Calculus - Integration by Parts

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Please help me solve this integration problem. I don't understand how to do it. Thank you!

  • Calculus - Integration by Parts -

    I do not believe you can do this in one integration-by-parts step. You must use the method twice.

    Let y = 2x and dx = dy/2, to simplify the problem to

    Next, let e^y = u and dv = siny dy
    du = e^y dy and v = -cos y

    (1/2)∫e^y*sin(y)dy = (1/2)*uv -(1/2)∫v*du
    = (1/2)[-u cos y +∫e^y cosy dy]

    Now you must apply integration by parts a second time on the
    +∫e^y cosy dy term

    This will give you an equation for
    ∫e^y*sin(y)dy that involves explicit functions of y.

    When you are all done, substitute 2x for y

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