Integration by Parts Math Example 2

Solution

1. 1
   Let $u = \sin x$, $dv = e^x\,dx$; $du = \cos x\,dx$, $v = e^x$.
2. 2
   $I = e^x\sin x - \int e^x\cos x\,dx$.
3. 3
   IBP again on $\int e^x\cos x\,dx$: $u=\cos x$, $dv=e^x\,dx$; gives $e^x\cos x + \int e^x\sin x\,dx$.
4. 4
   So $I = e^x\sin x - e^x\cos x - I$. Solve: $2I = e^x(\sin x - \cos x)$.
5. 5
   $I = \frac{e^x(\sin x - \cos x)}{2} + C$.