Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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

    User Avatar
    Homework Helper

    try u=x and dv=sinmx/m dx
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...