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Integration by substitution - I'm stuck

  1. Aug 25, 2012 #1
    θ1. The problem statement, all variables and given/known data

    I'm trying to do an integration by substitution, but I'm completely stuck at the moment

    ∫(1-sin2θ)cosθ dθ

    2. Relevant equations

    ∫u dv = uv - ∫v du

    3. The attempt at a solution

    u = 1 - sin2θ
    dv = cosθ dθ

    du = -2sinθcosθ or -sin(2θ)
    v = sin

    I found du as the derivative of (1 - sin2)
    = 0 - 2(sin)(cos) = -2sinθcosθ

    Now when I insert I get

    (1 - sin2θ)(sinθ) - ∫(sinθ)(-2sinθcosθ)
    sinθ-sin3θ - ∫-2 sin2θcosθ
    sinθ-sin3θ - (-2) ∫sin2θcosθ
    sinθ-sin3θ - (-2) ∫(1-cos2θ)cosθ
    sinθ-sin3θ - (-2) ∫cosθ-cos3θ

    I think I'm on the wrong track here.
    Where did I go wrong? - any help is much appreciated.

    Thanks
     
  2. jcsd
  3. Aug 25, 2012 #2
    I know the answer is

    [itex](sinθ - \frac{sin^3}{3}) + C[/itex]

    and I can see how to differentiate this

    [itex]cosθ - \frac{3}{3}sin^2θcosθ[/itex]
    = [itex](1- sin^2θ)cosθ[/itex]

    But I can't see what I'm doing wrong when trying to go the opposite direction
     
  4. Aug 25, 2012 #3

    dextercioby

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    Substitution or integration by parts ??

    [tex] \int (1-\sin^2 x) \cos x ~ dx = \int (1-\sin^2 x) ~ d(\sin x) = ... [/tex]

    Do you see the substitution ?
     
    Last edited: Aug 25, 2012
  5. Aug 25, 2012 #4

    ehild

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    Homework Helper
    Gold Member

    You tried to do integration by parts, instead of substitution. Substitute u=sinθ.

    ehild
     
  6. Aug 25, 2012 #5
    >>Substitute u=sinθ.

    Ah yes of course.

    ∫(1 - u2) du

    u - u3/3 + C

    Thank you very much.
     
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