Lesson 27: Integration by Substitution (Section 10 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.

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Lesson 27: Integration by Substitution (Section 10 version)

  1. 1. Section 5.5 Integration by Substitution V63.0121, Calculus I April 27, 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 Substitution for Definite Integrals Theory Examples . . . . . .
  3. 3. Office Hours and other help this week In addition to recitation Day Time Who/What Where in WWH M 1:00–2:00 Leingang OH 624 3:30–4:30 Katarina OH 607 5:00–7:00 Curto PS 517 T 1:00–2:00 Leingang OH 624 4:00–5:50 Curto PS 317 W 1:00–2:00 Katarina OH 607 2:00–3:00 Leingang OH 624 R 9:00–10:00am Leingang OH 624 5:00–7:00pm Maria OH 807 F 2:00–4:00 Curto OH 1310 . . . . . .
  4. 4. Final stuff Final is May 8, 2:00–3:50pm in CANT 101/200 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 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 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 f′ (x) = 2x and so d√ x 1 g′ (x) = √ g(x) = √ dx x2 + 1 2 g(x) . . . . . .
  18. 18. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. Then f′ (x) = 2x and so d√ x 1 g′ (x) = √ g(x) = √ dx x2 + 1 2 g(x) Thus ∫( ) ∫ d√ x √ dx = g(x) dx dx x2 + 1 √ √ = 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 ∫ ∫ ∫ 1( ) x 1 √ 1 du = √ √ du dx = 2 u 2u x2 + 1 . . . . . .
  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 ∫ ∫ ∫ 1( ) x 1 √ 1 du = √ √ du dx = 2 u 2u x2 + 1 ∫ 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 ∫ ∫ ∫ 1( ) x 1 √ 1 du = √ √ du dx = 2 u 2u x2 + 1 ∫ 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 14 1 2 u = (x + 3)4 = 2 2 . . . . . .
  27. 27. A polynomial example, the hard way Compare this to multiplying it out: ∫ ∫ (6 ) 2 3 x + 9x4 + 27x2 + 27 4x dx (x + 3) 4x dx = ∫ (7 ) 4x + 36x5 + 108x3 + 108x dx = 18 x + 6x6 + 27x4 + 54x2 = 2 . . . . . .
  28. 28. Compare We have ∫ 12 (x2 + 3)3 4x dx = (x + 3)4 2 ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 2 Now 1( 8 ) 12 (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 ∫ 12 (x2 + 3)3 4x dx = (x + 3)4 + C 2 ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C 2 Now 1( 8 ) 12 (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 ∫ tan x dx. Find . . . . . .
  31. 31. A slick example Example ∫ sin x tan x dx. (Hint: tan x = Find ) cos x . . . . . .
  32. 32. A slick example Example ∫ sin x tan x dx. (Hint: tan x = Find ) cos x Solution Let u = cos x. Then du = − sin x dx. . . . . . .
  33. 33. A slick example Example ∫ sin x tan x dx. (Hint: tan x = Find ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 dx = − tan x dx = du cos x u . . . . . .
  34. 34. A slick example Example ∫ sin x tan x dx. (Hint: tan x = Find ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 dx = − tan x dx = du cos x u = − ln |u| + C . . . . . .
  35. 35. A slick example Example ∫ sin x tan x dx. (Hint: tan x = Find ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 dx = − tan x 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 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 cos2 x sin x dx. Compute 0 . . . . . .
  39. 39. Example ∫ π cos2 x sin x dx. Compute 0 Solution (Slow Way) ∫ cos2 x sin x dx and then First compute the indefinite integral evaluate. . . . . . .
  40. 40. Example ∫ π cos2 x sin x dx. Compute 0 Solution (Slow Way) ∫ cos2 x sin x dx and then First compute the indefinite integral 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 = 2. 3 3 0 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 ∫ ∫ −1 π cos2 x sin x dx = −u2 du 0 1 ∫ 1 u2 du = −1 2 1 31 3 u −1 = = . 3 . . . . . .
  44. 44. An exponential example Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 . . . . . .
  45. 45. An exponential example Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x , so du = 2e2x dx. We have √ ∫ ∫ √ 8√ ln 8 1 e2x e2x + 1 dx = u + 1 du √ 2 ln 3 3 . . . . . .
  46. 46. An exponential example Example√ ∫ √ ln 8 e2x e2x + 1 dx Find √ ln 3 Solution Let u = e2x , so du = 2e2x dx. We have √ ∫ ∫ √ 8√ ln 8 1 e2x e2x + 1 dx = u + 1 du √ 2 ln 3 3 Now let y = u + 1, dy = du. So ∫ ∫ ∫ 8√ 9 9 √ 1 1 1 y1/2 dy u + 1 du = y dy = 2 2 2 3 4 4 9 12 1 19 = · y3/2 = (27 − 8) = 23 3 8 4 . . . . . .
  47. 47. Example Find () () ∫ 3π/2 θ 2θ 5 dθ. cot sec 6 6 π . . . . . .
  48. 48. Solution 1 θ Let φ = . Then dφ = dθ. 6 6 () () ∫ 3π/2 ∫ π/4 5θ 2θ cot5 φ sec2 φ dφ dθ = 6 cot sec 6 6 π π/6 ∫ π/4 sec2 φ dφ =6 tan5 φ π/6 . . . . . .
  49. 49. Solution 1 θ Let φ = . Then dφ = dθ. 6 6 () () ∫ 3π/2 ∫ π/4 5θ 2θ cot5 φ sec2 φ dφ dθ = 6 cot sec 6 6 π π/6 ∫ π/4 sec2 φ dφ =6 tan5 φ π/6 Now let u = tan φ. So du = sec2 φ dφ, and ∫ ∫ 1 sec2 φ dφ π/4 u−5 du 6 =6 √ tan5 φ 1/ 3 π/6 ( ) 1 1 − u−4 =6 √ 4 1/ 3 3 [9 − 1] = 12. = 2 . . . . . .
  50. 50. 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 ex and ex ax and ax (fudge the coefficient) √ 1 x and √ x 1 ln x and x . . . . . .

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