Results 1 to 2 of 2

Thread: Integration by parts

  1. #1
    Member
    Joined
    May 2008
    Posts
    138

    Integration by parts

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

    =

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

    $\displaystyle u = cos(t) $
    $\displaystyle du = -sin(t)$

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

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

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

    $\displaystyle -2e^u(u-1)$ from 1 to -1

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

    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; Feb 17th 2009 at 01:52 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    $\displaystyle -2\left[-2e^{-1} - 0\right] = 4e^{-1}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Jan 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: Apr 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jan 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: Feb 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum