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# EVEN07 - CHAPTER 7 Integration Techniques L’Hôpital’s...

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Unformatted text preview: CHAPTER 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1 Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Section 7.8 Basic Integration Rules . . . . . . . . . . . . . . . . . . . 308 Integration by Parts . . . . . . . . . . . . . . . . . . . . . 312 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 321 Trigonometric Substitution . . . . . . . . . . . . . . . . . 328 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . 336 Integration by Tables and Other Integration Techniques . . 343 Indeterminate Forms and L’Hôpital’s Rule Improper Integrals . . . . . . . . 348 . . . . . . . . . . . . . . . . . . . . . 353 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 CHAPTER 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.1 Basic Integration Rules Solutions to Even-Numbered Exercises d ln x2 dx d 2x dx x2 1 d arctan x dx d ln x2 dx x x2 1 1 2 2. (a) (b) (c) (d) 1 C C C C 1 2x 2 x2 1 x2 1 1 x2 2x x2 1 1 2 x x2 x2 1 1 2x _ 21 x2 3x2 13 2x 2 x2 14 2 dx matches (a). 4. (a) (b) (c) (d) d 2x sin x2 dx d dx 1 sin x2 2 1 1 1 1 C C C C 2x cos x2 1 cos x2 2 1 cos x2 2 1 2x 1 2x 2 sin x2 1 2 2x2 cos x2 1 1 sin x2 1 x cos x2 1 d1 sin x2 dx 2 d dx 2x sin x2 1 2x x cos x2 1 2x 2x cos x2 2 sin x2 1 2 2x2 cos x2 1 sin x2 1 x cos x2 1 dx matches (c). 6. u 2t t2 t2 t 1 2 t du . u dt 2, du 2t 1 dt 8. u 2t 2t 2 12 4 dt 2dt, a 2 10. 2x x2 u Use x2 4 dx 1 2 1, du du a2 . 4, du 2x dx, n Use Use u2 un du. 12. u sec 3x tan 3x dx 3x, du 3 dx 14. u 1 x x2 x, du 4 dx dx, a 2 . Use sec u tan u du. Use u du u2 a2 308 S ection 7.1 16. Let u 6x x 4, du 6 x dx. x 4 6 Basic Integration Rules dt. 2 t 9 2 309 18. Let u 4 5 dx C 6 x 6 4 6 t 2 t 9 9, du 2 4 5 dx C dt dt t 2 9 C 20. Let u 4 x4 2x2, du 2x2 dx 4x dx. 1 4 1 4 6 4 2x2 2x2 12 22. 4x d x C x 3 2x 3 2 dx x dx x2 2 x2 2 3 2 2x 3 1 1 3 C C 2 2 dx 3 2x 2 32 3 2 2x 3 24. Let u x2 x x2 2x 1 2x 4, du 4 dx 1 2 2x x2 x2 1 dx. 2x 2x 4 4 12 2x 1 dx C 26. 2x x 4 1 3x dx 2 dx 8 x 4 dx 2x 8 ln x 4 C 28. 1 1 3x 1 dx 1 1 3 dx 3 3x 1 1 ln 3x 3 1 1 1 3 dx 3 3x 1 1 C 1 3x ln 3 3x 1 1 C 1 ln 3x 3 30. x1 1 x 3 x1 3 x 3 x2 1 dx x3 x 3 3 x 1 dx x2 12 x 2 3x 3 ln x 1 x C 32. sec 4x dx 1 4 sec 4x 4 dx tan 4x C 34. Let u cos x, du sin x dx cos x sin x dx. cos x 2 cos x 12 1 ln sec 4x 4 sin x dx C e e x x 36. Let u cot x, du csc2 xecot x dx csc x dx. ecot x csc2 x dx ecot x C 2 38. 5 3ex 2 dx 5 5 1 3ex e 3 x 2 2e x dx dx 2e x x 5 23 5 ln 3 2 1 2e 2e x dx C 310 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals sin x dx cos x tan x dx 1 cos sin 40. Let u ln cos x , du 42. d csc ln csc d cot cot d ln sin C tan x ln cos x dx ln cos x ln cos x 2 2 tan x dx C 44. 2 3 sec x 1 dx 1 2 3 sec x 1 sec x sec x 1 dx 1 46. 3 t2 1 dt 3 arctan t C 2 sec x 1 dx 3 tan2 x 2 sec x dx 3 tan2 x 2 cos x dx 3 sin2 x 2 3 1 sin x 2 csc x 3 2 cot2 x dx 3 2 3 csc2 x 2 x 3 x C 1 dx C 2 cot x 3 cot x 48. Let u 1 4 3 x, du 3x2 dx 3 dx. 1 34 1 arctan 23 3 3x 2 50. Let u dx C 1 , du t e1 t dt t2 1 dt. t2 e1 t 1 dt t2 e1 t C 3x 2 52. x 1 1 4x2 8x 3 dx 2x 1 2 2x 1 2 1 dx arcsec 2 x 1 C 54. 1 dy dx (a) 1 4x x2 dx 1 5 x2 4x 4 dx 5 1 x 2 2 dx arcsin x 2 5 C a 5 56. tan2 2x , 0, 0 y 1 (b) tan2 2x dx sec2 2x 1 dx 1 tan 2x 2 1.2 x C 0, 0 : 0 −0.6 x 0.6 C 1 tan 2x 2 − 1.2 1.2 y −1 x − 1.2 58. (0, 1) 5 60. r 1 et e t et 2 dt 1 2et et e t e2t dt −2 −2 2 2 et dt 2t et C S ection 7.1 62. Let u y 2x, du 1 x 4x2 arcsec 2x 2 dx. dx C 1 dx. x e 2 Basic Integration Rules cos t dt. 13 sin t 3 0 311 64. Let u 2 2x 2 sin t, du sin2 t cos t dt 1 2x 1 dx 0 0 66. Let u e 1 1 1 x ln x, du ln x dx 1 68. 1 x x 2 2 dx 1 1 x 2 dx x 2 1 1 1 2 ln x ln x 2 1 dx x e 1 2 ln x 1 1 ln 4 0.386 1 2 4 70. 0 1 25 x x2 2 dx arcsin x 5 4 arcsin 0 4 5 0.927 6 72. x2 4x 13 dx 1 ln x2 2 4x 13 4 x2 arctan 3 3 C ` −10 10 The antiderivatives are vertical translations of each other. −6 74. ex 2 e x3 dx 1 3x e 24 9ex 9e x e 3x C 76. sec u tan u du sec u C The antiderivatives are vertical translations of each other. 5 −5 5 −5 78. Arctan Rule: du a2 u2 4 1 arctan a a C 80. They differ by a constant: sec2 x 2 C1 tan2 x 1 2 C1 tan2 x C. 82. f x 5 13 x 5 7x2 10x 84. 0 4 x2 1 dx 4 86. A 0 sin 2x dx 1 cos 2x 2 y f x dx < 0 because 0 Matches (d). y 2 0 1 more area is below the x-axis than above. 3 5 2 1 0 5 x 1 −5 2 3 4 1 2 1 x π 4 π 2 312 2 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 88. 0 2 x2 dx (a) Let f x 2 x2 over the interval 0, 2 . 2 2 x over the (b) Let f x interval 0, 2 . Revolve this region about the x-axis. 2 x over the interval (c) Let f x 0, 2 . Revolve this region about the y-axis. 2 A 0 y 2 x2 dx V 2 x dx 0 2 2 V 2 0 2 x x dx 2 x2 dx 25 20 15 10 f ( x) = 2π x 2 0 2 y x2 dx y 0 3 5 x 1 2 2 f ( x) = 2x 2 f ( x) = x 1 1 x x 1 2 3 1 2 90. (a) (b) 1 n 1 3 3 n 0 sin nx dx 0 3 31 1 0 n sin nx n dx 3 3 1 n cos nx 2 0 1 x2 dx 1 arctan x 6 1 arctan 3 3 x2 3 92. y y 1 y 2 2x 1 x 1 2 0 9 94. y y 2 3x1 1 8 3 1 x 9 x x 1 x x 1 dx 1 32 9 0 1 1 dx y 2 4 9x 2 1 3 S 2x 2x 0 s 1 4 dx 9x2 3 7.6337 2 4 2 x 3 8 10 10 3 y 12 9 6 3 x 3 6 9 12 1 256.545 Section 7.2 2. d2 x sin x dx Integration by Parts 2x cos x 2 sin x x2 cos x 2x sin x 2x sin x 2 cos x 2 cos x x2 cos x. Matches (d) S ection 7.2 d dx 1 x Integration by Parts 313 4. x x ln x 1 x ln x ln x. Matches (a) 6. u x2 e2x dx x2, dv e2x dx 8. u ln 3x dx ln 3x, dv dx 10. u x2 cos x dx x2, dv cos x dx 12. dv u 2 e x x dx ex x dx ⇒ v e dx x dx e x 14. e1 t dt t2 e1 t 1 dt t2 e1 t C ⇒ du 2 xe 2 xe 2xe x x dx x e 2e x5 5 1 dx x x x dx 2 xe x e x C C 1 dx. x ln x 3 16. dv u x 4 dx ⇒ ln x v 18. Let u ln x, du 1 dx x ln x 3 ⇒ du x5 ln x 5 x5 ln x 5 1 dx x 1 2 ln x 2 C x 4 ln x dx x5 1 dx 5x 15 x 25 1 dx x2 1 dx x 1 dx x2 ln x x C x5 ln x 5 x5 5 ln x 25 14 x dx 5 1 C 20. dv u 1 dx ⇒ x2 ln x v 1 x ⇒ du ln x x x ln x dx x2 1 x C 22. dv u x2 x2 x2e x3ex 2 1 x2 2 dx ⇒ v x2 2x3e 2 1 x2 2 x dx x2 1 2 x2 2xe x2 2 1 1 dx ex 2 2 ⇒ du 2 2xe 2 dx x2 1 dx x2ex 2 x2 1 1 dx x2 1 dx x 1 dx x2 xex dx x2ex 2 x2 1 C ex 2 x2 2 1 C 24. dv u 1 dx ⇒ v x2 ln 2x ⇒ du 1 x ln 2x dx x2 ln 2x x ln 2x x 1 x C ln 2x x 1 C 314 Chapter 7 1 2 x x 2 3x dx 3x Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 2 3 26. dv u dx ⇒ v dx 3x 3x 2 3x 1 2 dx 2 3x ⇒ du 2x 2 3 2x 2 3 2 3 4 2 27 2 3x dx 3x 32 C 2 2 3x 9x 27 22 3x C 2 2 3x 3x 27 4 C 28. dv u sin x dx ⇒ x v cos x dx cos x dx x cos x sin x C ⇒ du x cos x x sin dx 30. Use integration by parts twice. (1) u x2, du x2 cos x dx 2x dx, dv x2 sin x cos x dx, v 2 x sin x dx sin x (2) u x, du x2 cos x dx dx, dv sin x dx, v 2 x cos x cos x cos x dx 2 sin x C x2 sin x x2 sin x 2x cos x 32. dv u sec tan d ⇒ v d sec tan d sec 34. dv u dx ⇒ v dx 1 1 x dx x 1 1 x2 x2 ⇒ du sec tan d sec sec arccos x ⇒ du 4 x arccos x 4 x arccos x x2 sec d 4 arccos x dx ln sec tan C dx C 