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# Lesson 29: Integration by Substition (worksheet solutions)

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### Lesson 29: Integration by Substition (worksheet solutions)

1. 1. Section 5.5 Integration by Substitution V63.0121.034, Calculus I December 9, 2009 Announcements Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50 Practice ﬁnals on the website. Solutions Friday . . . . . .
2. 2. Schedule for next week Monday, class: review, evaluations, movie! Tuesday, 8:00am: Review session for all students with 8:00 recitations (Tuesday or Thursday) in CIWW 109 Tuesday, 9:30am: Review session for all students with 9:30 recitations (Tuesday or Thursday) in CIWW 109 Ofﬁce Hours continue Friday, 2:00pm: ﬁnal in Tisch UC50 . . . . . .
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: Scott Beale / Laughing Squid . . . . . .
4. 4. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
5. 5. 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 . . . . . .
6. 6. 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 . . . . . .
7. 7. 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 . . . . . .
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 Some are pretty particular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 What are we supposed to do with that? . . . . . .
9. 9. So far we don’t have any way to ﬁnd ∫ 2x √ dx x2 + 1 or ∫ tan x dx. . . . . . .
10. 10. 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. . . . . . .
11. 11. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
12. 12. Substitution for Indeﬁnite Integrals Example Find ∫ x √ dx. x 2+1 . . . . . .
13. 13. Substitution for Indeﬁnite Integrals Example Find ∫ x √ dx. x 2+1 Solution Stare at this long enough and you notice the the integrand is the √ derivative of the expression 1 + x2 . . . . . . .
14. 14. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. . . . . . .
15. 15. 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 . . . . . .
16. 16. 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. . . . . . .
17. 17. Leibnizian notation FTW Solution (Same technique, new notation) Let u = x2 + 1. . . . . . .
18. 18. Leibnizian notation FTW Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. . . . . . .
19. 19. Leibnizian notation FTW Solution (Same technique, new notation) √ √ 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 . . . . . .
20. 20. Leibnizian notation FTW Solution (Same technique, new notation) √ √ 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 . . . . . .
21. 21. Leibnizian notation FTW Solution (Same technique, new notation) √ √ 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. . . . . . .
22. 22. Useful but unsavory variation Solution (Same technique, new notation, 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 √ √ = u + C = 1 + x2 + C . Mathematicians have serious issues with mixing the x and u like this. However, I can’t deny that it works. . . . . . .
23. 23. 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 That is, if F is an antiderivative for f, then ∫ f(g(x))g′ (x) dx = F(g(x)) In Leibniz notation: ∫ ∫ du f(u) dx = f(u) du dx . . . . . .
24. 24. A polynomial example Example ∫ Use the substitution u = x2 + 3 to ﬁnd (x2 + 3)3 4x dx. . . . . . .
25. 25. 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 . . . . . .
26. 26. A polynomial example, by brute force 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 . . . . . .
27. 27. A polynomial example, by brute force 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 Which would you rather do? . . . . . .
28. 28. A polynomial example, by brute force 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 Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do substitution. . . . . . .
29. 29. Compare We have the substitution method, which, when multiplied 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 this a problem? . . . . . .
30. 30. Compare We have the substitution method, which, when multiplied 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 this a problem? No, that’s what +C means! . . . . . .
31. 31. A slick example Example ∫ Find tan x dx. . . . . . .
32. 32. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . .
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. . . . . . .
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 . . . . . .
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 . . . . . .
36. 36. 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 . . . . . .
37. 37. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . .
38. 38. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution du Let u = sin x. Then du = cos x dx and so dx = . cos x . . . . . .
39. 39. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution 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 = = 2 = cos2 x 1 − sin x 1 − u2 At this point, although it’s possible to proceed, we should probably back up and see if the other way works quicker (it does). . . . . . .
40. 40. For those who really must know all Solution (Continued, with algebra help) ∫ ∫ ∫ ( ) u du 1 1 1 tan x dx = = − du 1 − u2 2 1−u 1+u 1 1 = − ln |1 − u| − ln |1 + u| + C 2 2 1 1 = ln √ + C = ln √ +C (1 − u)(1 + u) 1 − u2 1 = ln + C = ln |sec x| + C |cos x| . . . . . .
41. 41. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indeﬁnite Integrals Theory Examples Substitution for Deﬁnite Integrals Theory Examples . . . . . .
42. 42. 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 ) . . . . . .
43. 43. 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 ) Why the change in the limits? The integral on the left 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 . . . . . .
44. 44. Example ∫ π Compute cos2 x sin x dx. 0 . . . . . .
45. 45. 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. . . . . . .
46. 46. 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 ∫ ∫ cos x sin x dx = − u2 du 2 = − 1 u3 + C = − 1 cos3 x + C. 3 3 Therefore ∫ π 1 π 1( ) 2 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3 . . . . . .
47. 47. Solution (Fast Way) Do both the substitution and the evaluation at the same time. . . . . . .
48. 48. 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. . . . . . .
49. 49. 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 1 1( ) 2 = u3 = 1 − (−1) = 3 −1 3 3 . . . . . .
50. 50. 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 1 1( ) 2 = u3 = 1 − (−1) = 3 −1 3 3 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 x variable. But the slow way is just as reliable. . . . . . .
51. 51. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 . . . . . .
52. 52. 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 . . . . . .
53. 53. 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 . . . . . .
54. 54. About those limits √ √ 2 e2(ln 3) = eln 3 = eln 3 = 3 . . . . . .
55. 55. About those fractional powers 93/2 = (91/2 )3 = 33 = 27 43/2 = (41/2 )3 = 23 = 8 . . . . . .
56. 56. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1 . . . . . .
57. 57. 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. . . . . . .
58. 58. 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 . . . . . .
59. 59. 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 . . . . . .
60. 60. 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 . . . . . .
61. 61. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 . . . . . .
62. 62. 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 . . . . . .
63. 63. 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 . . . . . .
64. 64. Example Find ∫ ( ) ( ) 3π/2 5 θ 2 θ cot sec dθ. π 6 6 . . . . . .
65. 65. 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? . . . . . .
66. 66. 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 . . . . . .
67. 67. 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 . . . . . .
68. 68. 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 . . . . . .
69. 69. 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 ( ) 1 √ 1 −4 3 [ −4 ]1 √ 3 [ −4 ]1/ 3 6 − u √ = −u 1 / 3 = u 1 4 1/ 3 2 2 3 [ −1/2 −4 ] = (3 ) − (1−1/2 )−4 2 3 3 = [32 − 12 ] = (9 − 1) = 12 2 2 . . . . . .
70. 70. 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 The areas of these two regions are the same. . . . . . .
71. 71. Graphs ∫ π/4 ∫ 1 . 6 cot5 φ sec2 φ dφ . √ 6u−5 du π/6 1/ 3 y . y . . . . . φ . .. u ππ 1 .1 . . .√ 64 3 The areas of these two regions are the same. . . . . . .
72. 72. Final Thoughts Antidifferentiation is a “nonlinear” problem that needs practice, intuition, and perserverance Worksheet in recitation (also to be posted) The whole antidifferentiation story is in Chapter 6 . . . . . .