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# Lesson 28: Integration by Subsitution

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• 1. Section 5.5 Integration by Substitution V63.0121.002.2010Su, Calculus I New York University June 22, 2010Announcements 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 SquidV63.0121.002.2010Su, 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 &#x222B; x d f(t) dt = f(x) dx a 2. Let f be continuous on [a, b] and f = F&#x2032; for some other function F. Then &#x222B; b f(x) dx = F(b) &#x2212; 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 &#x222B; &#x222B; &#x222B; [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 &#x222B; &#x222B; &#x222B; [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pretty particular, like &#x222B; 1 &#x221A; dx = arcsec x + C. x x2 &#x2212; 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 &#x222B; &#x222B; &#x222B; [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pretty particular, like &#x222B; 1 &#x221A; dx = arcsec x + C. x x2 &#x2212; 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&#x2019;t have any way to find &#x222B; 2x &#x221A; dx x2 + 1 or &#x222B; 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&#x2019;t have any way to find &#x222B; 2x &#x221A; dx x2 + 1 or &#x222B; tan x dx. Luckily, we can be smart and use the &#x201C;anti&#x201D; 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 &#x222B; x &#x221A; 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 &#x222B; x &#x221A; dx. x2 + 1 Solution Stare at this long enough and you notice the the integrand is the &#x221A; 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&#x2032; (x) = 2x and so d&#x221A; 1 x g(x) = &#x221A; g&#x2032; (x) = &#x221A; 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&#x2032; (x) = 2x and so d&#x221A; 1 x g(x) = &#x221A; g&#x2032; (x) = &#x221A; dx 2 g(x) x2 + 1 Thus &#x222B; &#x222B; ( ) x d&#x221A; &#x221A; dx = g(x) dx x2 + 1 dx &#x221A; &#x221A; = 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) &#x221A; &#x221A; 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) &#x221A; &#x221A; Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into &#x222B; &#x222B; 1 &#x222B; &#x221A; x dx 2 du &#x221A; 1 &#x221A; 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) &#x221A; &#x221A; Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into &#x222B; &#x222B; 1 &#x222B; &#x221A; x dx 2 du &#x221A; 1 &#x221A; du = = x2 +1 u 2 u &#x222B; 1 &#x2212;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) &#x221A; &#x221A; Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into &#x222B; &#x222B; 1 &#x222B; &#x221A; x dx 2 du &#x221A; 1 &#x221A; du = = x2 +1 u 2 u &#x222B; 1 &#x2212;1/2 = 2u du &#x221A; &#x221A; = 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) &#x221A; &#x221A; Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. &#x201C;Solve for dx:&#x201D; du dx = 2x So the integrand becomes completely transformed into &#x222B; &#x222B; &#x222B; x x du 1 &#x221A; dx = &#x221A; &#xB7; = &#x221A; du x2+1 u 2x 2 u &#x222B; 1 &#x2212;1/2 = 2u du &#x221A; &#x221A; = u + C = 1 + x2 + C. Mathematicians have serious issues with mixing the x and u like this. However, I can&#x2019;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 &#x222B; &#x222B; &#x2032; f(g(x))g (x) dx = f(u) du That is, if F is an antiderivative for f, then &#x222B; f(g(x))g&#x2032; (x) dx = F(g(x)) In Leibniz notation: &#x222B; &#x222B; 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 &#x222B; 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 &#x222B; 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 &#x222B; (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 &#x222B; 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 &#x222B; &#x222B; &#x222B; (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 &#x222B; 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 &#x222B; &#x222B; &#x222B; (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 &#x222B; 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 &#x222B; &#x222B; &#x222B; (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: &#x222B; (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: &#x222B; &#x222B; ( ) 