36. Use integration by parts twice. (1) dv u e x dx ⇒ v e x dx ex (2) dv u ex cos 2x 2 ex sin 2x e x dx ⇒ v e x dx ex cos 2x ⇒ du ex cos 2x ex cos 2x ex cos 2x 5 v x 1 dx x 2 sin 2x dx 2 ex sin 2x dx 2ex sin 2x 2 sin 2x C sin 2x ⇒ du 2 cos 2x dx ex cos 2x dx 5 e x cos 2x dx e x cos 2x dx 2 ex cos 2x dx 38. dv u y y dx ⇒ ln x ⇒ du ln x ln x dx x ln x x 1 dx x x ln x x C x 1 ln x C S ection 7.2 40. Use integration by parts twice. (1) dv u (2) dv u y x2 x x x2 x 22 xx 3 22 xx 3 1 1 1 dx 32 Integration by Parts 315 x 1 dx ⇒ v x 2x dx v dx x 1 1 2 dx 2 x 3 1 32 ⇒ du 1 3 2dx ⇒ 1 3 2 dx 2 x 5 1 52 ⇒ du 4 xx 3 8 xx 15 1 1 3 2 dx 22 xx 3 16 x 105 1 1 72 32 42 xx 35 C 2x 1 105 1 52 2 5 x 12x 1 5 2 dx 32 32 52 15x2 8 C 42. dv u y dx arctan ⇒ v dx x 1 x2 x 2 1 dx 2 2x 4 dy dx y x2 2 4 dx x2 dx x 2 18 37 ln 4 x2 C x ⇒ du 2 x dx 2 1 2 arctan x arctan x arctan 44. (a) 4 y (b) e e x3 sin 2x, 0, sin 2x dx x3 −6 x 4 Use integration by parts twice. (1) u sin 2x, du e x3 x3 2 cos 2x 3e 3e x3 −4 dv e (2) u dv e dx, v sin 2x dx x3 sin 2x 6e x3 cos 2x dx cos 2x, du e x3 x3 2 sin 2x 3e 3e x3 dx, v sin 2x dx x3 sin 2x 6 3e x3 cos 2x 6e x3 sin 2x dx C 37 e y e x3 sin 2x dx sin 2x dx 3e 1 37 1 0 37 18e x3 sin 2x x3 18e x3 cos 2x x3 C C x3 3e sin 2x 18e cos 2x 0, y 18 18 : 37 37 x3 18 x3 C⇒C 0 1 3e 37 sin 2x cos 2x 316 dy dx Chapter 7 x sin x, y 0 y Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 46. 4 8 (0, 4) 48. See Exercise 3. 1 1 x2e x dx 0 −5 −2 10 x2e x 2xe x 2e x 0 e 2 0.718 50. dv u sin 2x dx ⇒ x v sin 2x dx dx 1 2 1 cos 2x 2 52. dv u x dx ⇒ v x dx x2 2 ⇒ du arcsin x2 ⇒ du x sin 2x dx 1 x cos 2x 2 1 x cos 2x 2 1 sin 2x 4 2x dx 1 x4 x3 1 1 21 4 1 x4 dx 12 cos 2x dx x arcsin x2 dx C 1 sin 2x 4 x2 arcsin x2 2 x2 arcsin x2 2 x4 x4 1 C C x4 1 0 2x cos 2x 1 sin 2x 4 C 12 x arcsin x2 2 Thus, 0 x sin 2x dx 2x cos 2x 0 2 . Thus, 1 x arcsin x2dx 0 12 x arcsin x2 2 1 4 2. 54. Use integration by parts twice. (1) dv e (2) dv e x x e x, v cos x dx e x e x, u e x cos x, du cos x e x sin x dx sin x dx cos x dx x dx, v e x, u e x sin x, du e cos x dx cos x sin x e x cos x dx ⇒ 2 e x cos x dx e x sin x e x cos x Thus, 2 e 0 x cos x dx e e 2 x sin x 2 2 e x cos x 2 0 sin 2 cos 2 1 2 56. dv u dx ln 1 ⇒ v dx 2x 1 x2 x2 2 x2 x dx 2x2 dx 1 x2 1 1 1 x2 dx x ln 1 x2 2x 2 arctan x C x2 ⇒ du x ln 1 x ln 1 ln 1 x2 dx 1 1 Thus, 0 ln 1 x2 dx x ln 1 x2 2x 2 arctan x 0 ln 2 2 2 . S ection 7.2 58. u x, du x sec2 x dx dx, dv x tan x sec2 x dx, v tan x dx 0 Integration by Parts 317 tan x Hence, 4 4 x sec x dx 2 x tan x ln ln cos x 0 4 4 1 e 2 1 e 8 2x 2 2 0 1 ln 2 2 60. x 3e 2x dx x3 2x 3x2 6x2 1 e 4 2x 6x 3 C 1 e 8 2x 6 1 e 16 2x C Alternate signs u and its derivatives x3 3x2 6x 6 0 v and its antiderivatives e 2x 4x3 6x 1 2x 2e 1 2x 4e 1 2x 8e 1 2x 16 e 62. x3 cos 2x dx x3 1 sin 2x 2 3x2 1 cos 2x 4 6x 1 sin 2x 8 3 cos 2x 8 3 cos 2x C 6 1 cos 2x 16 C Alternate signs u and its derivatives x3 3x2 6x 6 0 1 16 13 x sin 2x 2 13 4x sin 2x 8 32 x cos 2x 4 6x2 cos 2x 3 x sin 2x 4 6x sin 2x v and its antiderivatives cos 2x 1 2 C sin 2x 1 4 1 8 cos 2x sin 2x cos 2x 64. x2 x 2 32 dx 22 xx 5 2 x 315 2 2 52 8 xx 35 35x2 40x 2 72 16 x 315 C 2 92 C Alternate signs u and its derivatives x2 2x 2 0 v and its antiderivatives x 2 5 4 35 8 315 52 32 2 2 32 52 72 92 x x x 2 2 66. Answers will vary. See pages 488, 493. 1 5 68. Yes. u ln x, dv 70. No. Substitution. x dx 72. No. Substitution. 74. 4 sin d 4 cos 4 3 sin 12 2 cos 24 sin 24 cos C 5 76. 0 x 4 25 x2 3 2 dx 1,171,875 arcsin x 5 128 14,381.0699 x 2x2 25 25 16 x2 52 625x 25 x2 64 32 46,875x 25 128 x2 5 0 318 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals x dx ⇒ 2 4 3 78. (a) dv u x x 4 4 v dx x x 32 4 x 12 dx x 32 ⇒ du x dx 2 x4 3 2 x4 3 2 3 4 x x 3 2 dx 32 4 4 15 du u3 C x 52 C 2 4 15 x 32 3x 8 C (b) u 4 x4 x⇒x x dx u u 25 u 5 2 4 and dx 4 u1 2du 83 u 3 x 32 2 2 4u1 2 2 du 20 2 4 15 C x 32 23 u 15 20 3u C 2 4 15 34 3x 8 C 80. (a) dv 4 x dx ⇒ v 4 2 4 3 x 1 2 dx 82. n n n 0: 1: 2: ex dx xex dx x2ex dx ex xex C ex C xex 2ex C ex dx x 32 u x x dx ⇒ du 2 x4 3 2 x4 3 2 4 15 2 4 15 dx x2ex x2ex x 52 2xex x4 x x x x 32 2 3 4 x 3 2 dx 2 xex dx 3x2ex 6xex 6ex C 32 4 4 15 5x 3x 24 8 du C n 3: x3ex dx x3ex x3ex C 3 x2e x dx 32 x C 32 (b) u 4 x4 x⇒x x dx n u and dx 4: x 4ex dx x 4ex 4x3ex 12x2ex 24xex 24ex C 4 4 u 2 u du u3 25 u 5 2 x 4ex du C C 34 8 x C C 4 x3ex dx xnex n xn 1ex dx. 4u1 83 u 3 2 In general, x nex dx (See Exercise 86) 2 23 u 15 2 4 15 2 4 15 2 20 x x 32 3u 20 3x 32 84. dv u xn cos x dx ⇒ xn cos x dx v sin x nxn n 1 86. dv dx 1 eax dx ⇒ v x n ⇒ du 1 ax e a nx n n a 1 ⇒ du xn sin x u sin x dx dx 1 ax xn x neax dx x neax a xn e dx S ection 7.2 88. Use integration by parts twice. (1) dv u eax dx ⇒ v 1 ax e a b sin bx b a eax sin bx dx b2 a2 eax cos bx a eax cos bx dx (2) dv u b eax sin bx a a eax dx ⇒ v Integration by Parts 319 1 ax e a b cos bx cos bx ⇒ du eax cos bx a eax cos bx a sin bx ⇒ du b a eax cos bx dx eax cos bx dx beax sin bx a2 Therefore, 1 b2 a2 eax cos bx dx eax cos bx dx eax a cos bx b sin bx a2 eax a cos bx b sin bx a2 b2 C. 90. n 2 (Use formula in Exercise 84.) x2 cos x dx x2 sin x x2 sin x 2 x sin x dx (Use formula in Exercise 83.) n 2 x cos x cos x dx x2 sin x 2x cos x 1 2 sin x C 92. n 3, a 2 (Use formula in Exercise 86 three times.) x3e2x 2 x3e2x 2 x3e2x 2 e2x 3 4x 8 3 2 x2e2x dx n 3, a n 1 2 C 2 2, a 2 x3e2x 2 3x2e2x 4 3xe2x 4 3e2x 8 Cn 1, a 2 x3e2x dx 3 x2e2x 22 3x2e2x 4 6x2 xe2x dx 3 xe2x 22 6x 3 e2x dx 94. dv u A e x 1 9 1 9 1 9 x3 dx ⇒ v 3e dx x3 96. A 0 x sin x dx x cos x sin x 0 ⇒ du 3 See Exercise 83. 3 xe 0 x3 dx x3 3 3 3xe 9 e 1 e 1 e 3 0 x3 3 0 e x3 dx −1 −1 4 9e 1 0 1 2 e 0.264 0.4 −1 4 − 0.1 320 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals y 98. In Example 6, we showed that the centroid of an equivalent region was 1, 8 . By symmetry, the centroid of this region is 8, 1 . You can also solve this problem directly. 1 π 2 A 0 2 arcsin x dx 2 2 x x arcsin x 0 1 1 1 x2 1 Example 3 0 1 x 2 8 x y My A Mx A 1 x 0 1 0 2 2 arcsin x dx arcsin x 2 2 2 arcsin x dx 1 2 100. (a) Average 1 4 1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t 1 4 3.2 ln 2 0.2 2.018 (b) Average 3 1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t 3 12.8 ln 4 7.2 ln 3 1.8 8.035 102. c t 5 30,000 30,000 0 500t, r 500t e t, dv 60 t 0.07t 7%, t1 dt 500 5 5 P 60 0 te 0.07t dt 0.07t. Let u P 60 500 e 0.07t dt, du 0.07t dt, v 5 0 5 100 e 7 100 7 5 100 e 7 100 e 7 e 0 0.07t dt 5 500 60 t 0.07t 0 10,000 e 49 2 sin nx n3 n 0.07t 0 \$131,528.68 104. x 2 cos nx dx x2 sin nx n 2 cos n n2 4 cos n n2 4 4 1 n4 n2 2x cos nx n2 2 cos n2 n2 , if n is even n2 , if n is odd 106. For any integrable function, f x dx C f x dx, but this cannot be used to imply that C 2 2 0. 