Practice Problems: Integration by Parts

Transcription

1 Practice Problems: Integration by Parts Answers. (a) Neither term will get simpler through differentiation, so let s try some choice for u and dv, and see how it works out (we can always go back and try the other choice if it does not). u = e x dv = cos(x)dx du = e x dx v = sin(x) e x cos(x) dx = e x sin(x) e x sin(x)dx = e x sin(x) + e x sin(x)dx We ll do another integration by parts for the integral on the right and hope we can solve the result for the integral we re looking for: u = e x du = e x dx dv = sin(x)dx v = cos(x) Integrating by parts again, e x cos(x) dx = e x sin(x) + [e ( x ) cos(x) Now add = e x sin(x) e x cos(x) to each side: e x cos(x) e x cos(x) e x cos(x) ] 5 e x cos(x) = e x sin(x) e x cos(x) Finally, multiply both sides by 5, and add our +C integration constant: e x cos(x) = [ e x ] sin(x) e x cos(x) + C 5 NOTE: you can (and should) check your work on any integral problem by differentiating your result and making sure it gives you back the thing you were trying to integrate. (b) The x term will get simpler with differentiation, and we know how to integrate the sec (x) term, so let s choose: u = x dv = sec (x)dx du = dx v = tan(x) x sec (x) dx = x tan(x) x tan(x) tan(x)dx = tan(x)dx

2 We can write tangent as sine over cosine: x sec (x) dx = x tan(x) sin(x) cos(x) dx and then we can use a substitution for the integral on the right: u = cos(x), du = sin(x)dx This leaves: x sec (x) dx = x tan(x) = x tan(x) = x tan(x) u du + ln u + C 6 + ln cos(x) + C 6 (c) The polynomial term will get simpler with differentiation, and we can integrate the exponential term, so let s choose: u = x + x + du = (x + )dx dv = e x dx v = ex (x + x + )e x dx = (x + x + )e x (x + )ex dx = (x + x + )e x (x + )e x dx We ll repeat the process once more for the remaining integral: u = x + du = dx dv = e x dx v = ex (x + x + )e x dx = (x + x + )e x [ ] (x + )ex ex dx = (x + x + )e x 9 (x + )ex + e x dx 9 = (x + x + )e x 9 (x + )ex + 7 ex + C We could combine these by factoring out e x and combining like terms over a common denominator: [ 9 (x + x + )e x dx = 7 (x + x + ) ] (x + ) + e x + C 7 7 = 7 (9x + x + 8)e x + C

3 (d) The x term would get simpler with differentiation, but we cannot easily integrate the natural log term. We can differentiate natural log easily, so let s choose: u = ln x du = x dx dv = x dx v = x x ln x dx = x ln x (x ) x dx = x ln x x dx = x ln x x + C = x ln x x 6 + C (e) Let s simplify this a little before we get started with a substitution, but I will want to use u later, so I ll use w: w = x + 9 dw = dx So the integral becomes e w dw Still not very illuminating, so let s try another substitution: z = w dz = w dw dw = wdz = zdz ze z dz Now it looks better - we can do integration by parts: e x+9 dx = [ ze z u = z du = dz dv = e z dx v = e z ] e z dz = [zez e z ] + C = (z )ez + C Now, we reverse the earlier substitutions: e x+9 dx = ( ) x + 9 e x+9 + C

4 . (a) The x term would get simpler with differentiation, and we can integrate the sine term, so let s choose: u = x dv = sin xdx du = xdx v = cos x π/ x sin x dx = [ x ] π/ π/ cos x x cos x dx = ( ) π cos π π/ ( cos ) + x cos x dx = π 8 + π/ x cos x dx We need a second integration by parts to finish it up: π/ u = x du = dx { [x x sin x dx = π ] π/ 8 + sin x { (π = π 8 + dv = cos xdx v = sin x π/ sin x dx } ) sin π ( sin ) [ cos x]π/ = π 8 π ( cos π + cos ) = 8 } (b) We can differentiate inverse secant but cannot integrate it directly, so choose: / u = sec x du = x dx x [ x x sec x dx = sec x ] = ( sec ) = π π 9 / dv = x dx v = x / x x dx ) sec ( / x x x dx / x x dx In the second line, I used the fact that the domain over which we re integrating is all positive values of x to remove the absolute values, then cancel an x. Now, do a u-substitution to evaluate the integral: u = x, du = x dx x sec x dx = π / π 9 du = π / u π 9 [ ] u / = π 9 ( ) = π 9

5 (c) Let s start with a substitution to clean the integral up a bit first. I ll use w rather then u in case integration by parts is needed later. which makes the integral w = x / dw = xdx sin w dw Now, since I don t know an antiderivative for inverse sine, but I do know its derivative, I ll use integration by parts: and now the integral is / u = sin w du = w dw sin w dw = [ w sin w ] / = ( sin = π / / dv = dw v = w ) ( sin ) w w dw For the remaining integral, we do another substitution z = w dz = wdw w w dw / so (changing limits of integration to being in terms of z), we have / / sin w dw = π + dz z = π + [ ] / z ( = π + ) = π + w w dw (d) We could split this into two integrals, but let s try keeping it together and integrating by parts. Neither e x or (sin x cos x) are going to get simpler with differentiation, so we make an arbitrary choice (assuming we bring the coefficient of outside the integral): So the integral becomes: π u = e x du = e x dx dv = sin x cos xdx v = cos x sin x = [ e x ( cos x sin x) ] π π ( e x )( cos x sin x) dx = [ e π ( cos π sin π) e ( cos sin ) ] = (e π + ) π e x (cos x + sin x) dx π e x (cos x + sin x) dx

6 Now we use integration by parts again on the remaining integral: u = e x du = e x dx dv = cos x + sin xdx v = sin x cos x So the integral becomes: π { [e = (e π + ) x (sin x cos x) ] π } π ( e x )(sin x cos x) dx = (e π + ) [ e π (sin π cos π) e (sin cos ) ] = (e π + ) (e π + ) = π Adding the integral to both sides, 8 π π = and so, dividing by, and moving the back into the integral, π = π