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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?
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?

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so you're saying you got \[\Large \int e^x\sin(x)dx = C\] ??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if so, then it is incorrect

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Bah! I had a feeling it was wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is the right equation though, yes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this one of those where you go around in a circle and get the same thing back again ? then divide by 2?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1hopefully you have this so far? \[\Large \int udv = uv  \int v du\] \[\Large \int e^x\sin(x)dx = e^x\sin(x)  \int e^x \cos(x)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, that's what I got halfway through. Am I supposed to stop there or did I integrate wrong?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you have to use one more iteration of integration by parts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ooooh. OH. I should've realized. WELP, let's see if I can make this work!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1u = 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 \[\large \int e^x\sin(x)dx = e^x\sin(x)  \color{red}{[wz  \int z dw]}\] \[\large \int e^x\sin(x)dx = e^x\sin(x)  \color{red}{[e^x\cos(x)  \int e^x(\sin(x)dx)]}\] \[\large \int e^x\sin(x)dx = e^x\sin(x)  \color{black}{[e^x\cos(x) + \int e^x\sin(x)dx]}\] \[\large \int e^x\sin(x)dx = e^x\sin(x)  e^x\cos(x)  \int e^x\sin(x)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, 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))?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, wait, that would just cancel out again.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Let \[M = \int e^x\sin(x)dx\] \[\large \int e^x\sin(x)dx = e^x\sin(x)  e^x\cos(x)  \int e^x\sin(x)dx\] \[\large M = e^x\sin(x)  e^x\cos(x)  M\] solve for M and resubstitute

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is it e^xsin(x)  e^xcos(x) + C?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A right Ouroboros of pain and trig, this problem. Okay. What next?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Let \[M = \int e^x\sin(x)dx\] \[\large \int e^x\sin(x)dx = e^x\sin(x)  e^x\cos(x)  \int e^x\sin(x)dx\] \[\large M = e^x\sin(x)  e^x\cos(x)  M\] \[\large M+M = e^x\sin(x)  e^x\cos(x)  M+M\] \[\large 2M = e^x\sin(x)  e^x\cos(x)\] \[\large M = \frac{e^x\sin(x)  e^x\cos(x)}{2}+C\] \[\large \int e^x\sin(x)dx = \frac{e^x\sin(x)  e^x\cos(x)}{2}+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Bless you, kind sir.
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