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# Lesson 27: Integration by Substitution (slides)

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Integration by substitution is the chain rule in reverse.

NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200

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### Lesson 27: Integration by Substitution (slides)

1. 1. Sec on 5.5 Integra on by Subs tu on V63.0121.011: Calculus I Professor Ma hew Leingang New York University May 4, 2011.
2. 2. Announcements Today: 5.5 Monday 5/9: Review in class Tuesday 5/10: Review Sessions by TAs Wednesday 5/11: TA oﬃce hours: Adam 10–noon (WWH 906) Jerome 3:30–5:30 (WWH 501) Soohoon 6–8 (WWH 511) Thursday 5/12: Final Exam, 2:00–3:50pm, CANT 200
3. 3. Resurrection Policy If your ﬁnal score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight.Image credit: Sco Beale / Laughing Squid
4. 4. Objectives Given an integral and a subs tu on, transform the integral into an equivalent one using a subs tu on Evaluate indeﬁnite integrals using the method of subs tu on. Evaluate deﬁnite integrals using the method of subs tu on.
5. 5. Outline Last Time: The Fundamental Theorem(s) of Calculus Subs tu on for Indeﬁnite Integrals Theory Examples Subs tu on for Deﬁnite Integrals Theory Examples
6. 6. Diﬀerentiation and Integration asreverse processes Theorem (The Fundamental Theorem of Calculus) 1. Let f be con nuous on [a, b]. Then ∫ d x f(t) dt = f(x) dx a 2. Let f be con nuous on [a, b] and f = F′ for some other func on F. Then ∫ b f(x) dx = F(b) − F(a). a
7. 7. Techniques of antidiﬀerentiation? So far we know only a few rules for an diﬀeren a on. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx
8. 8. Techniques of antidiﬀerentiation? So far we know only a few rules for an diﬀeren a on. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pre y par cular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1
9. 9. Techniques of antidiﬀerentiation? So far we know only a few rules for an diﬀeren a on. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pre y par cular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 What are we supposed to do with that?
10. 10. No straightforward system ofantidiﬀerentiation So far we don’t have any way to ﬁnd ∫ 2x √ dx x2 + 1 or ∫ tan x dx.
11. 11. No straightforward system ofantidiﬀerentiation So far we don’t have any way to ﬁnd ∫ 2x √ dx x2 + 1 or ∫ tan x dx. Luckily, we can be smart and use the “an ” version of one of the most important rules of diﬀeren a on: the chain rule.
12. 12. Outline Last Time: The Fundamental Theorem(s) of Calculus Subs tu on for Indeﬁnite Integrals Theory Examples Subs tu on for Deﬁnite Integrals Theory Examples
13. 13. Substitution for IndeﬁniteIntegrals Example Find ∫ x √ dx. x2 + 1
14. 14. Substitution for IndeﬁniteIntegrals Example Find ∫ x √ dx. x2 + 1 Solu on Stare at this long enough and you no ce the the integrand is the √ deriva ve of the expression 1 + x2 .
15. 15. Say what? Solu on (More slowly, now) Let g(x) = x2 + 1.
16. 16. Say what? Solu on (More slowly, now) Let g(x) = x2 + 1. Then g′ (x) = 2x and so d√ 1 x g(x) = √ g′ (x) = √ dx 2 g(x) x2 + 1
17. 17. Say what? Solu on (More slowly, now) Let g(x) = x2 + 1. Then g′ (x) = 2x and so d√ 1 x g(x) = √ g′ (x) = √ dx 2 g(x) x2 + 1 Thus ∫ ∫ ( √ ) x d √ dx = g(x) dx x2 + 1 dx √ √ = g(x) + C = 1 + x2 + C.
18. 18. Leibnizian notation FTW Solu on (Same technique, new nota on) Let u = x2 + 1.
19. 19. Leibnizian notation FTW Solu on (Same technique, new nota on) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u.
