Integration by Parts (IBP)

Review: Integration by Parts (IBP) We typically see integration by parts as: Z Z u dv = uv − v du And this work well for problems where we only need to do IBP once. For example, Z Z x u = ln(x) dv = dx ln(x) dx ⇒ ⇒ x ln(x) − dx = x ln(x) − x + C du = 1/x dx v=x x As an alternate form, we could write: Z f (x) g 0 (x) dx ⇒ sign + − Diff Int f (x) g 0 (x) f 0 (x) g(x) sign + − + Diff Int f (x) g 00 (x) f 0 (x) g 0 (x) f 00 (x) g(x) Z ⇒ f (x)g(x) − g(x)f 0 (x) dx In fact, we can go farther: Z 00 f (x) g (x) dx ⇒ ⇒ 0 Z 0 f (x)g (x) − f (x)g(x) + g(x)f 00 (x) dx And further still: Z 000 f (x) g (x) dx ⇒ sign + − + − Diff Int f (x) g 000 (x) f 0 (x) g 00 (x) f 00 (x) g 0 (x) f 000 (x) g(x) 00 0 0 00 Z ⇒ f (x)g (x) − f (x)g (x) + f (x)g(x) − g(x)f 000 (x) dx This technique works very well when the derivatives of f are 0 (like for polynomials): Z x2 e3x dx For this, we take f (x) = x2 : sign + − + − Diff x2 2x 2 0 Int e3x 1 3x 3e 1 3x 9e 1 3x 27 e ⇒ e 3x  2 1 2 2 x − x+ 3 9 27  1 − 27 Z 0 · e3x dx That last integral is just zero (it’s included so you see the pattern). Below, we’ll work out a couple more examples, then have some sample problems with solutions. Worked Examples 1. t2 sin(2t) dt Just as in the last example, we take f (t) = t2 so that the derivative is eventually 0. sign + − + − Diff t2 2t 2 0 Int sin(2t) − 12 cos(2t) − 14 sin(2t) 1 8 cos(2t) ⇒ − t2 t 1 cos(2t) + sin(2t) + cos(2t) + C 2 2 4 1 Z 2. e2t cos(3t) dt For this problem, we compute IBP twice in order to get the same integral to appear on both sides of the equation. Let’s start and see what that means: sign Diff + e2t − 2e2t + 4e2t Int cos(3t) 1 3 sin(3t) 1 − 9 cos(3t) We interpret this to mean:   Z Z 1 2 4 2t 2t e2t cos(3t) dt e cos(3t) dt = e sin(3t) + cos(3t) − 3 9 9 R Add 49 e2t cos(3t) dt to both sides, and solve for the integral:   Z 13 1 2 2t 2t e cos(3t) dt = e sin(3t) + cos(3t) 9 3 9   Z 2 3 sin(3t) + cos(3t) + C e2t cos(3t) dt = e2t 13 13 R 3. x ln(x) dx We can’t really use the shortcuts here, since the antiderivatives of ln(x) are themselves IBP problems. In that case, we’ll put f (x) = ln(x) instead: sign Diff + ln(x) − 1/x Int x 1 2 x 2 1 2 1 x ln(x) − 2 2 ⇒ Z x dx = 1 2 1 x ln(x) − x2 + C 2 4 Practice Problems Z 1. Z 2. √ Z x ln(x) dx 3. Z 2 x cos(3x) dx 4. Z t3 e−2t dt −2t e 5. tan−1 (1/t) dt Z sin(2t) dt 6. t sin(t) dt Solutions to the Practice (online) Z 1. √ x ln(x) dx sign + − Z 2. Diff ln(x) 1/x Int x1/2 2 3/2 3x ⇒ 2 3/2 2 x ln(x) − 3 3 Z x1/2 dx = 2 3/2 4 x ln(x) − x3/2 + C 3 9 x2 cos(3x) dx sign + − + − Diff x2 2x 2 0 Int cos(3x) 1 3 sin(3x) 1 − 9 cos(3x) 1 − 27 sin(3x) ⇒ 1 2 2x 2 x sin(3x) + cos(3x) − sin(3x) + C 3 9 27 2 Z 3. t3 e−2t dt sign + − + − + Z 4. Diff t3 3t2 6t 6 0 Int e−2t − 12 e−2t 1 −2t 4e − 18 e−2t 1 −2t 16 e ⇒ e −2t   1 2 3 2 3 3 − t − t − t− +C 2 4 4 8 e−2t sin(2t) dt In the example, we had put the exponential in the middle, but we could have switched that with the sine, which we’ll do below (either way is correct): sign + − + Diff Int sin(2t) e−2t 2 cos(2t) − 12 e−2t 1 −2t −4 sin(2t) 4e Therefore, Z e −2t −2t sin(2t) dt = e  Z  1 1 − sin(2t) − cos(2t) − e−2t sin(2t) dt 2 2 so that (add the integral to both sides, divide both sides by 2) Z 1 e−2t sin(2t) dt = − e−2t (sin(2t) + cos(2t)) + C 4 Z 5. tan−1 (1/t) dt In the old style, we let u = tan−1 (1/t). Differentiate to get du: du = 1 −1 −1 · = 1 + (1/t)2 t2 1 + t2 Further, if dv = dt, then v = t, and we get: Z 1 −t t arctan(1/t) − dt = t arctan(t) + ln(t2 + 1) + C 2 1+t 2 Z 6. t sin(t) dt + t sin(t) − 1 − cos(t) + 0 − sin(t) ⇒ 3 −t cos(t) + sin(t) + C