108. On 0, 2 , sin x ≤ 1 ⇒ x sin x ≤ x ⇒ 0 x sin x dx ≤ 0 x dx. Section 7.3 110. f x cos x, f 0 2 (b) You obtain the points n 0 1 2 3 xn 0 0.05 0.10 0.15 2 2.05 2.098755 2.146276 0 0 Trigonometric Integrals 321 (a) It cannot be solved by integration. yn 4 4 80 4.0 2.8403565 Section 7.3 2. (a) y Trigonometric Integrals sec x tan x sin x sec2 x. (b) y cos x sec x ⇒ y sin x sin x sin x 1 sec x tan x sec2 x sin x sec2 x sec x ⇒ y Matches (iii) sin x tan2 x Matches i (c) y x tan x 13 tan x ⇒ y 3 1 sec2 x tan2 x tan2 x sec2 x tan2 x 1 tan2 x tan4 x Matches iv (d) y y 3x 3 3 2 sin x cos3 x 2 cos x cos3 x 3 sin x cos x ⇒ 6 sin x cos2 x cos2 x sin x 3 cos2 x 3 cos2 x 31 3 sin2 x cos2 x 2 cos4 x 6 cos2 x 1 8 cos4 x Matches ii 4. cos 3 x sin4 x dx cos x 1 sin4 x sin5 x 5 sin2 x sin4 x dx sin6 x cos x dx sin7 x 7 C 6. Let u cos x, du sin3 x dx sin x dx. sin x 1 cos2 x 1 cos3 x 3 cos2 x dx sin x dx cos x C sin x dx 8. Let u x sin , du 3 cos3 x dx 3 3 1 x cos dx. 3 3 cos 1 x 3 x 3 x 3 1 sin2 x 3 sin2 x dx 3 1 x dx cos 3 3 C C 3 sin 3 sin 1 3x sin 3 3 sin3 x 3 322 Chapter 7 sin5 t dt cos t Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 10. sin t 1 sin t 1 cos t cos2 t 2 cos t 12 dt 12 2 cos2 t 12 cos4 t cos t 32 dt sin t dt 2 cos t 12 2 cos t cos t 72 4 cos t 5 1 1 4 1 4 1 4 13 42 3 8 52 2 cos t 9 1 92 C 12. sin2 2x dx 1 1 4x 8 cos 4x dx 2 sin 4x 1 x 2 C 1 sin 4x 4 C 14. sin4 2 d cos 4 2 1 1 3 2 cos 4 d 2 cos2 4 1 d d d C C 2 cos 4 2 cos 4 2 cos 4 1 sin 4 2 1 sin 4 8 cos 8 2 1 cos 8 2 1 sin 8 16 1 sin 8 64 16. Use integration by parts twice. dv u dv u sin2 x dx x2 ⇒ du sin 2x dx ⇒ x 1 cos 2x ⇒v 2 2x dx v 1 cos 2x 2 dx sin 2x 12 x sin 2x 4 12 x sin 2x 4 12 x sin 2x 4 6x2 sin 2x 1 2 13 x 3 1 2 2x2 1 2 x sin 2x dx x sin 2x dx 1 2 cos 2x dx C C 2 2 x 2 sin 2x 4 1 2x 4 sin 2x ⇒ du 12 x 2x 4 13 x 2 13 x 6 13 x 6 1 4x3 24 x2 sin2 x dx 1 x cos 2x 2 1 x cos 2x 4 6x cos 2x 1 sin 2x 8 3 sin 2x 18. Let u sin x, du 2 cos x dx. 2 20. 0 sin2 x dx 1 2 1 0 cos 2x dx 2 0 cos5 x dx 0 0 2 1 1 0 sin2 x 2 cos x dx 2 sin2 x 23 sin x 3 sin4 x cos x dx 15 sin x 5 2 0 1 x 2 1 sin 2x 2 4 sin x 8 15 S ection 7.3 1 tan 2x 2 Trigonometric Integrals 323 22. sec2 2x 1 dx 1 C 24. sec6 3x dx 1 1 tan2 3x 2 sec2 3x dx 2 tan2 3x tan4 3x sec2 3x dx 1 tan5 3x 15 C 1 tan 3x 3 x x sec2 dx 2 2 23 tan 3x 9 1 x tan4 2 2 26. tan2 x dx sec2 x 1 dx tan x x C 28. tan3 C 30. Let u tan3 2t sec 2t, du sec3 2t dt 2 sec 2t tan 2t sec2 2t sec4 2t sec5 2t 10 1 sec3 2t tan 2t dt 32. tan5 2x sec2 2x dx 1 tan6 2x 12 C sec2 2t sec 2t tan 2t dt sec3 2t 6 C 34. sec2 x x tan dx 2 2 2 sec sec2 x 2 x1 x x sec tan dx 22 2 2 C x1 x sec2 dx 22 2 C 36. tan3 3x dx sec2 3x 1 tan 3x dx 1 3 C 3 sin 3x dx cos 3x 1 tan 3x 3 sec2 3x dx 3 12 tan 3x 6 1 ln cos 3x 3 or sec2 x x tan dx 2 2 2 tan tan2 x 2 38. tan2 x sec5 x sin2 x cos2 x sin2 x sin2 x 1 sin2 x sin3 x 3 cos5 x dx cos3 x dx sin2 x cos x dx sin4 x cos x dx sin5 x 5 C 40. s sin2 1 1 4 1 8 1 2 16 2 cos2 2 d 1 1 8 C C cos 2 1 d cos 2 d 1 cos2 4 d cos 2 d sin2 sin 2 2 sin 2 42. y tan x sec4 x dx tan1 2 x tan2 x tan5 2 x 2 72 tan x 7 1 sec2 x dx tan1 2 x sec2 x dx 2 32 tan x 3 C 324 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals dy dx y 1 4 tan x, du sec2 x dx 44. (a) 1 y (b) sec2 x tan2 x, 0, sec2 x tan2 x dx u tan3 x 3 C 1 4 −1 x 1 () 0, 4 1 −1 y 0, 1 : 4 C⇒y 13 tan x 3 1 4 46. dy dx 3 y tan2 x, y 0 8 3 48. cos 4 cos 3 d 1 2 cos 4 cos 3 d cos 7 sin 2 cos C d (0, 3) −1 −2 1 sin 7 14 50. sin 4x cos 3x dx 1 2 1 2 sin 4x cos 3x dx sin x cos x sin 7x dx 1 cos 7x 7 cos 7x C C 52. Let u x tan , du 2 tan4 x 1 sec2 dx. 2 2 tan4 2 x x tan2 2 2 x 2 tan4 1 sec2 x 2 x dx 2 x x sec4 dx 2 2 tan6 x 1 sec2 dx 2 2 C 1 7 cos x 14 2 7x tan 7 2 cot3 t dt csc t cos3 t dt sin2 t cos t dt sin2 t 1 sin t csc t sin t sin t 1 2 5x tan 5 2 54. u cot 3x, du 3 csc2 3x dx 1 3 cot 3x 3 csc2 3x dx C 56. sin2 t cos t dt sin2 t csc2 3x cot 3x dx cos t dt C C 12 cot 3x 6 58. sin2 x cos2 x dx cos x 1 sec t dt cos t 1 1 2 cos 2 x dx cos x sec x 2 cos x dx ln sec x tan x 2 sin x C 60. cos t 1 dt cos t 1 cos t sec t dt ln sec t tan t C 62. 0 3 3 tan2 x dx 0 sec2 x 3 1 dx tan x x 0 3 3 64. Let u tan t, du 4 sec2 t dt. 2 32 tan t 3 4 0 66. 2 3 sin 3 cos d 1 2 sin 4 11 cos 4 24 sin 2 d 1 cos 2 2 0 sec2 t tan t dt 0 S ection 7.3 2 2 Trigonometric Integrals 325 68. 2 sin2 x 1 dx 2 2 2 1 3 2 cos 2x 2 1 dx 3 x 2 1 sin 2x 4 2 2 1 cos 2x dx 2 3 2 tan2 1 2 x 70. sin2 x cos2 x dx 2 1 4x 32 sin 4x C 72. tan3 1 x dx 2 ln cos 1 x C −6 6 −3 3 −2 −2 74. sec4 1 x tan 1 2 x dx sec4 1 4 x 2 C 76. 0 1 cos 2 d 3 2 3 4 2 sin 2 1 sin 2 4 2 0 − 3.5 3.5 −2 2 78. 0 sin6 x dx 1 5x 82 2 sin 2x 3 sin 4x 8 13 sin 2x 6 2 0 5 32 80. See guidelines on page 500. 82. (a) Let u tan x, du sec2 x tan x dx Or let u sec x, du sec2 x d x. 12 tan x 2 C1 −4 −2 4 (b) 8 sec x tan x dx. 1 sec2 x 2 1 C C 12 tan x 2 y sec x sec x tan x dx (c) 1 sec2 x 2 C 1 tan2 x 2 1 2 C 12 tan x 2 C2 84. Disks Rx rx V tan x 0 4 1 1 2 2 0 4 tan x dx sec x 0 4 2 2 −1 2 −1 π 8 π 4 x 2 2 2 1 dx tan x 1 x 0 4 1.348 326 Chapter 7 2 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 2 86. (a) V 0 2 cos2 x dx cos x dx 0 2 1 0 2 cos 2x d x 1 2 x 1 sin 2x 2 2 0 2 4 (b) A Let u x sin x 0 x, dv 2 cos x dx, du dx, v 2 sin x. 2 2 x cos x dx 0 x sin x 0 0 sin x dx x sin x cos x 0 y 2 1 2 2 y 1 2 1 4 2 cos2 x dx 0 2 1 1 0 cos 2x d x 2 0 1 2 − (π 2 2 , π ( 8 1 x 4 x, y 2 1 sin 2x 2 2 , 8 8 π 4 π 2 x 88. dv u cos x dx ⇒ cosn 1 v sin x n n n n 1 x ⇒ du cosn cosn cosn 1 1 cosn 1 1 1 x sin x 2 x sin x dx 2 cosn x dx x sin x x sin x x sin x cosn cosn cosn cosn cosn n n x sin2 x dx x1 x dx 1 1 cos2 x dx n cosn cosn 2 1 2 1 2 1 x dx cosn x dx Therefore, n cosn x dx cosn x dx 1 x sin x n 2 2 n x dx. 90. Let u secn 2 x, du n secn secn secn 2 2 secn x tan x x tan x x tan x x tan x secn 2 x tan x dx, dv n n n n 2 2 2 n n 2 secn secn sec2 x dx, v 2 tan x. secn x dx x tan2 x dx 1 dx secn 2 2 2 x sec2 x 2 secn x dx secn 2 1 2 x dx n 1 secn x dx secn x dx secn 1 n 2 x dx 2 1 x tan x secn x dx 92. cos4 x dx cos3 x sin x 4 1 cos3 x sin x 4 3 4 cos2 x dx 3 cos x sin x 8 cos3 x sin x 4 3 x 8 C 3 cos x sin x 4 2 1 2 cos3 x sin x 8 1 2 dx 3x C 3 cos x sin x S ection 7.3 cos3 x sin3 x 6 cos3 x sin3 x 6 1 cos3 x sin3 x 6 1 8 cos3 x sin3 x 48 96. (a) n is odd and n ≥ 3. 2 Trigonometric Integrals 327 94. sin4 x cos2 x dx 1 2 1 2 cos2 x sin2 x dx cos3 x sin x 4 1 cos3 x sin x 8 6 cos3 x sin x 1 4 cos2 x dx x 2 3x C C 1 cos x sin x 8 2 3 cos x sin x cosn x dx 0 cosn n n n n n n n n n n n n 1 2 3 2 3 4 5 1 1 1 1 1 1 x sin x n 2 0 n n 2 0 1 0 2 cosn n n 3 2 2 0 6 2 x dx 2 cosn 3 x sin x n2 n n n n n n 1 n n 4 5 n n 3 2 3 2 3 2 3 2 3 2 n n cosn 0 4 x dx 2 cosn 5 x sin x n4 n n n n n n 5 4 2 n n x dx 5 4 cosn 0 6 x dx cosn 0 2 5... 4 cos x dx 0 2 0 5... sin x 4 5... 1 4 (Reverse the order) 6 ... n 1 7 n 6 ... n 1 7 n 2 (b) n is even and n ≥ 2. 2 cosn x dx 0 n n n 1 1 n 1 n 2 1 2 1 2 3 4 n n n n n n 3 4 3 2 3 2 3 2 n n n n n n 5... 4 cos2 x dx 0 (From part (a).) 2 0 5... x 4 2 5... 4 4 1 sin 2x 4 n (Reverse the order) 5 ... n 1 6 n 2 5 ... n 1 6 n 328 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.4 2. d 8 ln dx x2 Trigonometric Substitution 16 x 1 x x2 2 16 C 8 x x2 16 1 2 x 16 x 16 x2 x2 16 16 x 1 x 2 x x 2 16 x2 2 1 2 x2 x2 16 16 8x x 16 2 2x 16 2 x2 x2 16 2 x2 16 x2 2 x Indefinite integral: x2 x2 x 3 16 7 2 Matches (d) 16 4. x3 d 8 arcsin dx 4 6x x2 C 8 1 8 16 x x2 x x 6x 1 x 3 34 2 2 1 4 1 x 2 3 x x2 2 2 3 3 7 x 6x x 2 x2 3 1 2 2 7 6x x2 x 2 16 16 x3 16 3 2 16 2 16 2 16 16 7 Indefinite integral: 7 6x x2 dx Matches (c) 6x 9 2 16 32 x 32 3 x2 2 6x 9 6. Same substitution as in Exercise 5. 10 x2 25 x2 dx 10 5 cos 25 sin2 d 5 cos 2 csc2 5 d 2 5 cot C 2 25 5x x2 C 8. Same substitution as in Exercise 5 x2 dx 25 x2 25 sin2 5 cos 25 2 5 cos d C x 5 25 2 25 2 25 5 x2 1 cos 2 sin cos C d C 1 x 25 arcsin 2 5 x 25 x2 C 1 sin 2 2 25 x arcsin 2 5 10. Same substitution as in Exercise 9 x2 x 4 dx 2 tan 2 sec 2 tan 2 sec tan C 2 d x2 2 2 4 tan2 d x 2 2 sec2 C 1d x2 4 2 arcsec x 2 C arcsec Section 7.4 12. Same substitution as in Exercise 9 x3 x2 4 dx 8 8 3 8 sec3 2 tan 1 x2 2 2 sec tan tan2 4 sec2 3 x2 4 d 4 d 8 sec4 8 tan C 1 3 d tan3 3 x2 C 4 12 8 tan 3 x2 4 3 C Trigonometric Substitution 329 tan2 1 3 x2 C 4 x2 8 C 14. Same substitution as in Exercise 13. 9x3 dx 1 x2 9 tan3 sec sec2 sec2 d 3 9 C sec2 31 1 sec tan d x2 1 x2 3 9 sec3 3 C sec 31 x2 x2 C 2 C 3 sec 16. Same substitution as in Exercise 13 x2 dx 1 x2 2 1 2 x2 1 1 x2 dx 4 d x 1 x2 1 tan2 1 2 sec2 sec4 d sin2 d 1 2 C sin cos 1 arctan x 2 C x 1 x2 C cos 2 sin 2 2 1 x2 1 arctan x 2 18. Let u x, a 1 x 9 x2 1, and du x2 dx 1 2 9 24. Let u 16 x 16 26. Let u 1 1 t t2 4x2, du 4x2 dx t 2, du 32 dx. x2 ln x 1 x2 C 1 25 x2 1 x1 2 9 x2 x2 12 20. dx 12 2x dx (Power Rule) 22. dx arcsin x 5 C C 8x dx. 1 8 2t dt. 1 2 3 sec2 2 3 sec 8 9 cos sin4 1 t2 32 16 4x2 12 8x dx 1 16 12 4x2 32 C 2 4 3 x2 32 C dt 2t dt 1 1 3 sec . t2 C 28. Let 2x 3 tan , dx 4x2 x4 9 dx d, 4x2 9 4x 2 + 9 2x 3 2 sec2 d 3 2 4 tan4 d C C 32 θ 3 8 27 sin3 8 csc3 27 4x2 9 27x3 C 330 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 2 sec2 d, sec2 4x2 16 4 sec . 30. Let 2x 4 tan , dx 16 dx 1 x 4x2 2 2 tan 1 4 d 4 sec 1 4 cot csc d C 1 ln 4 x2 x 4 2 C θ x2 + 4 x sec tan d 2 1 ln csc 4 32. Let x x2 3 tan , dx 1 3 32 3 sec2 d , x2 3 sec2 d 3 3 sec3 1 3 cos d C C 3 3 sec2 . dx x2 + 3 x θ 3 1 sin 3 x 3 x2 34. Let u x x2 1 2x x2 2, du 2x 2x 2 dx 3 2 dx. 1 2 x2 2x 2 12 2x 2 dx 12 x 3 2x 2 32 C 36. Let x 1 sin , x x x dx sin2 , dx cos 2 sin cos d , 1 x cos . 2 sin cos d sin 1 x 2 cos2 d 1 cos 2 sin cos arcsin x 38. Let x x4 tan , dx x3 sec2 d , x2 1 4 4x3 x4 2x2 2x2 1 1 1 x1 1 d C x C θ 1− x sec2 . 4x 1 1 1 2 1 2 1 dx 1 x2 sec2 sec4 cos 2 sin cos x x2 1 d C C d 1 2 x1 dx 2x2 1 dx x2 + 1 x 1 ln x 4 4 1 ln x2 2 1 ln x2 2 1 ln x2 2 θ 1 arctan x Section 7.4 1 1 x2 1 2 1 1 2 1 4 x2 x2 2 Trigonometric Substitution 331 40. u arcsin x, ⇒ du x arcsin x dx dx, dv x dx ⇒ v x2 arcsin x 2 cos d , x2 arcsin x 2 x2 arcsin x 2 x2 arcsin x 2 x2 dx 1 x2 cos x2 arcsin x 2 C x2 1 4 1 1 4 cos 2 d sin cos 1 arcsin x C x1 x2 C sin2 cos d cos 1 sin 2 2 x1 x sin , dx x arcsin x dx x2 arcsin x 2 C 1 arcsin x 4 1 1 2 1 2x2 4 x2 cos . 42. Let x 1 sin , dx cos d , 1 1 x2 x x dx 1 2 2x x2 dx 2x x2 1 sin 2 cos d cos 2 sin 2 sin 2 cos 2 cos 1 1 sin2 d d C cos x2 x2 x C 1 x 2 3 1 C 2x x2 C x−1 θ 1 − (x − 1)2 1 3 2 3 2 3 2 1 cos 2 2 1 sin 2 4 1 sin 2 3 arcsin x 2 3 arcsin x 2 44. Let x 3 2 sec , dx x 6x dx 2 2x 1 2 2x 2 sec tan d , x 3 x dx 3 2 4 2 tan . 3 2 sec tan d x 2 5 x 2 4 2 sec 2 tan 2 sec2 2 tan 2 x 3 sec 3 ln sec 3 2 3 d tan 3 ln x 2 3 x2 6x C1 3 x 3 2 2 4 4 C1 x2 6x 5 3 ln x 5 C 332 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 46. Same substitution as in Exercise 45 (a) 1 1 t2 52 dt cos d cos5 13 tan 3 tan sec4 d C t3 31 t 232 tan2 1 3 t 1 t 1 32 14 t 2 0 3 1 sec2 d t 1 32 t 2 t2 C 32 Thus, 0 1 1 t2 52 dt 3 38 31 432 (b) When t 32 0 3 3 23 3.464. 0, 1 1 t2 0. When t 52 3 2, 13 tan 3 tan 0 3. Thus, 3 dt 1 3 3 3 3 23 3.464. 48. (a) Let 5x 9 3 sin , dx 25x2 dx 3 cos d , 5 3 cos 91 5 9 10 9 10 9 25x2 3 cos . 3 cos d 5 cos 2 d 2 1 sin 2 2 sin cos 5x 3 C C 9 3 5x 9 9 25x2 C 25x2 35 5x 9 arcsin 10 3 35 Thus, 0 9 25x2 dx 9 5x arcsin 10 3 3 , 5 9 10 . 0 9 10 2 9 . 20 (b) When x 35 0, 9 0. When x 25x2 dx 2 2 Thus, 0 sin cos 0 9 10 2 9 . 20 50. (a) Let x x2 x2 3 sec , dx 9 dx 3 sec tan d , 3 tan 9sec2 tan2 sec sin2 cos 1 d d d cos tan x2 3 9 d sin x2 9 3 tan . 3 sec tan d cos2 cos sec ln sec ln —CONTINUED— x 3 C x2 x 9 C Section 7.4 50. —CONTINUED— 6 Trigonometric Substitution 333 Hence, 3 x2 9 dx x2 3, ln x 3 6, x2 3 . 9 x2 x 9 6 ln 2 3 3 3 . 2 (b) When x 6 0; when x 3 Hence, 3 x2 9 dx x2 1 x 4 3 ln sec tan sin 0 ln 2 3 3 . 2 52. x2 2x 11 32 dx 1 x2 2x 26 x2 2x 11 75 ln 2 x2 2x 11 x 1 C 54. x2 x2 4 dx 13 2 xx 4 x2 4 1 x x2 2 2x dx sec 4 2 ln x x2 4 C 56. (a) Substitution: u 1, du (b) Trigonometric substitution: x 58. (a) x2 y k 2 25 5 50 5 y (b) Area square 25 1 4 1 circle 4 5 2 Radius of circle k2 k 52 52 52 (0, k) 5 25 1 r2 1 4 4 (c) Area 5 x r2 12 r 4 60. (a) Place the center of the circle at 0, 1 ; x2 d y 1 2 1. The depth d satisfies 0 ≤ d ≤ 2. The volume is V 3 6 2 0 1 y 1 1 2 dy d 1 arcsin y 2 1 y d 1 1 3d 1 1 1 1 d y 1 2d 2 1 2 0 (Theorem 7.2 (1)) 1 3 arcsin d 3 2 (b) 10 arcsin d 2. 3 arcsin d d (d) V dV dt 6 0 1 dd dt y 61 1 1 2 dy d 1 2 dV dd dt 1 4 0 0 2 ⇒d t 9.4248 cubic meters. The (e) 9 4 0 0 0.3 (c) The full tank holds 3 horizontal lines y 3 ,y 4 3 ,y 2 24 1 d 1 2 intersect the curve at d 0.596, 1. 0, 1.404. The dipstick would have these markings on it. 2 The minimum occurs at d part of the tank. 1, which is the widest 334 Chapter 7 h Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals r cos d , r2 x h 2 62. Let x r sin , dx r cos . y Shell Method: h r V 4 h r 2 x r2 h 2 x r sin h 2 dx h−r h h+r x 2 4 4 r2 h 2 r cos cos 2 2 r cos d r d 2 4 r2 2 h r sin cos2 d 2 1 2 sin cos2 d 2 r 2 x−h 2 r 2h 1 sin 2 2 4 2 r3 cos3 3 2 2 2 r 2h θ r 2 − (x − h)2 64. y s 12 x ,y 2 4 x, 1 x2 dx ln 4 y 2 1 x2 4 1 0 1 x x2 2 17 1 ln x x2 1 0 (Theorem 7.2) 1 4 17 2 9.2936 66. (a) Along line: d1 Along parabola: y a a2 a4 a1 2x a2 y x2, y (a, a 2) d2 0 1 4x2 dx a 1 2x 4x2 4 1 2a 4a2 4 (b) For a For a 1, d1 10, d1 y = x2 1 1 ln 2x ln 2a 5 2 4x2 4a2 1 0 (Theorem 7.2) (0, 0) x 1 5 1.4789. 