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husbishag

Compute ∫ x cos(x)dx. Would we use Integration by Parts and compute this as - -sin(x) = Cos(x) + C ????

  • one year ago
  • one year ago

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  1. husbishag
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    I wrote that incorrectly. -sin(x)+ cos(x) + C

    • one year ago
  2. TuringTest
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    close, but not quite

    • one year ago
  3. TuringTest
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    \[u=x\]\[dv=\cos xdx\]\[\int udv=uv-\int vdu\]look carefully and you should be able to see you dropped an x in there

    • one year ago
  4. husbishag
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    @TuringTest , I'm a little confused on your explanation, where have I dropped an x?

    • one year ago
  5. TuringTest
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    u=x, and there is a u on the right of the formula. There should be an x where the u is...

    • one year ago
  6. husbishag
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    @TuringTest I think I see. So, it would be -sin(x) + x cos(x) + C

    • one year ago
  7. TuringTest
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    what you have above is\[vdu-\int uv\]careful, identify your v and u, dv and du!

    • one year ago
  8. TuringTest
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    \[u=x\implies du=dx\]\[dv=\cos xdx\implies v=-\sin x\]\[\int udv=uv-\int vdu\]slowly and carefully plug in each for u, v, and du

    • one year ago
  9. husbishag
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    @TuringTest Thank you for your patience with me. Would it be x sin x + cos x + C?

    • one year ago
  10. RadEn
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    true

    • one year ago
  11. TuringTest
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    \[u=x\implies du=dx\]\[dv=\cos xdx\implies v=-\sin x\]\[\int udv=uv-\int vdu\]\[\int x\cos xdx=x(-\sin x)-\int(-\sin x)dx=-x\sin x+\int\sin xdx\]

    • one year ago
  12. TuringTest
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    crap I made up a negative sign

    • one year ago
  13. RadEn
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    int cosx = sinx not -sinx

    • one year ago
  14. TuringTest
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    \[u=x\implies du=dx\]\[dv=\cos xdx\implies v=\sin x\]\[\int udv=uv-\int vdu\]\[\int x\cos xdx=x(\sin x)-\int(\sin x)dx=x\sin x+\int\sin xdx\]yeah you were right, my bad :P

    • one year ago
  15. husbishag
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    If only one day I could understand math that way both of you do... thanks for your help!

    • one year ago
  16. TuringTest
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    welcome!

    • one year ago
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