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Integration by parts

  1. Sep 3, 2010 #1
    The problem statement, all variables and given/known data[/b]

    http://i324.photobucket.com/albums/k327/ProtoGirlEXE/Capture1.jpg [Broken]


    The attempt at a solution[/b]
    http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0635.jpg [Broken]
    Answer I got: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/Capture2.jpg [Broken]

    I thought my answer would be correct, but it was wrong. Where did I mess up at?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 3, 2010 #2
    It looks good until you get to 1/2∫cos2x(2x)dx
    You can't just integrate cos2x and 2x like that because they're multiplied together; you need to use integration by parts again since you have a product there. But before you do that, you can simplify it a bit by pulling the 2 from the (2x) out of the integrand and it will cancel with the 1/2 out front, giving you [STRIKE]∫xcosx dx[/STRIKE] ∫xcos2x dx. Then use u = x, [STRIKE]du = cosx dx[/STRIKE] dv = cos2x dx
     
    Last edited: Sep 3, 2010
  4. Sep 3, 2010 #3
    you mean ∫xcos(2x) dx, right?
    u=x dv=cos(2x)
    du=dx v=(1/2)sin(2x)
    (x^2 +1)((-1/2) cos(2x)) + x((1/2)sin(2x)) - ∫(1/2)sin(2x) dx
    (x^2 +1)((-1/2) cos(2x)) + x((1/2)sin(2x)) - (1/2) ∫sin(2x) dx
    (x^2 +1)((-1/2) cos(2x)) + x((1/2)sin(2x)) + (1/4) cos(2x) +c

    Is that right? did I make an error?
     
  5. Sep 3, 2010 #4
    Sorry about the errors, fixed now.
    And yes, you got it right. :smile:
     
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