20. 20. Leibnizian notation FTW Solu on (Same technique, new nota on) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ 1 ∫ x dx 2 du 1 √ = √ = √ du x2 + 1 u 2 u
21. 21. Leibnizian notation FTW Solu on (Same technique, new nota on) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ 1 ∫ x dx 2 du 1 √ = √ = √ du x2 + 1 ∫ u 2 u 1 −1/2 = 2u du
22. 22. Leibnizian notation FTW Solu on (Same technique, new nota on) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ 1 ∫ x dx 2 du 1 √ = √ = √ du x2 + 1 ∫ u 2 u 1 −1/2 = 2u du √ √ = u + C = 1 + x2 + C.
23. 23. Useful but unsavory variation Solu on (Same technique, new nota on, more idiot-proof) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:” du dx = 2x
24. 24. Useful but unsavory variation Solu on (Same technique, new nota on, more idiot-proof) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into ∫ ∫ x x du √ dx = √ · x2 + 1 u 2x
25. 25. Useful but unsavory variation Solu on (Same technique, new nota on, more idiot-proof) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into ∫ ∫ ∫ x x du 1 √ dx = √ · = √ du x2 + 1 u 2x 2 u
26. 26. Useful but unsavory variation Solu on (Same technique, new nota on, more idiot-proof) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into ∫ ∫ ∫ x x du 1 √ dx = √ · = √ du x2 + 1 u 2x 2 u ∫ 1 −1/2 = 2u du
27. 27. Useful but unsavory variation Solu on (Same technique, new nota on, more idiot-proof) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into ∫ ∫ ∫ x x du 1 √ dx = √ · = √ du x2 + 1 u 2x 2 u ∫ √ 1 −1/2 √ = 2 u du = u + C = 1 + x2 + C.
28. 28. Theorem of the Day Theorem (The Subs tu on Rule) If u = g(x) is a diﬀeren able func on whose range is an interval I and f is con nuous on I, then ∫ ∫ ′ f(g(x))g (x) dx = f(u) du
29. 29. Theorem of the Day Theorem (The Subs tu on Rule) If u = g(x) is a diﬀeren able func on whose range is an interval I and f is con nuous on I, then ∫ ∫ ′ f(g(x))g (x) dx = f(u) du That is, if F is an an deriva ve for f, then ∫ f(g(x))g′ (x) dx = F(g(x))
30. 30. Theorem of the Day Theorem (The Subs tu on Rule) If u = g(x) is a diﬀeren able func on whose range is an interval I and f is con nuous on I, then ∫ ∫ ′ f(g(x))g (x) dx = f(u) du In Leibniz nota on: ∫ ∫ du f(u) dx = f(u) du dx
31. 31. A polynomial example Example ∫ Use the subs tu on u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx.
32. 32. A polynomial example Example ∫ Use the subs tu on u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. Solu on If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So ∫ (x2 + 3)3 4x dx
33. 33. A polynomial example Example ∫ Use the subs tu on u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. Solu on If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So ∫ ∫ ∫ (x + 3) 4x dx = u 2du = 2 u3 du 2 3 3
34. 34. A polynomial example Example ∫ Use the subs tu on u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. Solu on If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So ∫ ∫ ∫ (x + 3) 4x dx = u 2du = 2 u3 du 2 3 3 1 = u4 2
35. 35. A polynomial example Example ∫ Use the subs tu on u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. Solu on If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So ∫ ∫ ∫ (x + 3) 4x dx = u 2du = 2 u3 du 2 3 3 1 1 = u4 = (x2 + 3)4 2 2
36. 36. A polynomial example (brute force) Solu on
37. 37. A polynomial example (brute force) Solu on ∫ (x2 + 3)3 4x dx
38. 38. A polynomial example (brute force) Solu on ∫ ∫ 2 3 ( ) (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
39. 39. A polynomial example (brute force) Solu on ∫ ∫ 2 3 ( ) (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx
40. 40. A polynomial example (brute force) Solu on ∫ ∫ 2 3 ( ) (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 = x8 + 6x6 + 27x4 + 54x2 2
41. 41. A polynomial example (brute force) Solu on ∫ ∫ 2 3 ( ) (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 = x8 + 6x6 + 27x4 + 54x2 2 Which would you rather do?
