 x →1 1 ln x x ln2 x x x2

lim 2xln x 0 x →1

Since ln y 0, y 1. 222 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

0.09 n 0.09 n 79. lim 10001 1000 lim 1 n → n n → n 0.09 n Let y lim 1 . n → n 0.09 0.09n2 ln1 0.09 n 1 0.09n 0.09 ln y lim n ln1 lim lim lim 0.09 n → n n → 1 n → 1 n → 0.09 1 n n2 n 0.09 n Thus,ln y 0.09 ⇒ y e0.09 and lim 10001 1000e0.09 1094.17. n → n

2 2 2x 2 2 ln x 80. lim lim x →1 ln x x 1 x →1 ln xx 1 2 2x lim x →1 x 11x ln x 2x 2 2 lim lim 1 x→1 x 1 x ln x x→1 1 1 ln x

16 1 4 16 32 1 6 b 81. dx lim x34 82. dx lim 6 lnx 1 4 → b→1 0 x b 0 3 b 3 0 x 1 0 Converges Diverges

x3 b e1x b 83. x2 ln x dx lim 1 3 ln x 84. dx lim e1x 1 0 1 → 2 a→0 1 b 9 1 0 x b → a Diverges

1 1 b 85. Let u ln x, du dx, dv x2 dx, v x1. 86. dx lim x14 dx 4 → x 1 x b 1 ln x ln x 1 ln x 1 4 b 34 2 dx 2 dx C lim x x x x x x b→ 3 1 ln x ln x 1 b 4 4 dx lim lim b 34 2 → → 1 x b x x 1 b 3 3 ln b 1 Diverges lim 1 b→ b b 0 1 1

t0 t 500,000 0 0.05t 0.05t 88. V xex 2dx 87. 500,000e dt e 0 0.05 0 0 500,000 0.05t 2 2x e 0 1 x e dx 0.05 0

0.05t 2x b 10,000,0001 e 0 e lim 2x2 2x 1 → b 4 0 4 (a) t0 20: $6,321,205.59 → (b) t0 : $10,000,000

1 2 2 89. (a) P13 ≤ x < ex12.9 20.95 dx 0.4581 0.952 13 1 2 2 (b) P15 ≤ x < ex12.9 20.95 dx 0.0135 0.952 15 Problem Solving for Chapter 8 223

Problem Solving for Chapter 8

1 x3 1 1 4 1. (a) 1 x2 dx x 21 1 3 1 3 3 1 1 2x3 x5 1 2 1 16 1 x22 dx 1 2x2 x4 dx x 21 1 1 3 5 1 3 5 15 (b) Let x sin u, dx cos u du, 1 x2 1 sin2 u cos2 u.

1 2 1 x2n dx cos2 un cos u du 1 2

2 cos2n1 u du 2 2 4 6 2n 2 . . . (Wallis’s Formula) 3 5 7 2n 1 22 42 62 . . . 2n2 2 2 3 4 5 . . . 2n2n 1 222nn!2 22n1n!2 2n 1! 2n 1!

1 1 2. (a) ln x dx lim x ln x b→0 0 b 1 lim b ln b b 1 b→0 ln b 1b Note: lim b ln b lim lim 0 b→0 b→0 b 1 b→0 1b2

1 1 ln x2 dx lim xln x2 2x ln x 2x → 0 b 0 b 2 lim bln b2 2b ln b 2b 2 b→0 (b) Note first that lim bln bn 0 (Mathematical induction). b→0

Also, ln xn1 dx xln xn1 n 1ln xn dx.

1 Assume ln xn dx 1n n!. 0 1 1 1 Then, ln xn1 dx lim xln xn1 n 1 ln xn dx → 0 b 0 b 0 0 n 11n n! 1n1n 1!. 224 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

x c x x c x 1 3. lim 9 4. lim x→ x c x→ x c 4 x c x c 1 lim x ln ln 9 lim x ln ln x→ x c x→ x c 4 lnx c lnx c lnx c lnx c lim ln 9 lim ln 4 x→ 1x x→ 1x

1 1 1 1 x c x c x c x c lim ln 9 lim ln 4 → → x 1 x 1 x2 x2 2c 2c lim x2 ln 9 lim x2 ln 4 x→ x cx c x→ x cx c

2 2 2cx 2cx lim ln 9 lim 2 2 ln 4 x→ x2 c2 x→ x c 2 c ln 9 2 c ln 4 2 c 2 ln 3 2 x 2 ln 2 c ln 3 c ln 2

PB 5. sin PB, cos OB OP AQ AP BR OR OB OR cos The triangles AQR and BPR are similar: AR BR OR 1 OR cos ⇒ AQ BP sin sin OR sin OR cos cos sin OR y sin Q cos sin P lim OR lim →0 →0 sin

sin cos cos θ lim x →0 cos 1 R OBA (1, 0) sin lim →0 cos 1 sin cos lim →0 sin cos cos sin lim →0 cos 2 Problem Solving for Chapter 8 225

