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Double checking my work.

  1. Jan 18, 2008 #1
    I have to evaluate the integral

    problem: [integral] x^2 cos mx dx

    my solution w/ steps: u = x^2, du = 2x. dv = cos mx, v = sin mx / m

    (x^2)(sin mx / m) - [integral] (sin mx / m)(2x)

    (x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x)

    (x^2)(sin mx / m) + 2 (cos mx / m) + c [is this correct?]

    any help would be appreciated.
     
  2. jcsd
  3. Jan 18, 2008 #2

    rock.freak667

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    Homework Helper

    you need to integrate (xsinmx)/m by parts again.
     
  4. Jan 18, 2008 #3
    ok when should I do that after which step?
     
  5. Jan 18, 2008 #4
    wait what would that look like?

    (xsinmx)/m

    u = ? dv = ?
     
  6. Jan 18, 2008 #5

    rock.freak667

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    Homework Helper

    try u=x and dv=sinmx/m dx
     
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