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: &#x222B; &#x222B; ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx &#x222B; ( ) = 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: &#x222B; &#x222B; ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx &#x222B; ( ) = 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: &#x222B; &#x222B; ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx &#x222B; ( ) = 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: &#x222B; &#x222B; ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx &#x222B; ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 Which would you rather do? It&#x2019;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: &#x222B; &#x222B; ( ) 2 3 (x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx &#x222B; ( ) = 4x7 + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 Which would you rather do? It&#x2019;s a wash for low powers But for higher powers, it&#x2019;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 &#x222B; 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 &#x222B; 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 &#x222B; 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 &#x222B; 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C 2 Is there a difference? Is this a problem? No, that&#x2019;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 &#x222B; 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 &#x222B; 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 &#x222B; sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = . Then du = . So &#x222B; &#x222B; 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 &#x222B; sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = . So &#x222B; &#x222B; 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 &#x222B; sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = &#x2212; sin x dx. So &#x222B; &#x222B; 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 &#x222B; sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = &#x2212; sin x dx. So &#x222B; &#x222B; &#x222B; sin x 1 tan x dx = dx = &#x2212; 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 &#x222B; sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = &#x2212; sin x dx. So &#x222B; &#x222B; &#x222B; sin x 1 tan x dx = dx = &#x2212; du cos x u = &#x2212; 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 &#x222B; sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = &#x2212; sin x dx. So &#x222B; &#x222B; &#x222B; sin x 1 tan x dx = dx = &#x2212; du cos x u = &#x2212; ln |u| + C = &#x2212; 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 &#x222B; 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 &#x222B; 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 &#x222B; 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 &#x222B; &#x222B; &#x222B; sin x u du tan x dx = dx = cos x cos x cos x &#x222B; &#x222B; &#x222B; u du u du u du = = = cos2 x 2 1 &#x2212; sin x 1 &#x2212; u2 At this point, although it&#x2019;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 &#x2212; u2 , so dy = &#x2212;2u du. Then &#x222B; &#x222B; &#x222B; u du u dy tan x dx = = 1 &#x2212; u2 y &#x2212;2u &#x222B; 1 dy 1 1 =&#x2212; = &#x2212; ln |y| + C = &#x2212; ln 1 &#x2212; u2 + C 2 y 2 2 1 1 = ln &#x221A; + C = ln &#x221A; +C 1 &#x2212; u2 1 &#x2212; 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&#x2032; is continuous and f is continuous on the range of u = g(x), then &#x222B; b &#x222B; g(b) &#x2032; 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&#x2032; is continuous and f is continuous on the range of u = g(x), then &#x222B; b &#x222B; g(b) &#x2032; 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 &#x201C;x-land&#x201D; The integral on the right happens in &#x201C;u-land&#x201D;, 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 &#x222B; &#x3C0; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = . Then du = and &#x222B; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = and &#x222B; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = &#x2212; sin x dx and &#x222B; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = &#x2212; sin x dx and &#x222B; &#x222B; cos2 x sin x dx = &#x2212; u2 du . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 23 / 37