2 and d2 10 101 d2 1 ln 2 4 100.4988 101.0473. (c) As a increases, d2 68. (a) 25 d1 → 0. (b) y 0 for x 72 − 10 −5 80 (c) y x 72 x2 ,y 72 1 0 1 x ,1 36 x 36 1 2 y 2 1 72 1 x 36 1 1 2 s 36 2 18 1 x 36 2 ln dx 36 0 x 36 2 1 dx 36 1 1 x 36 18 ln 2 72 0 1 1 x 36 2 ln 1 x 36 2 1 2 2 ln 1 36 2 2 2 1 1 82.641 Section 7.4 70. First find where the curves intersect. y2 16 16 2 16 2 16x 2 x4 xx A 0 Trigonometric Substitution y 6 335 x 4 2 14 x 16 4 2 (4, 4) 4 2 16 x 128x 16x2 4x 1 4 x4 x4 0 0, 4 4 x −2 −2 −4 −6 2 4 6 10 162 128x y= 16 − (x − 4) 2 4 x2 4 32 ⇒ x 4 2 12 x dx 4 x 13 x 12 x 16 4 0 16 3 2 4 4 My 0 12 x dx 4 4 0 8 8 x x 8 4 4 2 dx 8 4 x4 16 16 16 4 x 4 4 x 16 4 dx 4 4 16 x 4 112 3 4 x x 4 2 dx 8 1 16 3 13 16 3 112 x 24 x5 5 4 2 2 232 4 2 16 arcsin 16 x 64 3 4 2 4 16 x 4 2 4 2 16 8 2 1 16 2 x 6 4 16 dx 16 Mx 0 dx 4 1 32 32 5 x y x, y 72. Let r 1 R My A Mx A 38 4 8x 0 64 112 3 16 3 64 6 16 4 32 112 16 54 416 15 48 12 28 4 1.55 12 3 4.89 416 15 16 3 4 4.89, 1.55 L tan , dr R 0 104 3 L sec2 d , r 2 dr 2mL R 2m RL b b a L2 L2 sec2 . r 2 + L2 r 2mL r 2 L2 32 L sec2 d L3 sec3 θ L cos d a b a 2m sin RL 2m RL r r2 R L2 0 2m L R 2 L2 336 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 0.8 74. (a) Finside 48 1 0.8 0.8 y2 1 1 y2 dy y 2 dy 0.8 96 0.8 1 y1 1 y 2 dy 0.8 96 (b) Foutside 64 0.8 arcsin y 2 0.4 y1 1 y 2 dy y2 y 2 dy 0.4 1 1 3 y2 32 1 96 1.263 121.3 lbs 0.4 1 0.4 y2 1 1 128 0.4 128 y1 1 y 2 dy 0.4 0.4 arcsin y 2 15.8t2 y1 y2 1 1 3 y2 32 1 92.98 76. S (a) 60 1520.4 111.2t 78. False x2 x 1 dx tan sec tan2 sec tan d d 0 0 7 (b) S t S5 1 1520.4 2 2.71 1 2 111.2t 15.8t2 12 111.2 31.6t 12 (c) Average value S t dt 10 68.24 80. True 1 1 2 2 x2 1 1 x2 dx 2 0 x2 1 x2 dx 2 0 sin2 cos cos d 2 0 sin2 cos2 d Section 7.5 2. 4x 2 3 x 53 2x x x2 1 4x 2 9 1 When x When x 4x2 2x A 2x 3 2, Partial Fractions A B 5 A x Bx x2 1 3 2x 3 x 5 C 1 2 C x Dx x2 A 3 2x 3 3 2x 5 E 12 B 3 3 x 4. x x2 4x 2 3 x x2 1x A 3 x 1 x B 3 6. 1 1 2 8. 10. x x2 4x 1 3 dx x 1 x x1 1x 3 dx 3 dx 3 C B 2x 1 6. ln x 1 6A, A 3 6B, B 2, 1 9 dx 1 6 1 6. 1 1 2x 3 dx 1 2x ln 2x C 3 3 dx C 1 ln 2x 12 1 2x ln 12 2x 3 3 3 Section 7.5 5x2 xx 5x2 12 x 2x 12x 0, x3 x3 x2 x x 2x When x x3 x2 3 2 1 2, x x 12 2 12 12 A x Ax 2 Partial Fractions 337 12. B x 4 2 x C 2 2 Cx x 2, 2 16 4 x A x 2 x 2 8B ⇒ B dx 3 ln x 2. When x 2 ln x 2 2, 32 4 ln x 8C ⇒ C 2 C 4. Bx x When x 5x2 4A ⇒ A 3 dx x 3. When x x 2 dx 2 12 x 12 dx 4x 14. x Ax 3 1 1 x 2x 1 2x 1 Bx 2 x 1 B 1 3A, A x x2 2 x 1. When x 1 1 x ln x 2 2 1, 3 1 x 1 3B, B dx 1 C 1. 3 dx 2 ln x x2 2 2x x x ln x 2 x 2 C 16. x xx x 2 4 2 A x Ax 4, 6 0, 2 2 dx 4x 4 Bx B x 4 3 2. 1 2. 18. 3 12 3 A x Ax 1, B 3 dx 12 1 1 x B 1 B 1. When x 2 x 2 ln x 1 1 x 1 x 1 0, A 1 1 2 2 2x When x When x x x2 4A, A 4B, B When x 2x x 2. dx C 32 x4 3 ln x 2 4 12 dx x 1 ln x 2 C 20. x3 4x2 x2 x 4x2 1 4x2 x2 x Ax 1 1 2 x Bx 1 1x x2 1 4x2 1x Cx 1, 4 1 dx A 1 1 4A ⇒ A 2 x 2 x 1 1 2 B 1 x 1 x C 1 2 x When x x3 1, 4 4x2 x2 x 2C ⇒ C 1 dx 1 x ln x 1 2. When x dx 3 x 3 ln x 1. When x 0, 0 1 B 2⇒B 3. dx 1 1 C 22. 6x x3 8 6x When x 6x x3 8 x A x2 6x 2 x2 2x 2x 4 A 4 Bx x 2 x2 2 Cx 1. When x x2 1 ln x2 2 1 ln x2 2 Bx 2x C 4 2, 12 dx 12A ⇒ A 1 x ln x ln x 2 dx 0, 0 4 x 2C ⇒ C 1 2 dx x2 2. When x x1 dx 2x 4 C 1 C 1, 6 7 x2 B 3 2x 2 1 1 ⇒B 3 dx 1. x2 dx 2x 4 2x 2x 4 4 2 2 3 x1 arctan 3 3 3 arctan 3x 3 338 Chapter 7 x2 x 9 x2 9 2 x2 x 9 Ax x2 Ax Ax3 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals B 9 Cx x2 9 9A D 92 Cx Cx D 9B D 0, B 1, D 0, and C 1. 24. B x2 Bx2 By equating coefficients of like terms, we have A x2 x 9 x2 9 2 1 x2 9 dx x x2 1 2 x2 Bx 1 x2 x2 2x 3 0, 7 1 x 1 2x 9 C 3 Bx 3A 9 2 dx C x 1 arctan 3 3 x 2 4x 7 1 x 2 2x x When x A 2, B x2 x3 x2 2 26. A 3 7 x A x 4x 1, 12 1, C 4x x Cx 1 1, 4 2A 2B 2C. Solving these equations we have 6A. When x 1. 7 3 dx 2 C. When x x x2 2x 2x 1 3 3 dx dx C 2 ln x 1 1 ln x 2 2 28. x2 x 3 x2 3 2 x2 x 3 Ax x2 Ax Ax 3 B 3 Cx x2 3 3A D 32 Cx Cx D 3B D 0, B 1, 3A C 1, 3B D 3. Solving these equations we have B x2 Bx 2 By equating coefficients of like terms, we have A A 0, B 1, C 1, D 0. x2 x x4 6x 2 3 dx 9 1 x2 3 x2 x 1 arctan 3 3 x 3 2 dx C 1 2 x2 3 30. x1 x2 x 1 x 1 A x Ax x B x2 1 C x 1 Bx 1 Cx 2 1, C 5 1 When x 0, B 1. When x B 1, C 2. 5 1 2. When x 5 1, 0 2A 2B C. Solving these equations we have A 2, x x2 1 dx x1 5 2 1 1 dx x 1 x x x 1 4 5 1 dx x2 2 ln x 1 x 5 1 2 1 1 x 5 1 dx 2 ln x 2 ln 2 ln 5 3 1 1 S ection 7.5 1 Partial Fractions 339 32. 0 x2 x2 x x 1 1 1 dx 0 dx 0 2x x2 1 x x 1 1 2 x 1 dx x ln x 2 x 1 0 1 ln 3 34. 6x 2 x2 x 1 dx 13 1 2 1 2 3 ln 2 1 x x 7 2 1 7 1 2 2x 3 ln 1 1 2 2 C − 4.7 4 2, 1 : 3 ln C 1⇒ C (2, 1) 4.7 −4 36. x3 x2 3, 4 : 4 2 dx 2 5 1 ln x 2 2 C 4 4⇒C 2 x2 22 5 4 C 1 ln 5 2 8 1 ln 5 2 (3, 4) −8 −2 8 38. x3 x 2x 6x 2 1 9 12x 5 8 C dx 2 ln x 2 4 1 x 2 x 5 2 2 10 C (3, 2) − 10 −3 10 3, 2 : 0 2⇒C 40. x2 x3 2, 6 : x2 x x 2 1 dx 0 C arctan x 6⇒C ln x 6 1 C 10 arctan 2 arctan 2 −2 −2 (2, 6) 5 42. Let u uu cos x, du 1 1 1 A u Au sin x dx. B u 1 1 Bu 1, B 1 uu 1 u ln u ln u u 1 1 1 1 du ln u C 1, u cos x. 44. 1 uu 1 1 When u When u A u Au B u 1 1 ,u Bu tan x, du sec2 x dx 0, A 1. 1, 1 B⇒B 1 uu 1 u 1. 1 u ln u u C C du 1 1 du 1 C When u 0, A du sin dx. 1. When u sec2 x dx tan x tan x 1 sin x dx cos x cos2 x du 1 du u C ln u ln ln u 1 cos x 1 ln cos x ln 1 sec x tan x tan x 1 C C 340 Chapter 7 ex, du 1 1u Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals ex dx. A 1 1 u A u2 1 1 Bu u2 C 1 Bu Cu 1 46. Let u u2 When u When u When u e 2x 1, A 0, 1 1, 1 e 1 ex x 1 2. A 2A 1 C. 2B dx 1 2 2C. Solving these equations we have A u2 1 u 1 1 1 1 1u du 1 du u u2 1 ln u2 2 ln e 2x 1 du 1 1 1 arctan u 2 arctan e x C C 1 2, B 1 2, C 1 2, u e x, du e x dx. 1 ln u 2 1 2 ln e x 4 1 a2 x2 1 When x When x 1 a2 x2 a Aa A x x a B x Ba x 48. 50. 1 x2 a bx 1 When x When x When x A x Ax a B x2 bx C a bx Ba bx Cx 2 a, A 1 2a. a, B 1 2a. dx 1 2a 1 2a 1 a ln a x x x x C a 1 x ln a dx 0, 1 Ba ⇒ B 1 a. a b, 1 C a2 b2 ⇒ C b2 a2. 1, 1 a bA a bB C ⇒ A b a 2. b a2 x b ln x a2 1 ax 1 ax 1a x2 1 ax b2 a2 dx a bx b ln a a2 C C bx C 1 x C xa 2 bx dx 1 a ln 2a a a bx b ln a2 x x b ln a2 a bx A2 52. dy dx x2 8 4 2x 3 ,y 0 5 54. (a) (b) Nx Dx Nx Dx A1 px q px q 2 ... px Am q m A1 B1x ax2 bx c ... An Bn x ax2 bx c n −1 −2 3 3 y 56. A 2 0 3 1 dx 0 7 16 3 x2 1 16 x x dx dx x2 3 0 5 2 2 3 2 2 14 0 2x 6 14 4 ln 8 4 7 ln 7 4 x (From Exercise 46) −3 −2 −1 1 2 3 2.595 S ection 7.5 58. (a) 6 5 4 3 2 1 t 1 2 3 4 5 6 5 y Partial Fractions 341 (e) k 1, L 3 5: y 1 :y 2 5 15 2e 3t i y0 ii y 0 12 32 5 2e 3t 1 3 5e 3t (b) The slope is negative because the function is decreasing. (c) For y > 0, lim y t t→ 3. B L y y By ⇒ A 1 ,B L 1 L (f ) 0 0 5 (d) dy yL y 1 dy yL y 1 L 1 dy y 1 ln y L A y AL dy dt d 2y dt 2 ⇒y ⇒ ky L ky dy dt y y dy dt L L 2 y dy dt L y dy dt 0 k dt 1 L ln L ln y L y y dy y kt k Lt y L y y L y y0L y0 3 k dt C1 LC1 From the first derivative test, this is a maximum. C2ekLt When t 0, y0 L y0 C2 ⇒ y0 L y0 e y0 ekLt. Solving for y, you obtain y L x2 kLt . 