## anonymous one year ago Use integration in parts to integrate [e^x(sinx)]dx. I made u=sin(x), du=cos(x)dx, v=e^x and dv=e^xdx. I solved for the whole thing and everything cancelled out, leaving me with C. Is this correct?

1. jim_thompson5910

so you're saying you got $\Large \int e^x\sin(x)dx = C$ ??

2. jim_thompson5910

if so, then it is incorrect

3. anonymous

Bah! I had a feeling it was wrong.

4. anonymous

That is the right equation though, yes.

5. anonymous

is this one of those where you go around in a circle and get the same thing back again ? then divide by 2?

6. jim_thompson5910

hopefully you have this so far? $\Large \int udv = uv - \int v du$ $\Large \int e^x\sin(x)dx = e^x\sin(x) - \int e^x \cos(x)dx$

7. anonymous

Yeah, that's what I got halfway through. Am I supposed to stop there or did I integrate wrong?

8. jim_thompson5910

you have to use one more iteration of integration by parts

9. anonymous

Ooooh. OH. I should've realized. WELP, let's see if I can make this work!

10. jim_thompson5910

u = sin(x), du = cos(x)dx dv = e^x, v = e^x $\large \int udv = uv - \int v du$ $\large \int e^x\sin(x)dx = e^x\sin(x) - \int e^x \cos(x)dx$ $\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{\int e^x \cos(x)dx}$ w = cos(x), dw = -sin(x)dx dz = e^x, z = e^x $\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{[wz - \int z dw]}$ $\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{[e^x\cos(x) - \int e^x(-\sin(x)dx)]}$ $\large \int e^x\sin(x)dx = e^x\sin(x) - \color{black}{[e^x\cos(x) + \int e^x\sin(x)dx]}$ $\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx$

11. anonymous

Yep, just worked out that part! And when I solve out for the last part of the equation, will I end up with e^xsin(x) - 2(e^xcos(x))?

12. jim_thompson5910

very close

13. jim_thompson5910

but no

14. anonymous

Ah, wait, that would just cancel out again.

15. jim_thompson5910

Let $M = \int e^x\sin(x)dx$ $\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx$ $\large M = e^x\sin(x) - e^x\cos(x) - M$ solve for M and re-substitute

16. anonymous

Is it e^xsin(x) - e^xcos(x) + C?

17. jim_thompson5910

nope

18. anonymous

A right Ouroboros of pain and trig, this problem. Okay. What next?

19. jim_thompson5910

Let $M = \int e^x\sin(x)dx$ $\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx$ $\large M = e^x\sin(x) - e^x\cos(x) - M$ $\large M+M = e^x\sin(x) - e^x\cos(x) - M+M$ $\large 2M = e^x\sin(x) - e^x\cos(x)$ $\large M = \frac{e^x\sin(x) - e^x\cos(x)}{2}+C$ $\large \int e^x\sin(x)dx = \frac{e^x\sin(x) - e^x\cos(x)}{2}+C$

20. anonymous

Bless you, kind sir.

21. jim_thompson5910

no problem