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- 1. Section 5.5 Integration by Substitution V63.0121.002.2010Su, Calculus I New York University June 22, 2010 Announcements Tomorrow: Review, Evaluations, Movie Thursday: Final Exam . . . . . .
- 2. Announcements Tomorrow: Review, Evaluations, Movie Thursday: Final Exam roughly half-and-half MC/FR FR is all post-midterm MC might have some pre-midterm stuff on it . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 2 / 37
- 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 V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 3 / 37
- 4. Objectives Given an integral and a substitution, transform the integral into an equivalent one using a substitution Evaluate indefinite integrals using the method of substitution. Evaluate definite integrals using the method of substitution. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 4 / 37
- 5. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 5 / 37
- 6. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 6 / 37
- 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 7 / 37
- 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 7 / 37
- 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 What are we supposed to do with that? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 7 / 37
- 10. No straightforward system of antidifferentiation So far we don’t have any way to find ∫ 2x √ dx x2 + 1 or ∫ tan x dx. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 8 / 37
- 11. No straightforward system of antidifferentiation 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 8 / 37
- 12. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 9 / 37
- 13. Substitution for Indefinite Integrals Example Find ∫ x √ dx. x2 + 1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 10 / 37
- 14. 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 . . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 10 / 37
- 15. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 11 / 37
- 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 11 / 37
- 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 Thus ∫ ∫ ( ) x d√ √ dx = g(x) dx x2 + 1 dx √ √ = g(x) + C = 1 + x2 + C. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 11 / 37
- 18. Leibnizian notation FTW Solution (Same technique, new notation) Let u = x2 + 1. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 12 / 37
- 19. Leibnizian notation FTW Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 12 / 37
- 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 12 / 37
- 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 12 / 37
- 22. 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 12 / 37
- 23. 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 13 / 37
- 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 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 14 / 37
- 25. A polynomial example Example ∫ 2 Use the substitution u = x + 3 to find (x2 + 3)3 4x dx. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 15 / 37
- 26. A polynomial example Example ∫ 2 Use the substitution u = x + 3 to find (x2 + 3)3 4x dx. Solution If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So ∫ (x2 + 3)3 4x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 15 / 37
- 27. A polynomial example Example ∫ 2 Use the substitution u = x + 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 15 / 37
- 28. A polynomial example Example ∫ 2 Use the substitution u = x + 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 = u 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 15 / 37
- 29. A polynomial example Example ∫ 2 Use the substitution u = x + 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 15 / 37
- 30. A polynomial example, by brute force Compare this to multiplying it out: ∫ (x2 + 3)3 4x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 31. A polynomial example, by brute force Compare this to multiplying it out: ∫ ∫ ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 32. A polynomial example, by brute force Compare this to multiplying it out: ∫ ∫ ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 33. A polynomial example, by brute force Compare this to multiplying it out: ∫ ∫ ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 34. A polynomial example, by brute force Compare this to multiplying it out: ∫ ∫ ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 Which would you rather do? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 35. A polynomial example, by brute force Compare this to multiplying it out: ∫ ∫ ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 Which would you rather do? It’s a wash for low powers . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 36. A polynomial example, by brute force Compare this to multiplying it out: ∫ ∫ ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx ∫ ( ) = 4x7 + 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 16 / 37
- 37. 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 there a difference? Is this a problem? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 17 / 37
- 38. 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 there a difference? Is this a problem? No, that’s what +C means! . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 17 / 37
- 39. A slick example Example ∫ Find tan x dx. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 40. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 41. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = . Then du = . So ∫ ∫ sin x tan x dx = dx cos x . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 42. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = . So ∫ ∫ sin x tan x dx = dx cos x . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 43. 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 tan x dx = dx cos x . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 44. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 45. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 46. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 18 / 37
- 47. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 19 / 37
- 48. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 19 / 37
- 49. 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 = = = cos2 x 2 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). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 19 / 37
- 50. For those who really must know all Solution (Continued, 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| There are other ways to do it, too. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 20 / 37
- 51. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 21 / 37
- 52. Substitution for Definite Integrals 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) . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 22 / 37
