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Math Help - integration by parts

  1. #1
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    integration by parts

    Hi the question is evaluate the integral
    integral sign (S) cos^-1 (2x) dx using integration by parts

    SO I let t = 2x so dt = 2 dx so dx= dt/2

    So the original equation becomes S cos^-1 t (dt/2)

    Then I let u = cos^-1 t so du = -dt/(1-t^2))^(1/2)

    then dv = dt so v = t

    So the original equation is now (1/2) S cos^-1 t dt
    = (1/2) ((u cos^-1 u ) - S u (-du/sqrt(1-u^2)))
    = (1/2) (u cos-1u - (1/2)u^2 (-du/sqrt (1-u^2)))

    Is there any way to simplify this further or do I just substitute 2x and 2 dx back into the equation?

    Thanks
    Calculus beginner
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  2. #2
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    u = \arccos(2x)

    dv = dx

    du = -\frac{2}{\sqrt{1 - (2x)^2}} \, dx

    v = x

    \int \arccos(2x) \, dx = x\arccos(2x) + \int \frac{2x}{\sqrt{1 - (2x)^2}} \, dx

    now use the substitution t = 2x for the last integral
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