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Practice Problems: Integration by Parts (Solutions)

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.

The following are solutions to the Integration by Parts practice problems posted November 9. 1. R exsinxdx Solution: Let u= sinx, dv= exdx. Then du= cosxdxand v= ex. Then Z exsinxdx= exsinx Z excosxdx Now we need to use integration by parts on the second integral. Let u= cosx, dv= exdx. Then du= sinxdxand v= ex. Then Z exsinxdx= exsinx excosx Z ...

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