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Math Help - How do you solve this integral?

  1. #1
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    How do you solve this integral?

    \int{x*cotx^2 dx}

    thanks
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  2. #2
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    Quote Originally Posted by mojo0716 View Post
    \int{x*cotx^2 dx}

    thanks
    \int{x*cotx^2 dx}=\int{\frac{x \cos{x^2}}{\sin{x^2}}\,dx}

    Let u=\sin{x^2} which gives \frac{du}{dx}=2x\cos{x^2}.

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

    =\frac{1}{2}\int{\frac{1}{u}\frac{du}{dx}\,dx}

    =\frac{1}{2}\int{\frac{1}{u}\,du}

    =\frac{1}{2}[\ln|u|+c]

    =\frac{1}{2}\ln|\sin{x^2}|+C
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  3. #3
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    Quote Originally Posted by Prove It View Post
    \int{x*cotx^2 dx}=\int{\frac{x \cos{x^2}}{\sin{x^2}}\,dx}

    Let u=\sin{x^2} which gives \frac{du}{dx}=2x\cos{x^2}.

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

    =\frac{1}{2}\int{\frac{1}{u}\frac{du}{dx}\,dx}

    =\frac{1}{2}\int{\frac{1}{u}\,du}

    =\frac{1}{2}[\ln|u|+c]

    =\frac{1}{2}\ln|\sin{x^2}|+C
    ah thanks!
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