Transcription of Practice Problems: Integration by Parts (Solutions)
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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. Letw= sin 1x. Thendw=1 1 x2dxandx= sinw. Just looking at the last integral, we have: 2xsin 1x 1 x2dx= 2wsinwdw1We can use Integration by Parts on this last integral by lettingu= 2wanddv= method makes it rather quick: 2wsinwdw= 2wcosw+ 2 sinwAt this point you can plug back inw: 2wsinwdw= 2 sin 1xcos (sin 1x) + 2 sin (sin 1x)OR you can look at the triangle formed by our substitution forw.
vtkala@math.ucsb.edu November 25, 2014 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.
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