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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.

Practice Problems: Integration by Parts (Solutions) Written by Victoria Kala ... I will show you two solutions. Solution I: First do the substitution w= lnx. Then dw= 1 x dxand x= ew. Then Z 2 1 (lnx)2 x3 dx= Z ln2 0 w e2w dw= Z ln2 0 w2e 2wdw Tabular is easy on this guy: Z ln2 0 w2e 2wdw= w 2 2 e 2w w 2 e 2w 1 4 e 2w ln2ln2 0 = e w 2 w2 + w+ 1 ...

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