6. sin BD, cos OD 1 1 Area DAB DABD 1 cos sin 2 2 1 1 Shaded area 1BD sin 2 2 2 2 DAB 121 cos sin R Shaded area 12 sin 1 cos sin 1 cos cos sin2 lim R lim lim →0 →0 sin →0 1 cos 1 cos sin cos sin 2 sin cos lim →0 sin sin 4 cos sin 4 cos 1 lim lim 3 →0 sin →0 1

7. (a)0.2 Area 0.2986

0 4 0

(b) Let x 3 tan , dx 3 sec2 d, x2 9 9 sec2 . x2 9 tan2 dx 3 sec2 d 2 32 2 32 x 9 9 sec x2 + 9 x tan2 d sec θ 3 sin2 d cos 1 cos2 d cos lnsec tan sin C 4 x2 tan143 Area dx ln sec tan sin 2 32 0 x 9 0 4 x2 9 x x ln 3 3 x2 9 0 5 4 4 4 ln ln 3 3 3 5 5 (c) x 3 sinh u, dx 3 cosh u du, x2 9 9 sinh2 u 9 9 cosh2 u

1 1 4 x2 sinh 4 3 9 sinh2 u sinh 4 3 A dx 3 cosh u du tanh2 u du 2 32 2 32 0 x 9 0 9 cosh u 0 sinh 143 sinh143 1 sech2 u du u tanh u 0 0 4 4 4 16 4 16 sinh1 tanhsinh1 ln 1 tanh ln 1 3 3 3 9 3 9 4 5 4 5 ln tanhln ln 3 tanhln 3 3 3 3 3 3 13 4 ln 3 ln 3 3 13 5 226 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

x 1 u2 1 u2 3 u2 8. u tan , cos x , 2 cos x 2 2 1 u2 1 u2 1 u2 2 du dx 1 u2 2 1 1 1 u2 2 dx du 2 2 0 2 cos x 0 3 u 1 u 1 2 du 2 0 3 u 1 1 u 2 arctan 3 3 0 2 1 arctan 3 3 2 3 0.6046 3 6 9

2x 9. y ln1 x2, y 1 x2 4x2 1 2x2 x4 4x2 1 x2 2 1 y2 1 1 x22 1 x22 1 x2

12 Arc length 1 y2 dx 0 1 2 1 x2 dx 2 0 1 x 1 2 2 1 dx 2 0 1 x 12 1 1 1 dx 0 x 1 1 x 12 x ln1 x ln1 x 0 1 3 1 ln ln 2 2 2

1 ln 3 ln 2 ln 2 2

1 ln 3 0.5986 2 Problem Solving for Chapter 8 227

10. Let u cx, du c dx.

b cb cb 2 2 2 du 1 2 ec x dx eu eu du 0 0 c c 0 2 2 1 2 As b → , cb → . Hence, ec x dx ex dx. 0 c 0 x 0 by symmetry.

ec2x2 2 dx Mx 0 2 y m 2 2 2 ec x dx 0 e2c2x2 dx 1 0 2 2 2 ec x dx 0 1 2 ex dx 1 2c 0 2 1 2 ex dx c 0 1 2 22 4 2 Thus, x, y 0, . 4

1 11. Consider dx. ln x 1 1 1 eu Let u ln x, du dx, x eu. Then dx eu du du. x ln x u u 1 eu If dx were elementary, then du would be too, which is false. ln x u 1 Hence, dx is not elementary. ln x

12. (a) Let y f 1x, f y x, dx fy dy.

f 1x dx yfy dy

u y, du dy yfy f y dy dv fy dy, v f y

x f 1x f y dy

(b) f 1x arcsin x y, f x sin x

arcsin x dx x arcsin x sin y dy

1 x arcsin x cos y C x x arcsin x 1 x2 C y

1 − x2

—CONTINUED— 228 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

12. —CONTINUED— (c) f x ex, f 1x ln x y x 1 ⇔ y 0; x e ⇔ y 1

e e 1 ln x dx x ln x ey dy 1 1 0 1 e ey 0 e e 1 1

13. x4 1 x2 ax bx2 cx d x4 a cx3 ac b dx2 ad bcx bd a c, b d 1, a 2 x4 1 x2 2x 1x2 2x 1 1 1 1 Ax B 1 Cx D dx dx dx 4 2 2 0 x 1 0 x 2x 1 0 x 2x 1 1 2 1 2 x x 1 2 4 1 2 4 dx dx 2 2 0 x 2x 1 0 x 2x 1 2 1 2 1 arctan2x 1 arctan2x 1 lnx2 2x 1 lnx2 2x 1 4 0 8 0 2 2 2 2 arctan2 1 arctan2 1 ln2 2 ln2 2 0 4 8 4 4 4 8 0.5554 0.3116 0.8670