42. 42. A polynomial example (brute force) Solu on ∫ ∫ 2 3 ( ) (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 = x8 + 6x6 + 27x4 + 54x2 2 Which would you rather do? It’s a wash for low powers
43. 43. A polynomial example (brute force) Solu on ∫ ∫ 2 3 ( ) (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 = x8 + 6x6 + 27x4 + 54x2 2 Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do subs tu on.
44. 44. Compare We have the subs tu on method, which, when mul plied out, gives ∫ 1 (x2 + 3)3 4x dx = (x2 + 3)4 2 1( 8 ) = x + 12x6 + 54x4 + 108x2 + 81 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + 2 2 and the brute force method ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 2 Is there a diﬀerence? Is this a problem?
45. 45. Compare We have the subs tu on method, which, when mul plied out, gives ∫ 1 (x2 + 3)3 4x dx = (x2 + 3)4 + C 2 1( 8 ) = x + 12x6 + 54x4 + 108x2 + 81 + C 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + +C 2 2 and the brute force method ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C 2 Is there a diﬀerence? Is this a problem? No, that’s what +C means!
46. 46. A slick example Example ∫ Find tan x dx.
47. 47. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x
48. 48. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on Let u = cos x . Then du = − sin x dx . So ∫ ∫ sin x tan x dx = dx cos x
49. 49. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on Let u = cos x . Then du = − sin x dx . So ∫ ∫ sin x tan x dx = dx cos x
50. 50. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on Let u = cos x . Then du = − sin x dx . So ∫ ∫ sin x tan x dx = dx cos x
51. 51. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on Let u = cos x . Then du = − sin x dx . So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u
52. 52. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on Let u = cos x . Then du = − sin x dx . So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C
53. 53. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on Let u = cos x . Then du = − sin x dx . So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C = − ln | cos x| + C = ln | sec x| + C
54. 54. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x
55. 55. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on du Let u = sin x. Then du = cos x dx and so dx = . cos x
56. 56. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solu on du Let u = sin x. Then du = cos x dx and so dx = . cos x ∫ ∫ ∫ sin x u du tan x dx = dx = ∫ cos x ∫ cos x cos x ∫ u du u du u du = = = cos2 x 1 − sin2 x 1 − u2
57. 57. For those who really must know all Solu on (Con nued, with algebra help) Let y = 1 − u2 , so dy = −2u du. Then ∫ ∫ ∫ u du u dy tan x dx = = 1∫− u2 y −2u 1 dy 1 1 =− = − ln |y| + C = − ln 1 − u2 + C 2 y 2 2 1 1 = ln √ + C = ln √ +C 1 − u2 1 − sin2 x 1 = ln + C = ln |sec x| + C |cos x|
58. 58. Outline Last Time: The Fundamental Theorem(s) of Calculus Subs tu on for Indeﬁnite Integrals Theory Examples Subs tu on for Deﬁnite Integrals Theory Examples
59. 59. Substitution for Deﬁnite Integrals Theorem (The Subs tu on Rule for Deﬁnite Integrals) If g′ is con nuous and f is con nuous on the range of u = g(x), then ∫ b ∫ g(b) ′ f(g(x))g (x) dx = f(u) du. a g(a)
60. 60. Substitution for Deﬁnite Integrals Theorem (The Subs tu on Rule for Deﬁnite Integrals) If g′ is con nuous and f is con nuous on the range of u = g(x), then ∫ b ∫ g(b) ′ f(g(x))g (x) dx = f(u) du. a g(a) Why the change in the limits? The integral on the le happens in “x-land” The integral on the right happens in “u-land”, so the limits need to be u-values To get from x to u, apply g
61. 61. Example ∫ πCompute cos2 x sin x dx. 0
62. 62. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way)
63. 63. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) ∫ First compute the indeﬁnite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and ∫ cos2 x sin x dx
