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Math Help - Integration by Parts practice

  1. #1
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    Integration by Parts practice

    Hi I was wondering if I got this right so could someone check for me?

    \int x\sin x\cos xdx

    I used integration by parts with

    u=\sin x\cos x and dv=xdx

    and got... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}\int x^2\cos 2xdx

    Then, I used parts again with u=x^2 and dv=\cos 2xdx

    ... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}-\int x\sin 2xdx)

    Parts yet again with u=x and dv=\sin 2xdx

    ... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}+(\frac{x\cos 2x}{2}-\frac{1}{2}\int \cos 2xdx))

    =\frac{x^2\sin 2x}{4}-\frac{1}{2}(\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4})+C

    =\frac{x^2\sin 2x}{4}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C

    =\frac{\sin 2x-2x\cos 2x}{8}+C
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  2. #2
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    Quote Originally Posted by Keithfert488 View Post
    Hi I was wondering if I got this right so could someone check for me?

    \int x\sin x\cos xdx

    I used integration by parts with

    u=\sin x\cos x and dv=xdx

    and got... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}\int x^2\cos 2xdx

    Then, I used parts again with u=x^2 and dv=\cos 2xdx

    ... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}-\int x\sin 2xdx)

    Parts yet again with u=x and dv=\sin 2xdx

    ... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}+(\frac{x\cos 2x}{2}-\frac{1}{2}\int \cos 2xdx))

    =\frac{x^2\sin 2x}{4}-\frac{1}{2}(\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4})+C

    =\frac{x^2\sin 2x}{4}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C

    =\frac{\sin 2x-2x\cos 2x}{8}+C
    Rewrite the integral as \frac{1}{2}\int{x\sin{2x}\,dx}.

    Let u = x so that du = 1

    Let dv = \sin{2x} so that v = -\frac{1}{2}\cos{2x}.


    So \frac{1}{2}\int{x\sin{2x}\,dx} = -\frac{1}{2}x\cos{2x} - \int{-\frac{1}{2}\cos{2x}\cdot 1\,dx}

     = -\frac{1}{2}x\cos{2x} + \frac{1}{2}\int{\cos{2x}\,dx}

     = -\frac{1}{2}x\cos{2x} + \frac{1}{2}\sin{2x} + C.
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  3. #3
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    Quote Originally Posted by Keithfert488 View Post
    Hi I was wondering if I got this right so could someone check for me?

    \int x\sin x\cos xdx

    I used integration by parts with

    u=\sin x\cos x and dv=xdx

    and got... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}\int x^2\cos 2xdx

    Then, I used parts again with u=x^2 and dv=\cos 2xdx

    ... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}-\int x\sin 2xdx)

    Parts yet again with u=x and dv=\sin 2xdx

    ... \frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}+(\frac{x\cos 2x}{2}-\frac{1}{2}\int \cos 2xdx))

    =\frac{x^2\sin 2x}{4}-\frac{1}{2}(\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4})+C

    =\frac{x^2\sin 2x}{4}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C

    =\frac{\sin 2x-2x\cos 2x}{8}+C
    There are several ways you can check this, including:

    1. Using WolframAlpha.

    2. Differentiate your answer and see if it works.
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  4. #4
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    whoa I've never seen WolframAlpha. That's awesome!
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