Transcription of Practice Problems: Integration by Parts (Solutions)
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Practice Problems: Integration by Parts ( solutions )Written by Victoria 25, 2014 The following are solutions to the Integration by Parts Practice problems posted November exsinxdxSolution:Letu= sinx,dv=exdx. Thendu= cosxdxandv=ex. Then exsinxdx=exsinx excosxdxNow we need to use Integration by Parts on the second integral. Letu= cosx,dv= sinxdxandv=ex. Then exsinxdx=exsinx excosx exsinxdxThe right integral is the same as the one we started with! Move it over:2 exsinxdx=exsinx excosxAnd divide by 2: exsinxdx=12(exsinx excosx)This is our final solution, so make sure to add your constantC: exsinxdx=12(exsinx excosx) +C 2. (sin 1x)2dxSolution:Letu= (sin 1x)2,dv=dx. Thendu=2 sin 1x 1 x2dx,v=x. Then (sin 1x)2dx=x(sin 1x)2 2xsin 1x 1 x2dxWe need to use a substitution on the last integral.
This is the same as Problem #1, so Z ewsinwdw= 1 2 (ewsinw ewcosw) + C Plug back in w: Z sin(lnx)dx= 1 2 (xsin(lnx) xcos(lnx)) + C 13. R x3 p 1 + x2dx You can do this problem a couple di erent ways. I will show you two solutions. Solution I: You can actually do this problem without using integration by parts. Use the substitution w= 1 + x2 ...
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