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Lesson 27: Integration by Substitution (Section 4 version)

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The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples and graphics.

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Lesson 27: Integration by Substitution (Section 4 version)

1. 1. Section 5.5 Integration by Substitution V63.0121, Calculus I April 28, 2009 Announcements Quiz 6 this week covering 5.1–5.2 Practice ﬁnals on the website. Solutions Friday . . Image credit: kchbrown . . . . . .
2. 2. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
3. 3. Ofﬁce Hours and other help Day Time Who/What Where in WWH M 1:00–2:00 Leingang OH 624 5:00–7:00 Curto PS 517 T 1:00–2:00 Leingang OH 624 4:00–5:50 Curto PS 317 W 2:00–3:00 Leingang OH 624 R 9:00–10:00am Leingang OH 624 F 2:00–4:00 Curto OH 1310 . . . . . .
4. 4. Final stuff Final is May 8, 2:00–3:50pm in 19W4 101/102 Old ﬁnals online, including Fall 2008 Review sessions: May 5 and 6, 6:00–8:00pm, SILV 703 . . Image credit: Pragmagraphr . . . . . .
5. 5. 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: Scott Beale / Laughing Squid . . . . . .
6. 6. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
7. 7. Differentiation and Integration as reverse processes Theorem (The Fundamental Theorem of Calculus) 1. Let f be continuous on [a, b]. Then ∫ x d f(t) dt = f(x) dx a 2. Let f be continuous on [a, b] and f = F′ for some other function F. Then ∫ b f(x) dx = F(b) − F(a). a . . . . . .
8. 8. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx . . . . . .
9. 9. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pretty particular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 . . . . . .
10. 10. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pretty particular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 What are we supposed to do with that? . . . . . .
11. 11. So far we don’t have any way to ﬁnd ∫ 2x √ dx x2 + 1 or ∫ tan x dx. . . . . . .
12. 12. 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 “anti” version of one of the most important rules of differentiation: the chain rule. . . . . . .
13. 13. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
14. 14. Substitution for Indeﬁnite Integrals Example Find ∫ x √ dx. x2 +1 . . . . . .
15. 15. Substitution for Indeﬁnite Integrals Example Find ∫ x √ dx. x2 +1 Solution Stare at this long enough and you notice the the integrand is the √ derivative of the expression 1 + x2 . . . . . . .
16. 16. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. . . . . . .
17. 17. Say what? Solution (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 . . . . . .
18. 18. Say what? Solution (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. . . . . . .
19. 19. Leibnizian notation wins again Solution (Same technique, new notation) Let u = x2 + 1. . . . . . .
20. 20. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. . . . . . .
21. 21. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ ∫ x 1 ( ) 1 √ dx = √ 1 du = 2 √ du x2 + 1 u 2 u . . . . . .
22. 22. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ ∫ x 1 ( ) 1 √ dx = √ 1 du = 2 √ du x2 + 1 u 2 u ∫ 1 −1/2 = 2u du . . . . . .
23. 23. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ ∫ x 1 ( ) 1 √ dx = √ 1 du = 2 √ du x2 + 1 u 2 u ∫ 1 −1/2 = 2u du √ √ = u + C = 1 + x2 + C. . . . . . .
24. 24. Theorem of the Day Theorem (The Substitution Rule) If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫ ∫ ′ f(g(x))g (x) dx = f(u) du or ∫ ∫ du f(u) dx = f(u) du dx . . . . . .
25. 25. A polynomial example Example ∫ Use the substitution u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. . . . . . .
26. 26. A polynomial example Example ∫ Use the substitution u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. Solution 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 4 1 2 = u = (x + 3)4 2 2 . . . . . .
27. 27. A polynomial example, the hard way Compare this to multiplying it out: ∫ ∫ 2 3 ( 6 ) (x + 3) 4x dx = x + 9x4 + 27x2 + 27 4x dx ∫ ( 7 ) = 4x + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 . . . . . .
28. 28. Compare We have ∫ 1 2 (x2 + 3)3 4x dx = (x + 3)4 2 ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 2 Now 1 2 1( 8 ) (x + 3)4 = x + 12x6 + 54x4 + 108x2 + 81 2 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + 2 2 Is this a problem? . . . . . .