3 60. (a) V 0 2x x2 1 2 dx 4 0 3 x2 1 x2 1 1 2 dx 1 x2 1 2 4 0 dx x x2 2 1 3 (partial fractions) (trigonometric substitution) 0 4 2 —CONTINUED— arctan x arctan x 1 arctan x 2 x x2 1 3 arctan 3 0 3 10 5.963 342 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 60. —CONTINUED— 3 (b) A 0 2x x2 3 0 3 1 2x2 dx dx ln x2 1 ln 10 3 0 1 0 3 ln 10 2 y x 1 A x2 1 2 0 2 x2 1 dx 1.521 (1.521, 0.412) 1 1 2x ln 10 y 11 A2 2 ln 10 3 0 3 0 2 arctan x 2x x2 1 x2 1 x2 1 2 2 3 ln 10 2 ln 10 1 1 2 3 0 arctan 3 x2 x2 1 2 x 1 −1 2 3 dx dx (partial fractions) dx x x 3 2 3 2 arctan x ln 10 21 arctan x ln 10 2 x, y 1.521, 0.412 1 arctan x 2 x 2 x2 1 1 (trigonometric substitution) 0 0 1 arctan x ln 10 x x2 1 3 0 1 arctan 3 ln 10 3 10 0.412 62. (a) 1 z0 y0 y0 z0 1 y0 x 1 y0 1 x z0 1 z0 ln z0 y0 x A x dx x x C y0 kt x C z0 B x ,A 1 z0 y0 ,B 1 z0 y0 (Assume y0 z0) kt 1 z0 kt z0 e z0 C, when t ln z0 y0 0, x 0 y0 1 z0 y0 ln z0 y0 x x ln ln y0 z0 z0 y0 y0 z0 z0 y0 z0 y0 x x x x x y0 k t y0 kt y0z0 e z0 y0 kt 1 z0e z0 y0 kt y0 (c) If y0 z0, then the original equation is 1 y0 y0 x 0 when t 1 y0 y0 x x x As t → kt x 2 (b) (1) If y0 < z0, lim x t→ y0. z0. (2) If y0 > z0, lim x t→ dx 1 k dt kt 1 y0 C1 C1 1 x 0⇒ 1 y0 k t y0 y0 y0 kt y0 1 y0 y0 kt y0 1 x0. , x → y0 Section 7.6 Integration by Tables and Other Integration Techniques 343 Section 7.6 Integration by Tables and Other Integration Techniques 2, a 5 5 5 4x 2x 4 x ln 5 2x 5 5 5 C C 2. By Formula 13: b 2 1 dx 3 x2 2x 5 2 21 3 25 x x 8 ln 375 2x 5 2 4x 75 x 2x 4. By Formula 29: a 1 3 x2 x 9 dx 3 1 3 x2 9 arcsec x 3 C 6. By Formula 41: x 9 x4 dx 1 2 2x 32 x2 2 dx 1 x2 arcsin 2 3 1 dx 2x x 2 cos 3 x 1 dx 2x 2 sin x cos 2 3 x 2 C C 8. By Formulas 51 and 47: cos3 x x dx 2 2 cos3 cos2 x x sin 3 u 10. By Formula 71: 1 1 dx tan 5x 1 5 1 x, du 1 dx 2x 1 5 dx tan 5x ln cos u ln cos 5x sin u sin 5x C C 11 u 52 1 5x 10 u 5x, du 5 dx 1 ,b 2 ex 14 4 e 17 2 12. By Formula 85: e x2 a 2 1 sin 2x 2 1 sin 2x 2 2 cos 2x 2 cos 2x C C sin 2x dx 4 x2 14. By Formulas 90 and 91: ln x 3 dx x ln x x ln x 3 3 ln x 2 dx 2 ln x 2 3 3 3x 2 3 ln x ln x 6 x5 25 2 C C 15 x ln x 5 x ln x x5 52 6 ln x 16. (a) By Formula 89: x 4 ln x dx 1 4 1 ln x x 4 dx, v x5 25 C x5 5 C C (b) Integration by parts: u x 4 ln x dx x5 ln x 5 ln x, du 1 dx, dv x x5 ln x 5 x5 1 dx 5x 344 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 75, x 1 x ln 2 75 x 3x ln 30 x 75 75 u, and 75 75 C x x C (b) Partial fractions: 1 x2 75 1 75: 1 75: 1 1 x2 75 dx x Ax A 75 75 x Bx 1 2 75 B 75 75 1 10 3 3 30 x 75 75 3 30 dx 75 C 3 30 18. (a) By Formula 24: a 1 x2 75 dx 2A 75 ⇒ A 2B 75 ⇒ B 3 30 75 x 3x ln 30 x x 1 2 2 3 20. By Formula 21: x dx x 1 x C 22. By Formula 79: u arcsec 2x dx 2x, du 1 2x arcsec 2x 2 2 dx ln 2x 4x 2 1 C 24. By Formula 52: x sin x dx x2 3x 5 2 sin x x cos x C 26. By Formula 7: dx 1 3x 27 25 3x 5 10 ln 3x 5 C 28. By Formula 14: x2 1 2x 2 dx 2 2x 2 arctan 2 4 C arctan x 1 C 30. By Formula 56: 2 32. By Formula 71: d 1 31 1 tan 3 1 t1 u ln t 2 1 sin 3 1 sin 3 3 3 2 d C u 1 ex dx tan e x e x, du e x dx 1x e 2 ln cos e x sin e x C sec 3 34. By Formula 23: dt 1 dt t 1 1 ln t 2 1 dt t arctan ln t C ln t, du 36. By Formula 26: 3 x 2 dx 1 x x2 2 1 27 3 3 ln x x2 3 C 38. By Formula 27: x2 2 3x 2 dx 3x 2 2 2 3x 3 dx 2 9x 2 4 ln 3x 2 9x 2 C 2 1 3x 18x 2 8 27 2 S ection 7.6 2 3 Integration by Tables and Other Integration Techniques 3 2 2 345 40. By Formula 77: x arctan x3 2 dx arctan x3 2 x dx ln 1 x3 C 2 32 x arctan x3 3 ex e 2x e x, du ex 1 e 2x 42. By Formula 45: u 44. By Formula 27: 2x 3 2 1 32 dx C e x dx 2x 3 2 4 dx 1 2 2x 3 2 2x 3 2 3)2 2 4 2 dx 2x 3 2 1 2x 8 u 2x 3, du 2 dx 3 2x 4 ln 2x 3 2x 3 2 4 C 46. By Formula 31: u 3 3 x dx x 3 cos x sin2 x sin x, du 3 9 1 9 x 1 dx ln sin x sin2 x 1 C cos x dx 48. x dx x2 dx 2 9 x2 x 9 C x dx 2 50. By Formula 67: tan3 d tan2 2 tan2 2 tan d ln cos x C x 3 arcsin 3 52. Integration by parts: w un a bu du 2u n b 2u n b 2u n b 2u n b Therefore, 2n 1 a a a a u n, dw bu bu bu bu un a 2 bu du nun 1 du, dv 1 du ,v a bu bu du bu a a 2 b a bu 2n n u b 2n n u b a a 1 bu du bu 2n au n 1 b a 2na b un a 1 bu n du bu bu du 2n na un a un a 1 bu bu du 2n ua b bu du and un a bu 2n 1b un a bu na un a 1 bu du . 54. w un cos u du un, dv un sin u cos u du, dw n un nun 1 sin u du du, v sin u 1 346 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 56. ln u n du u ln u n n ln u n 1 1 u du u 1 du, v u u ln u n n ln u n 1 du w ln u n, dv du, dw n ln u n 1 u 58. x x2 0, 0 : 2x dx 1 3 ln 1 6 1 2 x2 6 C 2x 32 3x 0 1 x2 2x 15 3 ln x 1 x2 2x C 0⇒C −6 (0, 0) 6 − 15 3 60. 0, 2 2x x1 2: 2 x2 dx 3 ln 2 3 2x 2 x2 C 3 ln 3 x 2 1 2x x2 C 2 (0, 2 ) − 2.5 1 2⇒C 3 ln 3 −6 62. cos sin 1 sin d 1 sin 2 1 sin 1 sin 2 1 sin ln 1 sin cos 10 C 0, 1 : C 1⇒y ln 1 sin cos 1 −8 (0, 1) −2 8 64. sin 1 cos2 d 1 sin cos 2 2 d C 66. 0 3 1 2 cos 1 d 0 3 1 arctan cos 2u 1 u2 2 1 u2 1 u2 1 du 1 2 0 5u2 1 2 arctan 5 2 arctan 5 u tan 2 1 sec tan 1 1 cos cos sin sin d 5 5u 0 68. 1 cos cos d 1 cos 1 cos cos 1 cos cos2 sin2 d d 70. d sin cos d cos 1 cot2 csc2 C d u 1 sin , du 1d ln 1 csc cot csc cot csc cot C cos d S ection 7.6 2 Integration by Tables and Other Integration Techniques 347 72. A 0 x 1 2 y e x2 dx 1 2 1 2 0 2x dx 2 1 ex 2 1 4 12 x 2 1 4 2 ln 1 ex 2 x 1 2 0 ln 1 e4 1 ln 2 2 0.337 square units 1 du, u u 74. Log Rule: ex 1 76. Integration by parts 78. Formula 16 with u e2x 5 80. A reduction formula reduces an integral to the sum of a function and a simpler integral. For example, see Formula 50, 54. 82. W 0 500x dx 26 x 2 5 250 0 26 x2 5 12 2x dx 500 26 500 26 1 lbs x2 0 2049.51 ft 1 2 0 2 0 84. 1 5000 e4.8 1.9t dt 2500 1.9 2 0 1 1.9 dt e4.8 1.9t 2 2500 4.8 1.9 2500 1 1.9 2500 3.8 1.9 1.9t ln 1 e4.8 1.9t 0 ln 1 e 4.8 ln 1 e4.8 ln 1e 1 e4.8 401.4 k 20 86. (a) 0 6x 2e x 2 dx 50 5.51897. 0 5.52 −1 By trial and error, k 5.51897 (b) 0 6x 2e x2 dx 88. True 348 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals Section 7.7 2. lim 1 x 0.1 0.9516 ex 1 Indeterminate Forms and L’Hôpital’s Rule 1 x →0 −1 1 x fx 0.01 0.9950 0.001 0.9995 0.001 1.00005 0.01 1.005 0.1 1.0517 −2 4. lim x→ 6x 3x2 2x 1 6 10 3.4641 exact: 6 3 103 104 3.4642 105 3.4641 0 5 x fx 102 3.4757 3.5857 3.4653 100 0 6. (a) lim 2x2 x 2x2 x x 1 x 1 3 3 x→ 1 x→ 1 lim 2x x 3x 1 1 x→ 1 lim 2x lim 3 4x 1 1 5 5 (b) lim x→ 1 x→ 1 lim d dx 2x2 x 3 d dx x 1 x→ 1 8. (a) lim x →0 sin 4x 2x 2x 4x2 2x 4x2 x x 1 1 x 1 x 2 x →0 lim 2 sin 4x 4x 2x 4 21 2 (b) lim x →0 sin 4x 2x x →0 lim d dx sin 4x d dx 2x x →0 lim 4 cos 4x 2 2 10. (a) lim x→ x→ lim lim 1 x2 1x 1 x 0 4 x→ 0 lim 2 8x 1 0 (b) lim x→ x→ d dx 2x d dx 4x2 2x 1 1 12. lim x2 x→ 1 x→ 1 lim 3 14. lim x →2 4 x x2 2 x →2 lim x x 4 4 1 x2 x2 x →2 lim 16. lim x →0 ex 1 x3 x x →0 lim ex 1 3x2 ex 6x 2 18. lim ln x x → 1 x2 1 x→1 lim 2 ln x x2 1 2x 2x 1 x2 4 1 x →0 lim x→1 lim x→1 lim 20. lim sin ax x → 0 sin bx x x2 x2 ex 2x a cos ax x → 0 b cos bx lim 1 3 2x ex a b 1 2x 2 ex 2 22. lim arctan x x →1 x x3 x 1 1 3x2 1 x →1 lim 11 1 x2 1 2 24. lim x→ x→ lim 0 26. lim x→ x→ lim 28. lim x→ x→ lim x→ lim 0 Section 7.7 x2 x2 1 x 1 1x 2 Indeterminate Forms and L’ H ô pital ’s Rule sin x x 349 30. lim x→ x→ lim 32. lim x→ 0 . Note: Use the Squeeze Theorem for x > 1 x ln x 4 x3 4 ln x x3 4 3x3 0 4x 3x2 ex 2 x ≤ sin x 1 ≤ x x 1 2 ex 1 2 34. lim x→ x→ lim x→ lim 36. lim x→ x→ lim x→ lim 38. (a) lim x3 cot x x→0 0 x→0 40. (a) lim x tan x→ 1 x x→ 0 lim lim tan 1 x 1x 1 x2 sec2 1 x 1 x2 1 x 1 (b) lim x3 cot x x→0 1 lim x3 tan x x→0 lim 3x2 sec2 x 0 (b) lim x tan x→ 1 x (c) 0 x→ 3 x→ −1 lim sec2 (c) 2 1 10 −1 42. (a) lim e x x→0 x x→0 2x 1 x ex x 1 1 2 x. 44. (a) lim 1 x→ 1 x x→ x 1 1x . x 1 x x→ (b) Let y ln y lim lim ex 2 ln 2 ex x ex e4 x (b) Let y ln y 4 lim 1 lim x→0 x→ x ln 1 1 x2 1x 1 x2 lim ln 1 x→0 lim 1x 1x Thus, ln y (c) 60 4⇒y 54.598. x→ lim 1 x→ lim 1 1 1x 1 Thus, ln y x→ 0 0 2 1⇒y 1 x x e1 e. e. Therefore, lim 1 (c) 5 0 −1 10 350 Chapter 7 x x→ 1x Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 0 46. (a) lim 1 x→ 48. (a) lim 3 x x→4 4 x 4 00 4 x 4. (b) Let y ln y lim 1 lim ln 1 x x 1 x. (b) Let y ln y x→4 lim 3 x lim x x x→4 4 ln 3 x 4 4 4 4 4 4 x 4 2 4 x→ x→ lim 11 x 1 x→4 lim lim ln 3 x 1x 1x 1x x 0 1. 