- 53. Substitution for Definite Integrals 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 22 / 37
- 54. Example ∫ π Compute cos2 x sin x dx. 0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 55. 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 = . Then du = and ∫ cos2 x sin x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 56. 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 = and ∫ cos2 x sin x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 57. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 58. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 59. 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 = − 3 cos3 x + C. 3 1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 60. 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 = − 3 cos3 x + C. 3 1 Therefore ∫ π π 1 cos2 x sin x dx = − cos3 x 0 3 0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 61. 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 = − 3 cos3 x + C. 3 1 Therefore ∫ π 1 π 1( ) cos2 x sin x dx = − cos3 x =− (−1)3 − 13 0 3 0 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 62. 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 = − 3 cos3 x + C. 3 1 Therefore ∫ π 1 π 1( ) 2 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
- 63. Definite-ly Quicker 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) = and u(π) = . So ∫ π cos2 x sin x dx 0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 64. Definite-ly Quicker 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) = . So ∫ π cos2 x sin x dx 0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 65. Definite-ly Quicker 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 cos2 x sin x dx 0 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 66. Definite-ly Quicker 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 ∫ 1 2 cos x sin x dx = −u du = 2 u2 du 0 1 −1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 67. Definite-ly Quicker 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 ∫ 1 2 cos x sin x dx = −u du = 2 u2 du 0 1 −1 1 3 1 1( ) 2 = u = 1 − (−1) = 3 −1 3 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 68. Definite-ly Quicker 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 ∫ 1 2 cos x sin x dx = −u du = 2 u2 du 0 1 −1 1 3 1 1( ) 2 = u = 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 variable (x). . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 69. Definite-ly Quicker 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 ∫ 1 2 cos x sin x dx = −u du = 2 u2 du 0 1 −1 1 3 1 1( ) 2 = u = 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 variable (x). But the slow way is just as reliable. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 24 / 37
- 70. An exponential example Example ∫ ln √8 √ 2x Find √ e e2x + 1 dx ln 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 25 / 37
- 71. An exponential example Example ∫ ln √8 √ 2x Find √ e e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ 2x √ e e2x + 1 dx = u + 1 du ln 3 2 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 25 / 37
- 72. About those limits Since √ √ 2 e2(ln 3) = eln 3 = eln 3 = 3 we have √ ∫ ln 8 √ ∫ 8√ 2x 1 √ e e2x + 1 dx = u + 1 du ln 3 2 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 26 / 37
- 73. An exponential example Example ∫ ln √8 √ 2x Find √ e e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ 2x √ e 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 . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 27 / 37
- 74. 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 3/2 1 19 y1/2 dy = · y = (27 − 8) = 2 4 2 3 4 3 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 28 / 37
- 75. An exponential example Example ∫ ln √8 √ 2x Find √ e e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ 2x √ e 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 . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 29 / 37
- 76. Another way to skin that cat Example ∫ ln √8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 30 / 37
- 77. 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. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 30 / 37
- 78. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 30 / 37
- 79. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 30 / 37
- 80. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 30 / 37
- 81. A third skinned cat Example ∫ ln √8 √ 2x Find √ e e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 31 / 37
- 82. A third skinned cat Example ∫ ln √8 √ 2x Find √ e e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 31 / 37
- 83. A third skinned cat Example ∫ ln √8 √ 2x Find √ e 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 19 √ = u · u du = u3 = ln 3 2 3 2 3 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 31 / 37
- 84. A Trigonometric Example Example Find ∫ ( ) ( ) 3π/2 5θ 2 θ cot sec dθ. π 6 6 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 32 / 37
- 85. A Trigonometric Example Example Find ∫ ( ) ( ) 3π/2 5θ 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? . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 32 / 37
- 86. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 33 / 37
- 87. 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 −4 1 3 =6 − u √ = [9 − 1] = 12. 4 1/ 3 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 33 / 37
- 88. 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 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 34 / 37
- 89. 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 −4 1 3 [ −4 ]1 3 [ −4 ]1/ 3 6 − u √ = −u √ = u 4 1/ 3 2 1/ 3 2 1 3 [ ] = (3−1/2 )−4 − (1−1/2 )−4 2 3 3 = [32 − 12 ] = (9 − 1) = 12 2 2 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 34 / 37
- 90. Graphs ∫ 3π/2 ( ) ( ) ∫ π/4 θ 2 θ . cot 5 sec dθ . 6 cot5 φ sec2 φ dφ π 6 6 π/6 y . y . . . . . θ . . . φ 3π . π ππ . . . 2 64 The areas of these two regions are the same. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 35 / 37
- 91. Graphs ∫ ∫ π/4 1 5 2 −5 . 6 cot φ sec φ dφ . √ 6u du π/6 1/ 3 y . y . . . . . φ . .. u ππ 1 .1 . . .√ 64 3 The areas of these two regions are the same. . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 36 / 37
- 92. Summary If F is an antiderivative for f, then: ∫ f(g(x))g′ (x) dx = F(g(x)) If F is an antiderivative for f, which is continuous 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) Antidifferentiation in general and substitution in particular is a “nonlinear” problem that needs practice, intuition, and perserverance The whole antidifferentiation story is in Chapter 6 . . . . . . V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 37 / 37

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