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Math Help - Integration by parts

  1. #1
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    Integration by parts

     \int_0^\pi e^{cos(t)}sin(2t)dt

    =

     2\int_0^\pi e^{cos(t)}sin(t)cos(t)dt

     u = cos(t)
    du = -sin(t)

    When  t = \pi u = -1
    When  t = 0 u = 1

     -2\int_1^{-1} ue^u du

    Then integration by parts, differentiating u and integrating  e^u i end up with

     -2e^u(u-1) from 1 to -1

    Doing that I end up with 0 after plugging cos(t) back in for the u. but my calculator tells me the answer is 4e^-1

    What am I doin wrong, I know I'm close...i think :P

    PS: best part is, if i just substitute the -1 and 1 directly into the above equation...i get the right answer


    EDIT: ....im suppose to put -1 and 1 in aren't i....shrug haha
    Last edited by silencecloak; February 17th 2009 at 01:52 PM.
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  2. #2
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    -2\left[-2e^{-1} - 0\right] = 4e^{-1}
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