29. 29. Compare We have ∫ 1 2 (x2 + 3)3 4x dx = (x + 3)4 + C 2 ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C 2 Now 1 2 1( 8 ) (x + 3)4 = x + 12x6 + 54x4 + 108x2 + 81 2 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + 2 2 Is this a problem? No, that’s what +C means! . . . . . .
30. 30. A slick example Example ∫ Find tan x dx. . . . . . .
31. 31. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . .
32. 32. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. . . . . . .
33. 33. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u . . . . . .
34. 34. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C . . . . . .
35. 35. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution 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 . . . . . .
36. 36. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
37. 37. Theorem (The Substitution Rule for Deﬁnite Integrals) If g′ is continuous and f is continuous on the range of u = g(x), then ∫ ∫ b g(b) f(g(x))g′ (x) dx = f(u) du. a g(a) . . . . . .
38. 38. Example ∫ π Compute cos2 x sin x dx. 0 . . . . . .
39. 39. Example ∫ π Compute cos2 x sin x dx. 0 Solution (Slow Way) ∫ First compute the indeﬁnite integral cos2 x sin x dx and then evaluate. . . . . . .
40. 40. Example ∫ π Compute cos2 x sin x dx. 0 Solution (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 = − 1 cos3 x + C. 3 3 Therefore ∫ π π cos2 x sin x dx = − 1 cos3 x 3 0 = 2. 3 0 . . . . . .
41. 41. Solution (Fast Way) Do both the substitution and the evaluation at the same time. . . . . . .
42. 42. Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. . . . . . .
43. 43. Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So ∫ π ∫ −1 cos2 x sin x dx = −u2 du 0 1 ∫ 1 = u2 du −1 1 3 1 2 = 3 u −1 = 3 . . . . . . .
44. 44. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 . . . . . .
45. 45. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ √ e2x e2x + 1 dx = u + 1 du ln 3 2 3 . . . . . .
46. 46. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ √ e2x e2x + 1 dx = u + 1 du ln 3 2 3 Now let y = u + 1, dy = du. So ∫ 8√ ∫ 9 ∫ 9 1 1 √ 1 u + 1 du = y dy = y1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = · y3/2 = (27 − 8) = 2 3 4 3 3 . . . . . .
47. 47. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1 . . . . . .
48. 48. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. . . . . . .
49. 49. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 2x √ e e2x + 1 dx = u du ln 3 2 4 . . . . . .
50. 50. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 2x √ e e2x + 1 dx = u du ln 3 2 4 9 1 3/2 = u 3 4 . . . . . .
51. 51. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 2x √ e e2x + 1 dx = u du ln 3 2 4 1 3/2 9 = u 3 4 1 19 = (27 − 8) = 3 3 . . . . . .
52. 52. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 . . . . . .
53. 53. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx . . . . . .
54. 54. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus ∫ √ ∫ ln 8 3 3 1 3 19 √ = u · u du = u = ln 3 2 3 2 3 . . . . . .
55. 55. Example Find ∫ ( ) ( ) 3π/2 5 θ 2 θ cot sec dθ. π 6 6 . . . . . .
56. 56. Example Find ∫ ( ) ( ) 3π/2 5 θ 2 θ cot sec dθ. π 6 6 Before we dive in, think about: What “easy” substitutions might help? Which of the trig functions suggests a substitution? . . . . . .
57. 57. Solution θ 1 Let φ = . Then dφ = dθ. 6 6 ∫ 3π/2 ( ) ( ) ∫ π/4 5 θ 2 θ cot sec dθ = 6 cot5 φ sec2 φ dφ π 6 6 π/6 ∫ π/4 sec2 φ dφ =6 π/6 tan5 φ . . . . . .
58. 58. Solution θ 1 Let φ = . Then dφ = dθ. 6 6 ∫ 3π/2 ( ) ( ) ∫ π/4 5 θ 2 θ cot sec dθ = 6 cot5 φ sec2 φ dφ π 6 6 π/6 ∫ π/4 sec2 φ dφ =6 π/6 tan5 φ Now let u = tan φ. So du = sec2 φ dφ, and ∫ π/4 ∫ 1 sec2 φ dφ −5 6 =6 √ u du π/6 tan5 φ 1/ 3 ( ) 1 1 3 =6 − u−4 √ = [9 − 1] = 12. 4 1/ 3 2 . . . . . .
59. 59. Graphs ∫ 3π/2 ( ) ( ) ∫ π/4 θ 2 θ . 5 cot sec dθ . 6 cot5 φ sec2 φ dφ π 6 6 π/6 y . y . . . . . θ . . . φ . π 3π ππ . . . 2 64 . . . . . .
60. 60. Graphs ∫ π/4 ∫ 1 . 5 2 6 cot φ sec φ dφ . √ 6u−5 du π/6 1/ 3 y . y . . . . . φ . .. u ππ 1 . 1 . . .√ 64 3 . . . . . .
61. 61. Summary: What do we substitute? Linear factors (ax + b) are easy substitutions: u = ax + b, du = a dx Look for function/derivative pairs in the integrand, one to make u and one to make du: xn and xn−1 (fudge the coefﬁcient) sine and cosine (fudge the minus sign) ex and ex ax and ax (fudge the coefﬁcient) √ 1 x and √ (fudge the factor of 2) x 1 ln x and x . . . . . .