anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
so you're saying you got \[\Large \int e^x\sin(x)dx = C\] ??
jim_thompson5910
  • jim_thompson5910
if so, then it is incorrect
anonymous
  • anonymous
Bah! I had a feeling it was wrong.

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anonymous
  • anonymous
That is the right equation though, yes.
anonymous
  • anonymous
is this one of those where you go around in a circle and get the same thing back again ? then divide by 2?
jim_thompson5910
  • jim_thompson5910
hopefully you have this so far? \[\Large \int udv = uv - \int v du\] \[\Large \int e^x\sin(x)dx = e^x\sin(x) - \int e^x \cos(x)dx\]
anonymous
  • anonymous
Yeah, that's what I got halfway through. Am I supposed to stop there or did I integrate wrong?
jim_thompson5910
  • jim_thompson5910
you have to use one more iteration of integration by parts
anonymous
  • anonymous
Ooooh. OH. I should've realized. WELP, let's see if I can make this work!
jim_thompson5910
  • 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 \[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{[wz - \int z dw]}\] \[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{red}{[e^x\cos(x) - \int e^x(-\sin(x)dx)]}\] \[\large \int e^x\sin(x)dx = e^x\sin(x) - \color{black}{[e^x\cos(x) + \int e^x\sin(x)dx]}\] \[\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx\]
anonymous
  • 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))?
jim_thompson5910
  • jim_thompson5910
very close
jim_thompson5910
  • jim_thompson5910
but no
anonymous
  • anonymous
Ah, wait, that would just cancel out again.
jim_thompson5910
  • 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
anonymous
  • anonymous
Is it e^xsin(x) - e^xcos(x) + C?
jim_thompson5910
  • jim_thompson5910
nope
anonymous
  • anonymous
A right Ouroboros of pain and trig, this problem. Okay. What next?
jim_thompson5910
  • jim_thompson5910
Let \[M = \int e^x\sin(x)dx\] \[\large \int e^x\sin(x)dx = e^x\sin(x) - e^x\cos(x) - \int e^x\sin(x)dx\] \[\large M = e^x\sin(x) - e^x\cos(x) - M\] \[\large M+M = e^x\sin(x) - e^x\cos(x) - M+M\] \[\large 2M = e^x\sin(x) - e^x\cos(x)\] \[\large M = \frac{e^x\sin(x) - e^x\cos(x)}{2}+C\] \[\large \int e^x\sin(x)dx = \frac{e^x\sin(x) - e^x\cos(x)}{2}+C\]
anonymous
  • anonymous
Bless you, kind sir.
jim_thompson5910
  • jim_thompson5910
no problem

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