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Math Help - Indefinite integral , how to start

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    Indefinite integral , how to start

    \mbox{If someone tell me how to start in evaluating the indefinite integral}\\ \\ \phantom{xxxxxxx} \int{tan^{-1}\sqrt{\frac{1-x}{1+x}}
    Last edited by CaptainBlack; July 24th 2011 at 05:21 AM. Reason: sort out LaTeX post
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    Re: Indefinite integral , how to start

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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Indefinite integral , how to start

    Maybe with integration by parts:
    \int \arctan\left(\sqrt{\frac{1-x}{1+x}}\right)dx
    = \arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\cdot x - \int x \cdot d\left[\arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\right]

    If we calculate:
    d\left[\arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\right]
    =\frac{D\left(\sqrt{\frac{1-x}{1+x}}\right)}{1+\left(\frac{1-x}{1+x}\right)}dx
    =\frac{\frac{D\left(\frac{1-x}{1+x}\right)}{2\left(\sqrt{\frac{1-x}{1+x}}\right)}}{\frac{2}{1+x}}
    =...=\frac{-1}{2\sqrt{(1-x)(1+x)}}

    So you'll have:
    = \arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\cdot x + \frac{1}{2}\int \frac{xdx}{\sqrt{(1-x)(1+x)}}
    =\arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\cdot x + \frac{1}{2}\int \frac{xdx}{\sqrt{1-x^2}}

    Let 1-x^2=t then xdx=\frac{-1}{2}dt

    That means:
    \int \frac{xdx}{\sqrt{1-x^2}}=\frac{-1}{2} \int \frac{dt}{\sqrt{t}}=\frac{-1}{2} \frac{\sqrt{1-x^2}}{\frac{1}{2}}

    So:
    \int \arctan\left(\sqrt{\frac{1-x}{1+x}}\right)dx=\arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\cdot x-\frac{\sqrt{1-x^2}}{2}+C
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