• 59. Example &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = &#x2212; sin x dx and &#x222B; &#x222B; cos2 x sin x dx = &#x2212; u2 du = &#x2212; 1 u3 + C = &#x2212; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = &#x2212; sin x dx and &#x222B; &#x222B; cos2 x sin x dx = &#x2212; u2 du = &#x2212; 1 u3 + C = &#x2212; 3 cos3 x + C. 3 1 Therefore &#x222B; &#x3C0; &#x3C0; 1 cos2 x sin x dx = &#x2212; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = &#x2212; sin x dx and &#x222B; &#x222B; cos2 x sin x dx = &#x2212; u2 du = &#x2212; 1 u3 + C = &#x2212; 3 cos3 x + C. 3 1 Therefore &#x222B; &#x3C0; 1 &#x3C0; 1( ) cos2 x sin x dx = &#x2212; cos3 x =&#x2212; (&#x2212;1)3 &#x2212; 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 &#x222B; &#x3C0; Compute cos2 x sin x dx. 0 Solution (Slow Way) &#x222B; First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = &#x2212; sin x dx and &#x222B; &#x222B; cos2 x sin x dx = &#x2212; u2 du = &#x2212; 1 u3 + C = &#x2212; 3 cos3 x + C. 3 1 Therefore &#x222B; &#x3C0; 1 &#x3C0; 1( ) 2 cos2 x sin x dx = &#x2212; cos3 x =&#x2212; (&#x2212;1)3 &#x2212; 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 = &#x2212; sin x dx, u(0) = and u(&#x3C0;) = . So &#x222B; &#x3C0; 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 = &#x2212; sin x dx, u(0) = 1 and u(&#x3C0;) = . So &#x222B; &#x3C0; 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 = &#x2212; sin x dx, u(0) = 1 and u(&#x3C0;) = &#x2212; 1. So &#x222B; &#x3C0; 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 = &#x2212; sin x dx, u(0) = 1 and u(&#x3C0;) = &#x2212; 1. So &#x222B; &#x3C0; &#x222B; &#x2212;1 &#x222B; 1 2 cos x sin x dx = &#x2212;u du = 2 u2 du 0 1 &#x2212;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 = &#x2212; sin x dx, u(0) = 1 and u(&#x3C0;) = &#x2212; 1. So &#x222B; &#x3C0; &#x222B; &#x2212;1 &#x222B; 1 2 cos x sin x dx = &#x2212;u du = 2 u2 du 0 1 &#x2212;1 1 3 1 1( ) 2 = u = 1 &#x2212; (&#x2212;1) = 3 &#x2212;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 = &#x2212; sin x dx, u(0) = 1 and u(&#x3C0;) = &#x2212; 1. So &#x222B; &#x3C0; &#x222B; &#x2212;1 &#x222B; 1 2 cos x sin x dx = &#x2212;u du = 2 u2 du 0 1 &#x2212;1 1 3 1 1( ) 2 = u = 1 &#x2212; (&#x2212;1) = 3 &#x2212;1 3 3 The advantage to the &#x201C;fast way&#x201D; is that you completely transform the integral into something simpler and don&#x2019;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 = &#x2212; sin x dx, u(0) = 1 and u(&#x3C0;) = &#x2212; 1. So &#x222B; &#x3C0; &#x222B; &#x2212;1 &#x222B; 1 2 cos x sin x dx = &#x2212;u du = 2 u2 du 0 1 &#x2212;1 1 3 1 1( ) 2 = u = 1 &#x2212; (&#x2212;1) = 3 &#x2212;1 3 3 The advantage to the &#x201C;fast way&#x201D; is that you completely transform the integral into something simpler and don&#x2019;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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; e e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have &#x222B; &#x221A; &#x222B; ln 8 &#x221A; 1 8&#x221A; 2x &#x221A; 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 &#x221A; &#x221A; 2 e2(ln 3) = eln 3 = eln 3 = 3 we have &#x221A; &#x222B; ln 8 &#x221A; &#x222B; 8&#x221A; 2x 1 &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; e e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have &#x222B; &#x221A; &#x222B; ln 8 &#x221A; 1 8&#x221A; 2x &#x221A; e e2x + 1 dx = u + 1 du ln 3 2 3 Now let y = u + 1, dy = du. So &#x222B; 8&#x221A; &#x222B; 9&#x221A; &#x222B; 9 1 1 1 u + 1 du = y dy = y1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = &#xB7; y3/2 = (27 &#x2212; 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 &#x222B; 9 9 1 1 2 3/2 1 19 y1/2 dy = &#xB7; y = (27 &#x2212; 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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; e e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have &#x222B; &#x221A; &#x222B; ln 8 &#x221A; 1 8&#x221A; 2x &#x221A; e e2x + 1 dx = u + 1 du ln 3 2 3 Now let y = u + 1, dy = du. So &#x222B; 8&#x221A; &#x222B; 9&#x221A; &#x222B; 9 1 1 1 u + 1 du = y dy = y1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = &#xB7; y3/2 = (27 &#x2212; 