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How do you integrate cosx sin2x?

  1. Apr 13, 2010 #1
    If dy/dx = cosx sin2x and if y = 1 when x = pi/2, what is the value of y when x = 0?

    So, i separate the derivative into dy = cosx sin2x dx

    Then integrate both sides...this is where i'm stuck.

    How do you integrate cosx sin2x? I've tried u-sub and IBP but failed both times!
  2. jcsd
  3. Apr 13, 2010 #2


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    Re: Differentiation...

    Substitute u=sin(x)?
  4. Apr 13, 2010 #3
    Re: Differentiation...

    [tex]\int cosxsin^2x dx[/tex]

    u = sinx
    du = cosx dx

    [tex]\int u^2 du[/tex]

    = [tex]\frac{u^3}{3}[/tex]

    = [tex]\frac{sinx}{3}[/tex]

    plugging in 0 for x gets me [tex]\frac{1}{3}[/tex]


    CORRECTION: Nevermind, i got it. They gave me an initial condition.
    Last edited: Apr 13, 2010
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