1. x→4 Thus, ln y 0 ⇒ y e0 Therefore, lim 1 x 1 x x→ x→4 x→4 lim 0 1. (c) 5 Hence, lim 3 x (c) 2 0 −1 10 4 0 7 x 50. (a) lim x→0 cos 2 lim x cos 00 x 52. (a) lim x →2 1 x2 1 x2 4 4 x x2 x x2 1 4 1 4 x →2 (b) Let y ln y 0 x→0 2 2 x x . (b) lim x →2 lim lim 1 x2 x 4 12x 2x 1 1 1 8 x→0 lim x ln cos 0 x →2 0 x→0 x x →2 lim 1 4x x 1 Hence, lim (c) 2 cos 2 x 1. (c) 0.25 2 4 − 0.25 0 0 3 54. (a) lim x→0 10 x 10 x 3 x2 3 x2 x→0 (c) lim 10x 3 x2 0 10 5 (b) lim x→0 −20 56. (a) 2 (b) Let y x →0 sin x x, then ln y lim ln sin x 1x x →0 x ln sin x . cos x sin x 1 x2 x →0 x →0 lim lim − x2 tan x x x →0 lim 2x sec2 x 0 2 Therefore, since ln y −1 0, y 1 and lim sin x 1. Section 7.7 Indeterminate Forms and L’ H ô pital ’s Rule x3 e2x 3x2 2e2x 6x 4e2x 6 8e2x 351 58. (a) −1 1 5 (b) lim x→ x→ lim x→ lim x→ lim 0 −3 60. See Theorem 7.4. x3 e2x ln x x3 2 x→ 62. Let f x 6x 4e2x 6 8e2x x 25 and g x x. 64. lim x→ x→ lim 3x2 2e2x x→ lim x→ lim 0 66. lim x→ lim lim lim 2 ln x x 3x2 2 ln x 3x3 2x 9x2 x→ 68. lim x→ xm enx x→ lim lim mxm 1 nenx mm 1 xm n2enx m! nmenx 2 x→ x→ x→ lim 2 9x3 0 ... x→ lim 0 70. x ex x5 1 2.718 5 0.047 10 0.220 20 151.614 30 4.40 105 40 2.30 109 1.66 50 1013 100 2.69 1033 72. y x→ x x, x > 0 and lim x x x→0 74. y 1 lim x x ln x x Horizontal asymptote: y dy dx x1 x x2 x e ln x 1 0 (See Exercise 29) 1 ln x x2 0 No horizontal asymptotes ln y 1 dy y dx dy dx x ln x x 1 x ln x ln x x e 1 1 Critical number: Intervals: Sign of dy dx: 0 y 0, e e, Decreasing xx 1 f x : Increasing e, 1 e Critical number: Intervals: Sign of dy dx: y Relative maximum: e 1, 0 Increasing 1e 1 0, e 1 −1 4 f x : Decreasing (e, 1( e 1e −4 Relative minimum: e 1, e 4 11 , ee −1 −1 ( 1, 1 e ee (( 1 ( 4 76. lim x→ sin x x 1 0 (Numerator is bounded) 78. lim x→ e 1 x 0 x e 1 0 0 . Limit is not of the form 0 0 or L’Hôpital’s Rule does not apply. . Limit is not of the form 0 0 or L’Hôpital’s Rule does not apply. 352 Chapter 7 r n nt Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 80. A P1 ln A ln P nt ln 1 r n ln 1 ln P 1 nt 1 rn 1 n2t r n ln 1 n→ lim r n 1 nt n→ n→ lim r n2 1 n→ lim rt 1 1 r n rt Since lim ln A n→ ln P rt, we have lim A n→ e ln P r n rt eln Pert Pert. Pert. Alternatively, lim A n→ lim P 1 r n nt n→ n r rt lim P 1 82. Let N be a fixed value for n. Then x→ lim xN ex 1 x→ lim N 1 xN ex 2 x→ lim N 1N ex 2 xN 3 ... x→ lim N ex 1! 0. (See Exercise 68) 84. f x k k k lim 1, xk k 1 fx fx fx 1 x x 0.1 0.1 x 0.01 1 0.01 k →0 6 k=1 1 1 10 x 0.1 100 x 0.01 ln x 1 1 −2 −2 k = 0.1 k = 0.01 10 0.1, 0.01, xk k k →0 lim xk ln x 1 86. f x 1 ,g x x f2 g2 f1 g1 12 3 1 6 2c3 c x2 4, 1, 2 fc gc 1 c2 2c 1 2c3 6 3 88. f x ln x, g x f4 g4 f1 g1 ln 4 63 3c3 ln 4 c3 x3, 1, 4 fc gc 1c 3c 2 63 21 ln 4 3 1 3c 3 3 c 21 ln 4 2.474 90. False. If y y x2ex ex x2, then 2xex x 4 92. False. Let f x xex x 2 x4 ex x 2 . x3 x→ x and g x 1 x→ x 1. Then x 1 1. lim x x 1, but lim x S ection 7.8 e 0, gx lim x →0 x e 1 x2 1 x2 Improper Integrals 353 94. g x , x x 0 0 1.5 y g0 Let y g0 0 x →0 lim e 1 x2 x 0.5 x , then ln y ln x lim x → 0 1 x2 1 ln e 1 x2 x x →0 1 x2 1x 2 x3 ln x lim x2 2 e 1 x2 ln x x2 0 . Since x −2 −1 − 0.5 1 2 x →0 lim x2 ln x lim x →0 we have lim x2 ln x x2 x →0 . Thus, lim y x →0 0⇒g 0 0. Note: The graph appears to support this conclusion—the tangent line is horizontal at 0, 0 . gx 96. lim f x x →a 98. lim x ln 2 x →0 1 ln x 1 ln x y ln y x →a fx gx Let y x ln 2 ln y , then: ln x ln 2 ln x 1 ln x x →0 g x ln f x lim g x ln f x , and hence y . Thus, lim y x→ 1 ln 2 ln x As x → a, ln y ⇒ x →a . Thus, x →0 lim ln y x →0 lim ln 2 ln x 1 ln x ln 2 lim ln 2 x 1x lim f x gx x→0 lim ln 2 2. eln 2 Section 7.8 Improper Integrals 3. 4 32 2. Infinite discontinuity at x 4 3 x 1 3 dx b →3 lim b x 2x 2 3 3 3 2 dx 4 b →3 lim 12 b b →3 lim 2 b3 Diverges 4. Infinite discontinuity at x 2 0 1. 1 x 1 1 23 dx 0 x 1 1 b 2 23 dx 1 x 1 1 lim 23 2 dx 1 1 2 b→1 lim 0 x 1 1 1 23 b dx c→1 c x 1 23 dx 0 3 3 0 6 b→1 lim 33 x 0 c→1 lim 33 x c Converges 354 Chapter 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 6. Infinite limit of integration. 0 0 8. 0 e x dx b 0. You need to evaluate the limit. b e2x dx b→ lim b e2x dx 1 2x e 2 0 b b→ lim 0 e x dx b→ lim e e x 0 b b→ lim 1 2 0 1 2 b→ lim 1 1 Converges b b→ 10. 1 5 dx x3 lim 1 5 dx x3 5 x 2 b b 2 1 12. 1 4 4 b x dx b→ lim 1 4x 14 dx Diverges b→ lim 5 2 b→ lim 16 3 4 x 3 b 1 b x2 14. 0 xe x2 dx b→ lim 0 xe dx b→ lim e x2 2x 4 0 b→ lim e b2 2b 4 4 4 b b 16. 0 x 1e x dx b→ lim 0 x 1e x dx b→ lim xe x 0 b→ lim b eb 0 0 by L’Hôpital’s Rule. 18. 0 e ax sin bx dx c→ lim e ax a sin bx b cos bx a2 b2 b a2 b2 b c 0 20. 1 ln x dx x b b→ lim 1 ln x dx x 2b 0 b a2 b b2 x b→ lim ln x 2 Diverges 1 22. 0 x3 x2 1 2 dx b→ lim 0 x2 1 dx 1 b→ lim 0 x x2 1 1 b 2 dx 24. 0 ex 1 ex b dx b→ lim ln 1 ex 0 ln 2 b→ lim 1 ln x2 2 1 2 1 2 x2 Diverges 0 Diverges x dx 2 x 2 b 0 4 26. 0 sin b→ lim 2 cos 28. 0 8 dx x 4 b→0 lim b 8 dx x 4 b→0 lim 8 ln x b x Diverges since cos does not approach a limit as x → 2 6 . Diverges 30. 0 4 dx 6x b b→6 e e b→0 lim 0 46 86 86 x x 12 dx b 32. 0 ln x2 dx lim 0 2 ln x dx e b→6 lim 80 12 0 b→0 lim 2x ln x 2e 2e 2x b b→0 lim 2b ln b 2b 86 2 b b→ 2 0 1 4 x 2 b 0 34. 0 sec d Diverges lim 2 ln sec tan 0 , 36. 0 dx x2 b→2 lim arcsin 2 Section 7.8 2 Improper Integrals 355 38. 0 1 4 x2 b dx b→2 lim 0 11 42 x 1 2 x dx b→2 lim 12 ln 42 x x b 0 0 Diverges 3 40. 1 x 2 2 2 8 3 dx 3 2x 1 b b→2 2 2x 1 83 dx 2 83 2x dx b 2 83 dx 2 2 83 3 lim lim 2 2 c→2 lim c 2x 6 x 5 dx 3 b→2 6 x 5 53 1 c→2 lim 53 c Diverges 1 dx x ln x Thus, 1 dx x ln x b→1 1 42. ln ln x C 44. If p 1, 0 1 dx x 1, a→0 1 a→0 lim ln x a a→0 lim ln a . Diverges. If p 1 dx x ln x 1 e 1 c→ e 1 e 1 dx x ln x . e 1 0 1 dx xp lim 1 1 x1 p 1p p if 1 1 a a→0 lim 1 1 p a1 p . 1p lim ln ln x lim ln ln x This converges to p > 0 or p < 1. Diverges 46. (a) Assume a g x dx L (converges). ,0≤ a Since 0 ≤ f x ≤ g x on a, (b) a f x dx ≤ a g x dx L and a f x dx converges. f x dx. a g x dx diverges, because otherwise, by part (a), if a g x dx converges, then so does 1 48. 0 1 3 x dx 1 1 13 1 3. 3 converges. 2 50. 0 x 4e x dx converges. See Exercise 44, p 1 x 1 x1 1 (See Exercise 45.) 1 dx diverges by Exercise 43, x 1 2 52. Since ≥ 1 on 2, x and 2 x 1 dx diverges. 54. Since x ≤ 1 on 1, x3 2 and 1 1 dx converges by Exercise 43, x3 2 1 1 x1 x dx converges. 