64. 64. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) ∫ First compute the indeﬁnite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and ∫ cos2 x sin x dx
65. 65. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) ∫ First compute the indeﬁnite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and ∫ cos2 x sin x dx
66. 66. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) ∫ First compute the indeﬁnite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and ∫ ∫ cos2 x sin x dx = − u2 du
67. 67. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) ∫ First compute the indeﬁnite integral cos2 x sin x dx and then evaluate. Let u = cos x . Then du = − sin x dx and ∫ ∫ cos2 x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C. 3 1
68. 68. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) Let u = cos x . Then du = − sin x dx and ∫ ∫ cos x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C. 2 3 1 Therefore ∫ π π 1 cos2 x sin x dx = − cos3 x 0 3 0
69. 69. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) Let u = cos x . Then du = − sin x dx and ∫ ∫ cos x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C. 2 3 1 Therefore ∫ π 1( ) π 1 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 0 3 0 3
70. 70. Example ∫ πCompute cos2 x sin x dx. 0Solu on (Slow Way) Let u = cos x . Then du = − sin x dx and ∫ ∫ cos x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C. 2 3 1 Therefore ∫ π 1( ) 2 π 1 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3
71. 71. Deﬁnite-ly Quicker Solu on (Fast Way) Do both the subs tu on and the evalua on at the same me. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So ∫ π cos2 x sin x dx 0
72. 72. Deﬁnite-ly Quicker Solu on (Fast Way) Do both the subs tu on and the evalua on at the same me. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So ∫ π cos2 x sin x dx 0
73. 73. Deﬁnite-ly Quicker Solu on (Fast Way) Do both the subs tu on and the evalua on at the same me. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So ∫ π cos2 x sin x dx 0
74. 74. Deﬁnite-ly Quicker Solu on (Fast Way) Do both the subs tu on and the evalua on at the same me. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So ∫ π ∫ −1 ∫ 1 2 cos x sin x dx = −u du = 2 u2 du 0 1 −1
75. 75. Deﬁnite-ly Quicker Solu on (Fast Way) Do both the subs tu on and the evalua on at the same me. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So ∫ π ∫ −1 ∫ 1 2 cos x sin x dx = −u du = 2 u2 du 0 1 −1 1( ) 2 1 1 3 = u = 1 − (−1) = 3 −1 3 3
76. 76. Compare The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original variable (x). But the slow way is just as reliable.
77. 77. An exponential example Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3
78. 78. An exponential example Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 2x √ 1 8√ √ e e2x + 1 dx = u + 1 du ln 3 2 3
79. 79. About those limits Since √ √ 2 e2(ln 3) = eln 3 = eln 3 = 3 we have ∫ √ ∫ ln 8 √ 1 8√ √ e2x e2x + 1 dx = u + 1 du ln 3 2 3
80. 80. An exponential example Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Now let y = u + 1, dy = du. So ∫ ∫ 1 8√ 9 1 9√ 1 2 1 19 u + 1 du = y dy = · y3/2 = (27 − 8) = 2 3 2 4 2 3 4 3 3
81. 81. About those fractional powers We have 93/2 = (91/2 )3 = 33 = 27 43/2 = (41/2 )3 = 23 = 8 so ∫ 9 9 1 1 2 1 19 1/2 y dy = · y3/2 = (27 − 8) = 2 4 2 3 4 3 3
82. 82. An exponential example Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Now let y = u + 1, dy = du. So ∫ ∫ 1 8√ 9 1 9√ 1 2 1 19 u + 1 du = y dy = · y3/2 = (27 − 8) = 2 3 2 4 2 3 4 3 3
83. 83. Another way to skin that cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Let u = e2x + 1,
84. 84. Another way to skin that cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Let u = e2x + 1,so that du = 2e2x dx.
