4.6.INTEGRATION BY PARTS201Example 301FindRx2exdxIf we selectu=x2du= 2xdxv=exdv=exdxThen, applying formula 4.4 givesZx2exdx=x2ex°2ZxexdxAt this point, we are left with an integral that we still cannot do. However, wesee that it looks simpler than the one we started with. So, we try the integrationby parts formula again. In fact, we already did this integral in an example above,so we will simply use the result.Zx2exdx=x2ex°2 (xex°ex+C)=x2ex°2xex+ 2ex+CNote that we replaced°2CbyC. We only wish to denote that there is someconstant.Example 302FindRx2sinxdxIf we selectu=x2du= 2xdxv=°cosxdv= sinxdxThen, applying formula 4.4 givesZx2sinxdx=°x2cosx+ 2Zxcosxdx(4.6)We apply the integration by parts formula toRxcosxdx. If we selectu=xdu=dxv= sinxdv= cosxdxThen, applying formula 4.4 givesZxcosxdx=xsinx°Zsinxdx=xsinx+ cosx+CIf we use what we just found in equation 4.6, we obtainZx2sinxdx=°x2cosx+ 2 (xsinx+ cosx+C)=°x2cosx+ 2xsinx+ 2 cosx+CHere again, we replaced2CbyC.