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Thread: 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?

    $\displaystyle \int x\sin x\cos xdx$

    I used integration by parts with

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

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

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

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

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

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

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

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

    $\displaystyle =\frac{\sin 2x-2x\cos 2x}{8}+C$
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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?

    $\displaystyle \int x\sin x\cos xdx$

    I used integration by parts with

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

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

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

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

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

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

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

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

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

    Let $\displaystyle u = x$ so that $\displaystyle du = 1$

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


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

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

    $\displaystyle = -\frac{1}{2}x\cos{2x} + \frac{1}{2}\sin{2x} + C$.
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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?

    $\displaystyle \int x\sin x\cos xdx$

    I used integration by parts with

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

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

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

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

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

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

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

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

    $\displaystyle =\frac{\sin 2x-2x\cos 2x}{8}+C$
    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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    whoa I've never seen WolframAlpha. That's awesome!
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