85. 85. Another way to skin that cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Let u = e2x + 1,so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ √ e2x e2x + 1 dx = u du ln 3 2 4
86. 86. Another way to skin that cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Let u = e2x + 1,so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 1 9 √ e2x e2x + 1 dx = u du = u3/2 ln 3 2 4 3 4
87. 87. Another way to skin that cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on Let u = e2x + 1,so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 1 9 1 19 √ e2x e2x + 1 dx = u du = u3/2 = (27 − 8) = ln 3 2 4 3 4 3 3
88. 88. A third skinned cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on √ Let u = e2x + 1, so that u2 = e2x + 1
89. 89. A third skinned cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx
90. 90. A third skinned cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solu on √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus ∫ √ ∫ 3 ln 8 √ 1 3 19 √ e2x e2x + 1 dx = u · u du = u3 = ln 3 2 3 2 3
91. 91. A Trigonometric Example Example Find ∫ ( ) ( ) 3π/2 θ θ cot5 sec2 dθ. π 6 6
92. 92. A Trigonometric Example Example Find ∫ ( ) ( ) 3π/2 θ θ cot5 sec2 dθ. π 6 6 Before we dive in, think about: What “easy” subs tu ons might help? Which of the trig func ons suggests a subs tu on?
93. 93. Solu on θ 1Let φ = . Then dφ = dθ. 6 6 ∫ 3π/2 ( ) ( ) ∫ π/4 θ θ cot5 sec2 dθ = 6 cot5 φ sec2 φ dφ π 6 6 π/6 ∫ π/4 sec2 φ dφ =6 π/6 tan5 φ
94. 94. Solu on θ 1Let φ = . Then dφ = dθ. 6 6 ∫ 3π/2 ( ) ( ) ∫ π/4 θ θ cot5 sec2 dθ = 6 cot5 φ sec2 φ dφ π 6 6 π/6 ∫ π/4 sec2 φ dφ =6 π/6 tan5 φNow let u = tan φ. So du = sec2 φ dφ, and
95. 95. Solu onNow let u = tan φ. So du = sec2 φ dφ, and ∫ π/4 ∫ 1 sec2 φ dφ 6 5φ = 6 √ u−5 du π/6 tan 1/ 3 ( )1 1 −4 3 =6 − u √ = [9 − 1] = 12. 4 1/ 3 2
96. 96. The limits explained √ π sin(π/4) 2/2 tan = =√ =1 4 cos(π/4) 2/2 π sin(π/6) 1/2 1 tan = =√ =√ 6 cos(π/6) 3/2 3
97. 97. The limits explained ( ) 3 [ −4 ]1 √ 3 [ −4 ]1/√3 1 1 −4 6 − u √ = −u 1/ 3 = u 1 4 1/ 3 2 2 3 [ −1/2 −4 −1/2 −4 ] = (3 ) − (1 ) 2 3 3 = [32 − 12 ] = (9 − 1) = 12 2 2
98. 98. Graphs ∫ 3π/2 ( ) ( ) ∫ π/4 θ θ cot5 sec2 dθ 6 cot5 φ sec2 φ dφ π 6 6 π/6 y y . θ φ π 3π ππ 2 64 The areas of these two regions are the same.
99. 99. ∫ ∫Graphs π/4 5 2 6 cot φ sec φ dφ 1 √ 6u −5 du π/6 1/ 3 y y . φ u ππ 11 √ 64 3 The areas of these two regions are the same.
100. 100. u/du pairs When deciding on a subs tu on, look for sub-expressions where one is (a constant mul ple of) the deriva ve of the other. Such as: √ u xn ln x sin x cos x tan x x ex 1 1 constant × du xn−1 cos x sin x sec2 x √ ex x x
101. 101. Summary If F is an an deriva ve for f, then: ∫ f(g(x))g′ (x) dx = F(g(x)) If F is an an deriva ve for f, which is con nuous on the range of g, then: ∫ b ∫ g(b) ′ f(g(x))g (x) dx = f(u) du = F(g(b)) − F(g(a)) a g(a) An diﬀeren a on in general and subs tu on in par cular is a “nonlinear” problem that needs prac ce, intui on, and perserverance. The whole an diﬀeren a on story is in Chapter 6.