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Math Help - Integration

  1. #1
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    Integration

    Hello! Please help me with intergration:

    Find ∫x tan inverse (x^2) dx, for x from 0 to 1.

    Thank you in advance~
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle\int_0^1 x\arctan x^2dx=\int_0^1\left(\frac{x^2}{2}\right)'\arctan x^2dx=

    \displaystyle=\frac{x^2}{2}\arctan x^2|_0^1 -\int_0^1\frac{x^3}{1+x^4}dx=

    \displaystyle\frac{\pi}{8}-\frac{1}{3}\ln(1+x^4)\left|_0^1\right.=\frac{\pi}{  8}-\frac{1}{3}\ln 2
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  3. #3
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    Hello! Thank you for helping, may I know how did you get from

    \displaystyle\frac{x^2}{2}\arctan x^2|_0^1 -\int_0^1\frac{x^3}{1+x^4}dx to \displaystyle\frac{\pi}{8}-\frac{1}{3}\ln(1+x^4)\left|_0^1\right. ??

    Thank you!
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  4. #4
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    \frac{x^{2}}{2}\arctan x^{2}\bigg|_{0}^{1}=\frac{1^{2}}{2}\arctan 1-\frac{0^{2}}{2}\arctan 0=\frac{\pi }{8}.

    Those are things you should know since you're evaluating definite integrals.

    As for the last integral, it's an \frac14 instead an \frac13.

    ----

    Just for fun: \int_{0}^{1}{x\arctan x^{2}\,dx}=\int_{0}^{1}{\int_{0}^{x^{2}}{\frac{x}{  1+y^{2}}\,dy}\,dx}=\int_{0}^{1}{\int_{\sqrt{y}}^{1  }{\frac{x}{1+y^{2}}\,dx}\,dy}, hence \int_{0}^{1}{x\arctan x^{2}\,dx=\frac{1}{2}\int_{0}^{1}{\frac{1-y}{1+y^{2}}\,dy}=}\frac{\pi }{8}-\frac{\ln 2}{4}.
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