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husbishag

  • 3 years ago

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

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  1. husbishag
    • 3 years ago
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    I wrote that incorrectly. -sin(x)+ cos(x) + C

  2. TuringTest
    • 3 years ago
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    close, but not quite

  3. TuringTest
    • 3 years ago
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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

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

  5. TuringTest
    • 3 years ago
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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...

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

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

  8. TuringTest
    • 3 years ago
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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

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

  10. RadEn
    • 3 years ago
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    true

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

  12. TuringTest
    • 3 years ago
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    crap I made up a negative sign

  13. RadEn
    • 3 years ago
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    int cosx = sinx not -sinx

  14. TuringTest
    • 3 years ago
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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

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

  16. TuringTest
    • 3 years ago
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    welcome!

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