14. (a) Let x u, dx du. 2 2 0 sin u sin x 2 I dx du cos x sin x 0 2 cos u sin u 2 2 2 cos u du 0 sin u cos u Hence, 2 2 sin x cos x 2 I dx dx 0 cos x sin x 0 sin x cos x 2 1 dx ⇒ I . 0 2 4 n 0 sin u 2 (b) I du 2 cosn u sinn u 2 2 2 cosn u du n n 0 sin u cos u 2 Thus, 2I 1 dx ⇒ I . 0 2 4 Problem Solving for Chapter 8 229

15. Using a graphing utility, 1 (a) lim cot x x→0 x 1 (b) lim cot x 0 x→0 x 1 1 2 (c) lim cot x cot x . x→0 x x 3 Analytically, 1 (a) lim cot x x→0 x 1 x cot x 1 x cos x sin x (b) lim cot x lim lim x→0 x x→0 x x→0 x sin x cos x x sin x cos x x sin x lim lim x→0 sin x x cos x x→0 sin x x cos x sin x x cos x lim 0. x→0 cos x cos x x sin x 1 1 1 (c) cot x cot x cot2 x x x x2 x2 cot2 x 1 x2 x2 cot2 x 1 2x cot2 x 2x2 cot x csc2 x lim lim x→0 x2 x→0 2x cot2 x x cot x csc2 x lim x→0 1 cos2 x sin x x cos x lim x→0 sin3 x 1 sin2 xsin x x cos x lim x→0 sin3 x sin x x cos x lim 1 x→0 sin3 x sin x x cos x cos x cos x x sin x Now, lim lim x→0 sin3 x x→0 3 sin2 x cos x x lim x→0 3 sin x cos x x 1 1 lim . x→0 sin x 3 cos x 3 1 1 1 2 Thus, lim cot x cot x 1 . x→0 x x 3 3 The form 0 is indeterminant. 230 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

P P P N x 1 2 . . . n 16. D x x c1 x c2 x cn . . . N x P1 x c2 x c3 . . . x cn P2 x c1 x c3 . . . x cn Pn x c1 x c2 . . . x cn1 Let x c1: N c1 P1 c1 c2 c1 c3 . . . c1 cn Nc 1 P1 c1 c2 c1 c3 . . . c1 cn Let x c2: N c2 P2 c2 c1 c2 c3 . . . c2 cn Nc 2 P2 c2 c1 c2 c3 . . . c2 cn Let x cn: N cn Pn cn c1 cn c2 . . . cn cn1 Nc n Pn cn c1 cn c2 . . . cn cn1 If D x x c1 x c2 x c3 . . . x cn , then by the Product Rule . . . D x x c2 x c3 . . . x cn x c1 x c3 . . . x cn x c1 x c2 x c3 . . . x cn1 and D c1 c1 c2 c1 c3 . . . c1 cn D c2 c2 c1 c2 c3 . . . c2 cn D cn cn c1 cn c2 . . . cn cn1 . Thus, Pk N ck D ck for k 1, 2, . . ., n.

x3 3x2 1 P P P P 17. 1 2 3 4 ⇒ c 0, c 1, c 4, c 3 x 4 13x2 12x x x 1 x 4 x 3 1 2 3 4 Nx x3 3x2 1 Dx 4x3 26x 12

N0 1 P 1 D0 12 N1 1 1 P 2 D1 10 10 N4 111 111 P 3 D4 140 140 N3 1 P 4 D3 42

x3 3x2 1 112 110 111140 142 Thus, . x4 13x2 12x x x 1 x 4 x 3 Problem Solving for Chapter 8 231

50,000 18. st 32t 12,000 ln dt 50,000 400t

16t2 12,000ln 50,000 ln50,000 400t dt

400t 16t2 12,000t ln 50,000 12,000 t ln50,000 400t dt 50,000 400t 50,000 50,000 16t2 12,000t ln 12,000t1 dt 50,000 400t 50,000 400t 50,000 16t2 12,000t ln 12,000t 1,500,000 ln50,000 400t C 50,000 400t s0 1,500,000 ln 50,000 C 0 C 1,500,000 ln 50,000 50,000 50,000 400t st 16t2 12,000t1 ln 1,500,000 ln 50,000 400t 50,000 When t 100, s100 557,168.626 feet.

19. By parts,

b b b f xgx dx f xgx fxgx dxu fx, dv g x dx a a a b fxgx dx a

b b fxgx gx fx dxu fx, dv g x dx a a b fxgx dx. a

20. Let u x ax b, du x a x b dx, dv fx dx, v fx.

b b b x ax b dx x ax b fx x a x b fx dx a a a b u 2x a b 2x a b f x dx a dv f x dx

b b 2x a b f x 2f x dx a a

b 2 f x dx a

1 1 1 1 1 1 2 21. dx < dx < dx 5 10 15 5 5 10 15 2 x x x 2 x 1 2 x x x 1 1 1 b 1 1 1 1 b lim < dx < lim → 4 9 14 5 → 4 9 14 b 4x 9x 14x 2 2 x 1 b 4x 9x 7x 2 1 0.015846 < dx < 0.015851 5 2 x 1