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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?

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  1. jim_thompson5910
    • one year ago
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    so you're saying you got \[\Large \int e^x\sin(x)dx = C\] ??

  2. jim_thompson5910
    • one year ago
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    if so, then it is incorrect

  3. anonymous
    • one year ago
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    Bah! I had a feeling it was wrong.

  4. anonymous
    • one year ago
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    That is the right equation though, yes.

  5. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Yeah, that's what I got halfway through. Am I supposed to stop there or did I integrate wrong?

  8. jim_thompson5910
    • one year ago
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    you have to use one more iteration of integration by parts

  9. anonymous
    • one year ago
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    Ooooh. OH. I should've realized. WELP, let's see if I can make this work!

  10. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    very close

  13. jim_thompson5910
    • one year ago
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    but no

  14. anonymous
    • one year ago
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    Ah, wait, that would just cancel out again.

  15. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    Is it e^xsin(x) - e^xcos(x) + C?

  17. jim_thompson5910
    • one year ago
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    nope

  18. anonymous
    • one year ago
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    A right Ouroboros of pain and trig, this problem. Okay. What next?

  19. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    Bless you, kind sir.

  21. jim_thompson5910
    • one year ago
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    no problem

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