56. 1 1 ≥ since x x ln x x ln x < x on 2, . Since 2 1 dx diverges by Exercise 43, x 2 1 dx diverges. x ln x 58. See the definitions, pages 540, 543. 60. Answers will vary. (a) ex 1 e 2x 62. f t (b) x dx Diverges Fs t te 0 st dx (Example 4) dt b→ lim 1 s2 b st 1e st 0 Converges 1 ,s > 0 s2 356 64. f t Fs Chapter 7 eat eat e 0 st Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 66. f t dt 0 sin at e 0 st et a lim s dt b Fs et a s 0 sin at dt e st b 1 a 1 a s s b→ b→ lim s2 a s2 a2 a2 s sin at a s2 a2 a cos at 0 0 1 s a ,s > a 0 ,s > 0 68. f t Fs sinh at e 0 st sinh at dt 0 e et s a st eat 2 1 s e at dt et s a 1 2 b et 0 s a et 1 s s a dt 1 b→ lim 1 2 1 2 1 s 1 dx x2 1 s a a a a 0 0 1 2 a s a 1 s 1 x a s2 a2 ,s > a 70. (a) A 1 1 1 (b) Disk: V 1 1 dx x4 b b→ lim 3x3 1 3 (c) Shell: V 2 1 x 1 dx x2 b b→ lim 2 ln x 1 Diverges 72. x 2x y 1 S 4 1 2 2 2 y2 2yy 1 0 x y y 3 2 2 1 (Assume y > 0. y 3 1 3 x 2 2 y2 x dx y 4 4 1 1 x x x 2 2 2 2 dx 4 1 b x 1 a 2 x 40 2 2 1 2 x 2 2 dx 1 8 2 a→1 b→3 lim 1 2 arcsin x 2 2 arcsin 1 2 arcsin 74. (a) F x W (b) K ,5 x2 K ,K 4000 2 80,000,000 lim 80,000,000 x b 4000 b 80,000,000 dx x2 4000 W 2 10,000 10,000 8000 b→ 20,000 mi-ton 4000 80,000,000 x 80,000,000 b 20,000 80,000,000 b b Therefore, 4000 miles above the earth’s surface. Section 7.8 2 e 5 4 Improper Integrals 357 76. (a) (b) 0 2t 5 dt 0 2t 5 dt 2 e 5 e b 2t 5 dt e b→ lim 85 e 1 2t 5 0 1 2 e 5 4 2t 5 0 0.7981 (c) 0 79.81% te2t 5 t 2 e 5 2t 5 dt b→ lim 5 e 2 b 2t 5 0 5 2 5 78. (a) C 650,000 0 25,000 1 25,000 10 0.08t e 0.06t 0.06t dt t e 0.06 0.06t 650,000 1 e 0.06 0.08 1 e 0.06 2 5 0.06t 0 \$778,512.58 (b) C 650,000 0 25,000 1 25,000 0.08t e 0.06t 0.06t dt t e 0.06 0.06t 650,000 (c) C 650,000 1 e 0.06 0.08 0.06t 1 e 0.06 2 10 0.06t 0 \$905,718.14 25,000 1 0 0.08t e t e 0.06 dt 0.08 t e 0.06 0.06t 650,000 25,000 lim b→ 0.06t 1 e 0.06 2 b 0.06t 0 \$1,622,222.22 80. (a) 1 1 dx x 1 dx x2 b b→ lim lim ln x 1 (b) It would appear to converge. y 1 b→ 1 x b 1 1 1.00 0.75 1 1 dx will converge if n > 1 and will diverge if n ≤ 1. xn sin x dx ⇒ 1 x b→0 0.50 0.25 x −5 − 0.25 15 20 (c) Let dv u v cos x 1 dx x2 b 1 1 ⇒ du lim cos x x 1 sin x dx x cos x dx x2 cos 1 1 cos x dx x2 Converges 82. (a) Yes, the integral is not defined at x (c) As n → (d) In 0 2. 4 . (b) 5 , the integral approaches 4 2 1 4 dx tan x n 0 −2 2 I2 I4 I8 I12 3.14159 3.14159 3.14159 3.14159 358 Chapter 7 1 e 32 f x dx 50 0.4 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals 84. (a) f x 90 x 70 2 18 (b) P 72 ≤ x < (c) 0.5 0.2525 0.5 0.2475 0.2525 P 70 ≤ x ≤ 72 1.0 These are the same answers because by symmetry, P 70 ≤ x < and P 70 ≤ x < P 70 ≤ x ≤ 72 P 72 ≤ x < . 0.5 50 90 0.5 − 0.2 86. False. This is equivalent to Exercise 85. 88. True Review Exercises for Chapter 7 2. xe x 2 1 dx 1 x2 1 e 2x dx 2 1 x21 e 2 C 4. x 1 x2 dx 1 2 1 x2 12 12 2x dx C 1 1 x2 2 12 1 u5 5 1 x4 x4 x4 2x2 x 1 2x2 1 2x2 x x2 1 2 1 x2 C x 6. 2x 2x 3 dx u4 2 2x 5 u2 2 3 3u2 du 3 32 u3 C C 8. 1 dx x2 dx x 1 2 x u du 1 2 1 2 x2 2x x2 1 1 C 2 dx u 2x 3, x , dx 10. x2 (1) dv u (2) dv u 1 ex dx ex dx x2 x2 ⇒ v 1 ex ex 2 xex dx x2 1 ex 2xex 2 ex dx ex x2 2x 1 1 1 ⇒ du v ex ⇒ du 2x dx ex dx ⇒ x dx 12. u arctan 2x, du arctan 2x dx 2 1 4x2 dx, dv dx,v 2x dx 4x2 4x2 x 14. ln x2 1 dx 1 2 ln x2 1 dx 1 1 1 x x2 x2 dx 1 dx 1 x2 1x ln 2x 1 1 1 dx C x arctan 2x x arctan 2x 1 1 x ln x2 2 C 1 x ln x2 2 1 x ln x2 2 dv u dx ln x2 ⇒ v x 1 ln 1 4 1 ⇒ du 2x x2 1 dx Review Exercises for Chapter 7 e2x dx 1 e2x 1 ln 1 2 e2x C 359 16. ex arctan ex dx ex arctan ex ex arctan ex dv u ex dx ⇒ v ex ex 1 e2x dx arctan ex ⇒ du 18. sin2 x dx 2 1 1 2 cos x d x 1 x 2 1 sin x C 1 2 x sin x C 20. tan sec4 or tan sec4 d tan3 tan sec2 d 14 tan 4 12 tan 2 C1 d sec3 sec tan d 1 sec4 4 C2 22. cos 2 sin cos 2 d cos 2 sin sin2 cos 3 sin cos cos sin 2 d 1 sin 4 cos 4 d C 24. x2 x 9 dx 3 tan 3 sec 3 tan2 3 sec2 3 sec tan d 1d C 9 3 arcsec x2 x 3 d x x2 − 9 θ 3 3 tan x2 x 3 sec , dx C 9 3 tan 3 sec tan d , 26. 9 4x2 dx 1 2 1 2 9 2x 2 2 dx 2x 9 9 4x2 4x2 C C 28. 1 sin 2 cos2 d 1 2 1 1 2 cos2 2 cos 2 sin C d 1 2x 9 arcsin 2 3 x 2 1 arctan 2 u 2 cos , du 2 sin d 9 2x arcsin 4 3 360 Chapter 7 Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 30. (a) x4 x dx 64 tan3 64 sec4 sec3 d sec tan d 5 8 d, C C (b) x4 x dx 2 u4 4u2 du 20 32 sec2 3 sec3 2u3 2 3u 15 24 u2 4 x, dx x 15 C 8 C 64 sec3 15 24 x 4 (c) x4 4 tan2 , dx x 2 sec x dx u3 2 3x x 15 32 3x 2u du 8 tan sec2 4u1 2 du C 8 C (d) x4 x dx 2x 4 3 2x 4 3 24 x 15 v x x 32 2 3 4 x x C 3 2 dx 2u3 2 3u 15 24 u 4 x, du dx x 15 32 20 3x 32 4 4 15 8 x 32 52 C 32 3x 2 4 3 dx dv u x 4 x dx ⇒ ⇒ du 32. 2x3 2x3 5x2 x2 5x2 x2 4x x 4 4 2x 2x 3 3 4 x 4 x 3 x x 1 3 1 dx x2 3x 4 ln x 3 ln x 1 C 4x x A x 3A x 1 dx 34. 4x 3x 4x Let x Let x 2 12 2 B x 1 1 3B 2 3 4 3 dx 2 3 1 x 1 1 u C 1 2 2 1: 2 2: 6 4x 3x 3B ⇒ B 3A 3B ⇒ A 4 3 1 x 1 1 uu ln u 1 1 2 dx 12 dx 4 ln x 3 1 du u tan tan 1 1 2 3x 1 C 2 2 ln x 3 1 1 x 1 C 36. sec2 tan tan 1 d du ln u du ln C ln 1 cot C u uu tan , du 1 1 1 A u Au sec2 B u 1 d 1 Bu 1 Let u Let u 0: 1 1: 1 B A⇒ A R eview Exercises for Chapter 7 x 2 3x 2 4 3x 27 6x 8 27 2 x 1 ex 2 361 38. dx 2 3x 3x C C (Formula 21) 40. dx 1 1 dx 2 1 eu 1 u 2 12 x 2 ln 1 ln 1 eu ex 2 u C C x2 (Formula 84) 42. 3 2x 9x2 1 dx 3 2 3x 1 3x 2 1 C 3 dx u 3x 3 arcsec 3x 2 1 dx tan x 1 1 11 2 x 1 tan x (Formula 33) 44. 1 dx sin x C u x ln cos x (Formula 71) 46. tann x dx tann tann 1 n 1 2x sec2 x 1 dx tann tann 2 2 48. x dx u csc 2x dx x 2 csc 2x 1 dx 2x cot 2x C 2 x sec2 x dx 1 2 ln csc 2x 2x, du 1 dx 2x tann x x dx 50. u 3x3 x2 3x3 1 1 4x 12 4x x dx x, x Ax x2 Ax Ax3 B 1 u 4u3 u4 4u du 2u2 1, dx D 12 1 A Cx Cx 1, D B 4u4 4u3 4u2 du 4u du 4u5 5 4u3 3 C 4 1 15 x 32 3x 2 C 52. Cx x2 B x2 Bx2 C 0 x x2 1 D A B 3, B D 3x3 x2 0, A 0⇒D 4⇒C 4x dx 12 3 dx 1 x x2 1 2 x2 1 1 2 dx C 3 ln x2 2 54. sin cos 2 d sin2 1 2 sin cos sin 2 d cos2 1 cos 2 2 d C 1 2 2 cos 2 C 362 Chapter 7 4 x2 dx 2x Integration Techniques, L’ H ô pital ’s Rule, and Improper Integrals 2 cos 2 cos 4 sin sin d 56. y csc ln csc ln 2 d cos cos 4 2 x2 2 cos x2 C C 4 x 4 sin 1 cos x2 x 2 sin , dx 2 cos d, 58. y u 1 1 1 cos d d 1 cos 12 sin d 21 cos C cos , du x sin d 1 2 60. 0 x 2x 4 dx 2 ln x 2 ln 3 ln 9 8 x 4 2 ln 4 0.118 ln x ln 2 2 0 62. 0 xe3x dx e3x 3x 9 2 1 0 16 5e 9 1 224.238 3 64. 0 x 1 x dx 22 3 3 1 x 0 4 3 4 3 8 3 4 66. A 0 1 25 x2 1 x ln 10 x dx 5 5 4 0 1 1 ln 10 9 1 ln 9 10 0.220 68. By symmetry, y A x 1 17 5 x, y 3.4, 0 sin x sin 2 x lim x 4 0. 5 44 4 3.4 3 2 1 y 70. s 0 1 sin2 2x dx 3.82 (3.4, 0) x 1 3 4 5 −1 −2 −3 72. lim x →0 x →0 lim cos x 2 cos 2 x ln x 2 1 2 74. lim xe x→ x2 x→ lim x 2 ex x→ lim 1 2 2 xe x 0 76. y ln y x →1 1 x →1 lim ln x ln x 1 1 x1 1 1 x ln2 x 2 x →1 x →1 lim ln x 1 ln x 1 x →1 lim x →1 lim ln2 x x1 x lim 1 ln x x 1 x2 x →1 lim 2x ln x 0, y 1. 0 Since ln y ...
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