Indefinite integral , how to start

**waqarhaider** · July 24th 2011, 04:57 AM

$\mbox{If someone tell me how to start in evaluating the indefinite integral}\\ \\ \phantom{xxxxxxx} \int{tan^{-1}\sqrt{\frac{1-x}{1+x}}$

**General** · July 24th 2011, 05:38 AM

http://www.mathhelpforum.com/math-he...-s-144704.html

**Siron** · July 24th 2011, 06:54 AM

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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