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 &#x222B; ln &#x221A;8 &#x221A; Find &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; Find &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; Find &#x221A; e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1,so that du = 2e2x dx. Then &#x222B; &#x221A; &#x222B; ln 8 &#x221A; 1 9&#x221A; 2x &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; Find &#x221A; e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1,so that du = 2e2x dx. Then &#x222B; &#x221A; &#x222B; ln 8 &#x221A; 1 9&#x221A; 2x &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; Find &#x221A; e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1,so that du = 2e2x dx. Then &#x222B; &#x221A; &#x222B; ln 8 &#x221A; 1 9&#x221A; 2x &#x221A; e e2x + 1 dx = u du ln 3 2 4 1 3/2 9 = u 3 4 1 19 = (27 &#x2212; 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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; e e2x + 1 dx ln 3 Solution &#x221A; 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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; e e2x + 1 dx ln 3 Solution &#x221A; Let u = e2x + 1, so that u2 = e2x + 1 =&#x21D2; 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 &#x222B; ln &#x221A;8 &#x221A; 2x Find &#x221A; e e2x + 1 dx ln 3 Solution &#x221A; Let u = e2x + 1, so that u2 = e2x + 1 =&#x21D2; 2u du = 2e2x dx Thus &#x221A; &#x222B; ln 8 &#x222B; 3 3 1 19 &#x221A; = u &#xB7; 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 &#x222B; ( ) ( ) 3&#x3C0;/2 5&#x3B8; 2 &#x3B8; cot sec d&#x3B8;. &#x3C0; 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 &#x222B; ( ) ( ) 3&#x3C0;/2 5&#x3B8; 2 &#x3B8; cot sec d&#x3B8;. &#x3C0; 6 6 Before we dive in, think about: What &#x201C;easy&#x201D; 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 &#x3B8; 1 Let &#x3C6; = . Then d&#x3C6; = d&#x3B8;. 6 6 &#x222B; 3&#x3C0;/2 ( ) ( ) &#x222B; &#x3C0;/4 5 &#x3B8; 2 &#x3B8; cot sec d&#x3B8; = 6 cot5 &#x3C6; sec2 &#x3C6; d&#x3C6; &#x3C0; 6 6 &#x3C0;/6 &#x222B; &#x3C0;/4 sec2 &#x3C6; d&#x3C6; =6 &#x3C0;/6 tan5 &#x3C6; . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 33 / 37
• 87. Solution &#x3B8; 1 Let &#x3C6; = . Then d&#x3C6; = d&#x3B8;. 6 6 &#x222B; 3&#x3C0;/2 ( ) ( ) &#x222B; &#x3C0;/4 5 &#x3B8; 2 &#x3B8; cot sec d&#x3B8; = 6 cot5 &#x3C6; sec2 &#x3C6; d&#x3C6; &#x3C0; 6 6 &#x3C0;/6 &#x222B; &#x3C0;/4 sec2 &#x3C6; d&#x3C6; =6 &#x3C0;/6 tan5 &#x3C6; Now let u = tan &#x3C6;. So du = sec2 &#x3C6; d&#x3C6;, and &#x222B; &#x3C0;/4 &#x222B; 1 sec2 &#x3C6; d&#x3C6; &#x2212;5 6 =6 &#x221A; u du &#x3C0;/6 tan5 &#x3C6; 1/ 3 ( ) 1 &#x2212;4 1 3 =6 &#x2212; u &#x221A; = [9 &#x2212; 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 &#x221A; &#x3C0; sin &#x3C0;/4 2/2 tan = =&#x221A; =1 4 cos &#x3C0;/4 2/2 &#x3C0; sin &#x3C0;/6 1/2 1 tan = =&#x221A; =&#x221A; 6 cos &#x3C0;/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 &#x221A; &#x3C0; sin &#x3C0;/4 2/2 tan = =&#x221A; =1 4 cos &#x3C0;/4 2/2 &#x3C0; sin &#x3C0;/6 1/2 1 tan = =&#x221A; =&#x221A; 6 cos &#x3C0;/6 3/2 3 ( ) &#x221A; 1 &#x2212;4 1 3 [ &#x2212;4 ]1 3 [ &#x2212;4 ]1/ 3 6 &#x2212; u &#x221A; = &#x2212;u &#x221A; = u 4 1/ 3 2 1/ 3 2 1 3 [ ] = (3&#x2212;1/2 )&#x2212;4 &#x2212; (1&#x2212;1/2 )&#x2212;4 2 3 3 = [32 &#x2212; 12 ] = (9 &#x2212; 1) = 12 2 2 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 34 / 37
• 90. Graphs &#x222B; 3&#x3C0;/2 ( ) ( ) &#x222B; &#x3C0;/4 &#x3B8; 2 &#x3B8; . cot 5 sec d&#x3B8; . 6 cot5 &#x3C6; sec2 &#x3C6; d&#x3C6; &#x3C0; 6 6 &#x3C0;/6 y . y . . . . . &#x3B8; . . . &#x3C6; 3&#x3C0; . &#x3C0; &#x3C0;&#x3C0; . . . 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 &#x222B; &#x222B; &#x3C0;/4 1 5 2 &#x2212;5 . 6 cot &#x3C6; sec &#x3C6; d&#x3C6; . &#x221A; 6u du &#x3C0;/6 1/ 3 y . y . . . . . &#x3C6; . .. u &#x3C0;&#x3C0; 1 .1 . . .&#x221A; 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: &#x222B; f(g(x))g&#x2032; (x) dx = F(g(x)) If F is an antiderivative for f, which is continuous on the range of g, then: &#x222B; b &#x222B; g(b) f(g(x))g&#x2032; (x) dx = f(u) du = F(g(b)) &#x2212; F(g(a)) a g(a) Antidifferentiation in general and substitution in particular is a &#x201C;nonlinear&#x201D; 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