Calculating $\int \sin x e^x \Bbb d x$ by parts twice

Question

Integrate $e^x \sin x $. I know I need to integrate by parts 2 times, but I'm stuck at the second integration. For the first I get

$$-e^x \cos x - \int e^x\cdot (-\cos x) \,dx $$

Correct me if I'm wrong.

Can you show us what you did, and where you are stuck in the second integration? — Clement C., Dec 19 '15 at 14:38
Clement is right. Show us what you tried. The rest of you: wait until he does. — GEdgar, Dec 19 '15 at 14:40

score 2 · Answer 1 · answered Dec 19 '15 at 14:47

2

Use IBP again:

$u = e^x, dv = \cos x$

You should come up with an $\int e^x \sin(x) dx$ after that. Rearrange the equation to solve for $\int e^x \sin(x) dx$. It's loopy

answered Dec 19 '15 at 14:47

BCLC

14,197

score 2 · Answer 2 · edited Dec 19 '15 at 15:59

2

Start with: $\int e^{x}\sin(x)dx=e^{x}(A\sin(x)+B\cos(x))$

edited Dec 19 '15 at 15:59

Imago

2,142

answered Dec 19 '15 at 14:49

Dr. Sonnhard Graubner

97,058

score 0 · Answer 3 · answered Dec 19 '15 at 17:12

You started with $u=e^x, dv=\sin x dx$. The usual mistake is to take $u=\cos x, dv=e^xdx$ the second time. That leads to a circle and your integral cancels out. You need to take $u=e^x, v=\cos x dx$ the second time. Now things don't cancel, you get twice the original integral.

Calculating $\int \sin x e^x \Bbb d x$ by parts twice

3 Answers3