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Integration by parts I believe

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫x sin^2 x dx

    2. Relevant equations

    integration by parts ∫u dv= uv-∫ v du

    3. The attempt at a solution
    u=x dv=1-cos2x
    v= 1/2 sin 2x

    is that correct

    i substituted sin^2 x= 1-cos2x Am I allowed to do that.
  2. jcsd
  3. Aug 26, 2008 #2
    [tex]cos2x = 2cos^2x-1 = 1-2sin^2x=cos^2x-sin^2x[/tex]

    I believe you mean [tex]sin^2x = 1 - cos^2 x[/tex]?
  4. Aug 27, 2008 #3
    oh ok I understand.
  5. Aug 27, 2008 #4
    So it would be ∫x (1-cos^2x) dx

    and then i'd subsitute u= cos x
    du=-sin x

    so then it would be ∫x- u^2 x^2 dx

    is that correct
  6. Aug 27, 2008 #5


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    Homework Helper

    You should use this identity [tex]\sin^2 x = \frac{1}{2}(1-\cos (2x))[/tex].
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