Practice Problems: Integration by Parts (Solutions)

1 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.

2 Sincex= sinwthen thehypotenuse will be 1, the opposite side will bexand the adjacent side will be 1 x2. Then 2wsinwdw= 2 1 x2sin 1x+ 2xEither of these solutions is fine. So then our integral will look like either one of the solutionsbelow: (sin 1x)2dx=x(sin 1x)2 (2 sin 1xcos (sin 1x) + 2 sin (sin 1x)) +C (sin 1x)2dx=x(sin 1x)2 (2 1 x2sin 1x+ 2x) +C 3. xtan2xdxSolution:Use the identity tan2x= sec2x 1: xtan2xdx= x(sec2x 1)dx= xsec2xdx xdxThe last integral is no problemo. The first integral we need to use Integration by Parts . Letu=x,dv= sec2x. Thendu=dx,v= tanx, so: xsec2xdx=xtanx tanxdxYou can rewrite the last integral as sinxcosxdxand use the substitutionw= cosx. tanxdx= ln|cosx|, so: xsec2xdx=xtanx+ ln|cosx|Plug that into the original integral: xtan2xdx=xtanx+ ln|cosx| 12x2+C 24. 10tcoshtdtSolution:This is quick with tabular method.

3 Letu=t,dv= cosht: 10tcoshtdt=tsinht cosht 10= sinh(1) cosh(1) + cosh(0)You can leave your answer like this. If you want to evaluate it further, remember thatsinhx=ex e x2and coshx=ex+e x2. Then we see that sinh(1) =12(e1 e 1),cosh(1) =12(e1+e 1), and cosh 0 = 1. Then 10tcoshtdt= sinh(1) cosh(1) + cosh(0) = 1 1e 5. z3ezdxSolution:Tabular is the way to go with this baby. Letu=z3,dv=ezdz. Then z3ezdx=z3ez 3z2ez+ 6zez 6ez+C=ez(z3 3z2+ 6z 6) +C 6. 31arctan(1/x)dxSolution:Letu= arctan(1/x),dv=dx. Thendu= dxx2+1(using chain rule),v=x: 31arctan(1/x)dx=xarctan(1/x) 31+ 31xx2+ 1dxThe last integral you can use the substitutionw=x2+ 1. Then: 31arctan(1/x)dx=xarctan(1/x) +12ln (x2+ 1) 31= 3 arctan ( 3) +12ln 4 arctan (1) +12ln 2 = 3 3+12ln 2 4 7. cosxln (sinx)dxSolution:We first need to do a substitution. Letw= sinx, thendw= cosxdx: cosxln (sinx)dx= lnwdw3 Next use Integration by Parts withu= lnw,dv=dw.

4 Thendu=1wdw,v=w: lnwdw=wlnw dw=wlnw wWe need to plug back inw: cosxln (sinx)dx= sinxln(sinx) sinx+C 8. 21(lnx)2x3dxYou can do this problem a couple different ways. I will show you two I:First do the substitutionw= lnx. Thendw=1xdxandx=ew. Then 21(lnx)2x3dx= ln 20w2e2wdw= ln 20w2e 2wdwTabular is easy on this guy: ln 20w2e 2wdw= w22e 2w w2e 2w 14e 2w ln 20= e 2w2(w2+w+12) ln 20= 18((ln 2)2+ ln 2 +32)Solution II:Start of with Integration by Parts . Letu= (lnx)2,dv=1x3dx. Thendu=2 lnxxdx,v= 12x2: 21(lnx)2x3dx= (lnx)22x2 21+ 21lnxx3dxDo Integration by Parts again. Letu= lnx,dv=1x3dx. Thendu=1xdx,v= 12x2: lnxx3dx= lnx2x2 21+ 2112x3dx=( lnx2x2 14x2) 21 Plugging this into the original integral we get: 21(lnx)2x3dx=( (lnx)22x2 lnx2x2 14x2) 21= 12x2((lnx)2+ lnx+12) 21= 18((ln 2)2+ ln 2 +32) 49. cos xdxSolution:First do the substitutionw= x.

5 Thendw=12 xdx 2 xdw=dx 2wdw=dx: cos xdx= 2wcoswdwUsing tabular withu= 2w,dv= coswdwwe get: 2wcoswdw= 2wsinw+ 2 cosw+CPlug back inwto get the final solution: cos xdx= 2 xsin x+ 2 cos x+C 10. /2 3cos( 2)d Note:There was a typo on the original, it should bed instead :Rewrite: /2 3cos( 2)d = /2 2cos( 2)d . Then use the substitutionw= 2, so we havedw= 2 d : /2 3cos( 2)d =12 /2wcoswdwTabular makes this easy withu=w,dv= coswdw:12 /2wcoswdw=12(wsinw+ cosw) /2= 12 4 11. xln(1 +x)dxSolution:Use the substitutionw= 1 +x. Thendw=dxandx=w 1: xln(1 +x)dx= (w 1) lnwdwNext use Integration by Parts withu= lnw,dv= (w 1)dw. Thendu=1wdwandv=(12w2 w): (w 1) lnwdw=(12w2 w)lnw (12w 1)dw5 The right integral is straightforward, so (w 1) lnwdw=(12w2 w)lnw 14w2+w+CNext, plug back inw: xln(1 +x)dx=(12(1 +x)2 (1 +x))ln (1 +x) 14(1 +x)2+ 1 +x+CThis answer is fine.

6 You can simplify it a bit more for kicks and giggles: xln(1 +x)dx=12(x2 1) ln (1 +x) 14x2+12x+C 12. sin(lnx)dxSolution:Use the substitutionw= lnx. Thendw=1xdx xdw=dx ewdw=dxsincex=ewfrom our substitution. Then we have: sin(lnx)dx= ewsinwdwThis is the same as problem #1, so ewsinwdw=12(ewsinw ewcosw) +CPlug back inw: sin(lnx)dx=12(xsin (lnx) xcos (lnx)) +C 13. x3 1 +x2dxYou can do this problem a couple different ways. I will show you two I:You can actually do this problem without using Integration by Parts . Use thesubstitutionw= 1 +x2. Thendw= 2xdxandx2=w 1: x3 1 +x2dx= x x2 1 +x2dx=12 (w 1) wdw=12 (w3/2 w1/2)dw=15w5/2 13w3/2+C=15(1 +x2)5/2 13(1 +x2)3/2+CSolution II:You can use Integration by Parts as well, but it is much more complicated. Rewritethe integral: x3 1 +x2dx= 12x2 2x 1 +x2dx6 Letu=12x2,dv= 2x 1 +x2dx. Thendu=xdx,v=23(1 +x2)3/2(using a substitution ondv): 12x2 2x 1 +x2dx=13x2(1 +x2)3/2 23 x(1 +x2)3/2dxYou can use a substitution on the last integral: 12x2 2x 1 +x2dx=13x2(1 +x2)3/2 215(1 +x2)5/2+C 14.

7 Find the area between the given curves:y=x2lnx,y= 4 lnxSolution:We need to find when the two curves intersect, so set them equal to each other:x2lnx= 4 lnx (x2 4) lnx= 0 (x 2)(x+ 2) lnx= 0 The solutions to this equation arex= 2,2,1. But,x= 2 isn t in our domain (since lnxhas the domain (0, )), so we are going to toss that solution out. This means we are goingto integrate fromx= 1 tox= 2. You can just guess which function is on the top or bottom:A= 21(top function bottom function)dx= 21(4 lnx x2lnx)dx= 21(4 x2) lnxdxUsing Integration by Parts , letu= lnx,dv= (4 x2)dx. Thendu=1xdx,v= 4x 13x3: 21(4 x2) lnxdx=(4x 13x3)lnx 21 21(4 13x2)dx=[(4x 13x3)lnx 4x+19x3] 21=163ln 2 299 15. Use the method of cylindrical shells to the find the volume generated by rotating the regionbounded by the given curves about the specified axis:y=e x,y= 0,x= 1,x= 0 aboutx= :Draw a picture of what is happening.

8 Recall that the volume for a cylinder isV= 2 RH. In this scenario,R= 1 xandH=e x(sinceHis the top function minus thebottom function).xis going from -1 to 0:V= 0 12 (1 x)e xdxUsing tabular is pretty quick withu= 1 x,dv=e xdx: 0 12 (1 x)e xdx= 2 xe x 0 1= 2 e 7