1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook