Lesson 30: Integration by Parts

Slide 1: Section 5.6 Integration by Parts Math 1a Introduction to Calculus April 23, 2008 Announcements ◮ Midterm III is Wednesday 4/30 in class (covers §5.1–5.6) ◮ Friday 5/2 is Movie Day! ◮ Final (tentative) 5/23 9:15am ◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 . Image: Flickr user Powi...(ponanwi) ◮ . . . . . .

Slide 2: Announcements ◮ Midterm III is Wednesday 4/30 in class (covers §5.1–5.6) ◮ Friday 5/2 is Movie Day! ◮ Final (tentative) 5/23 9:15am ◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 ◮ Office hours Tues, Weds, 2–4pm SC 323 . . . . . .

Slide 3: Outline Last Time: Integration by Substitution Integration by Parts What to make u and what to make dv? Other dirty tricks Double IBP in the case of exps and sines/cosines Letting dv = dx Substitute, then IBP Reduction formulas: “simplifying” with IBP . . . . . .

Slide 4: Last Time: Integration by Substitution 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 For definite integrals, b g(b) f(g(x))g′ (x) dx = f(u) du. a g(a) . . . . . .

Slide 5: Outline Last Time: Integration by Substitution Integration by Parts What to make u and what to make dv? Other dirty tricks Double IBP in the case of exps and sines/cosines Letting dv = dx Substitute, then IBP Reduction formulas: “simplifying” with IBP . . . . . .

Slide 6: Introduction We can verify that ln x dx = x ln x − x by differentiating the right-hand side: d 1 (x ln x − x) = x · + ln x − 1 = ln x dx x . . . . . .

Slide 7: Introduction We can verify that ln x dx = x ln x − x by differentiating the right-hand side: d 1 (x ln x − x) = x · + ln x − 1 = ln x dx x How could we do that without knowing it ahead of time? . . . . . .

Slide 8: Every rule of antidifferentiation is a rule of differentiation in reverse . . . . . .

Slide 9: Every rule of antidifferentiation is a rule of differentiation in reverse Take the product rule: (uv)′ (x) = u′ (x)v(x) + u(x)v′ (x) . . . . . .

Slide 10: Every rule of antidifferentiation is a rule of differentiation in reverse Take the product rule: (uv)′ (x) = u′ (x)v(x) + u(x)v′ (x) So u(x)v′ (x) = (uv)′ (x) − v(x)u′ (x) . . . . . .

Slide 11: Every rule of antidifferentiation is a rule of differentiation in reverse Take the product rule: (uv)′ (x) = u′ (x)v(x) + u(x)v′ (x) So u(x)v′ (x) = (uv)′ (x) − v(x)u′ (x) Now integrate! . . . . . .

Slide 12: Theorem (Integration by Parts) Let u and v be differentiable functions. Then u(x)v′ (x) dx = u(x)v(x) − v(x)u′ (x) dx. Succinctly, u dv = uv − v du. . . . . . .

Slide 13: Theorem (Integration by Parts) Let u and v be differentiable functions. Then u(x)v′ (x) dx = u(x)v(x) − v(x)u′ (x) dx. Succinctly, u dv = uv − v du. Theorem (Integration by Parts, definite form) b b b u dv = uv − v du. a a a . . . . . .

Slide 14: Outline Last Time: Integration by Substitution Integration by Parts What to make u and what to make dv? Other dirty tricks Double IBP in the case of exps and sines/cosines Letting dv = dx Substitute, then IBP Reduction formulas: “simplifying” with IBP . . . . . .

Slide 15: Example xex dx. Solution A possible choice would be u = ex , dv = x dx. Then du = ex dx and v = 2 x2 . Thus 1 xex dx = 1 x2 ex − 2 1 2 x2 ex dx. This doesn’t make the integral any simpler, though. . . . . . .

Slide 16: Example xex dx. Solution A possible choice would be u = ex , dv = x dx. Then du = ex dx and v = 2 x2 . Thus 1 xex dx = 1 x2 ex − 2 1 2 x2 ex dx. This doesn’t make the integral any simpler, though. Instead, try u = x, dv = ex dx. Then du = dx, and v = ex . Thus xex dx = xex − ex dx = xex − ex + C . . . . . .

Slide 17: Example xex dx. Solution A possible choice would be u = ex , dv = x dx. Then du = ex dx and v = 2 x2 . Thus 1 xex dx = 1 x2 ex − 2 1 2 x2 ex dx. This doesn’t make the integral any simpler, though. Instead, try u = x, dv = ex dx. Then du = dx, and v = ex . Thus xex dx = xex − ex dx = xex − ex + C The moral of the story is to choose u and dv which make v du easier than u dv. Polynomials are a good choice for u. . . . . . .

Slide 18: Example Find x2 ex dx. . . . . . .

Slide 19: Example Find x2 ex dx. Solution Let u = x2 , du = 2x dx. Then x2 ex dx = u2 ex − 2 xex dx and we did the last part already! So x2 ex dx = x2 ex − 2xex + 2ex + C . . . . . .

Slide 20: Example Find x ln x dx. . . . . . .

Slide 21: Example Find x ln x dx. Solution Let u = ln x, dv = x dx. Then 1 2 1 1 x ln x dx = x ln x − x2 · dx 2 2 x 1 1 1 1 = x2 ln x − x dx = x2 ln x − x2 + C 2 2 2 4 . . . . . .

Slide 22: Worksheet 1–7 . . . . . .

Slide 23: Outline Last Time: Integration by Substitution Integration by Parts What to make u and what to make dv? Other dirty tricks Double IBP in the case of exps and sines/cosines Letting dv = dx Substitute, then IBP Reduction formulas: “simplifying” with IBP . . . . . .

Slide 24: Double IBP in the case of exps and sines/cosines Worksheet #8 Example Find e−2x sin(3x) dx . . . . . .

Slide 25: Double IBP in the case of exps and sines/cosines Worksheet #8 Example Find e−2x sin(3x) dx Solution Let u = sin(3x), dv = e−2x dx. Then du = 3 cos(3x) dx and 1 v = − e−2x . So 2 I = − 1 e−2x sin(3x) + 2 3 2 e−2x cos(3x) dx . . . . . .

Slide 26: To do the second integral, let u = cos(3x), dv = e−2x dx. Then I = − 2 e−2x sin(3x) + 1 3 2 − 1 e−2x cos(3x) − 2 3 2 e−2x sin(3x) dx = − 2 e−2x sin(3x) − 4 e−2x cos(3x) − 4 I 1 3 9 We are back where we started, but in a good way. 13 4 I = − 1 e−2x sin(3x) − 4 e−2x cos(3x) 2 3 =⇒ I = − 13 e−2x sin(3x) − 2 3 −2x 13 e cos(3x) . . . . . .

Slide 27: To do the second integral, let u = cos(3x), dv = e−2x dx. Then I = − 2 e−2x sin(3x) + 1 3 2 − 1 e−2x cos(3x) − 2 3 2 e−2x sin(3x) dx = − 2 e−2x sin(3x) − 4 e−2x cos(3x) − 4 I 1 3 9 We are back where we started, but in a good way. 13 4 I = − 1 e−2x sin(3x) − 4 e−2x cos(3x) 2 3 =⇒ I = − 13 e−2x sin(3x) − 2 3 −2x 13 e cos(3x) Danger! Don’t undo your work. The choice u = e−2x , dv = cos(3x) dx will cancel everything on the right-hand side but I. . . . . . .

Slide 28: Letting dv = dx Worksheet #9–10 Example Find arctan x dx . . . . . .

Slide 29: Letting dv = dx Worksheet #9–10 Example Find arctan x dx Solution 1 Let u = arctan x, dv = dx. So du = dx and v = x. Thus 1 + x2 x arctan x dx = x arctan x − dx 1 + x2 Now we can integrate the last one by substituting u = 1 + x2 , du = 2x dx. We get arctan x dx = arctan x − 1 2 ln(1 + x2 ) + C . . . . . .

Slide 30: Integral of the logarithm Example Find ln x dx. . . . . . .

Slide 31: Integral of the logarithm Example Find ln x dx. Solution Let u = ln x, dv = dx. Then ln x dx = x ln x − x + C as predicted. . . . . . .

Slide 32: One more Example arcsin x dx = 1 − x2 + x arcsin x + C . . . . . .

Slide 33: Substitute, then IBP Example Find z(ln z)2 dz . . . . . .

Slide 34: Substitute, then IBP Example Find z(ln z)2 dz Solution Let u = ln z, so z = eu and dz = eu du. Thus z(ln z)2 dz = e2u u2 du . . . . . .

Slide 35: Substitute, then IBP Example Find z(ln z)2 dz Solution Let u = ln z, so z = eu and dz = eu du. Thus z(ln z)2 dz = e2u u2 du Now integrate by parts. . . . . . .

Slide 36: Worksheet #11–13 . . . . . .

Slide 37: Reduction formulas: “simplifying” with IBP Example Find cosn x dx. . . . . . .

Slide 38: Reduction formulas: “simplifying” with IBP Example Find cosn x dx. Solution Let u = cosn−1 x, dv = cos x dx. Then v = sin x and du = (n − 1) cosn−2 (x)(− sin x) dx We get cosn x dx = cosn−1 x sin x + cosn−1 x sin2 x dx Now write sin2 x = 1 − cos2 x, expand, and collect the cosn x dx on the left. You get 1 n−1 cosn x dx = cosn−1 x sin x + cosn−2 x dx n n . . . . . .