Question

Solve $\int {\tan }^{-1}\sqrt{\frac{1-x}{1+x}}dx$

Answer 1

$\int ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}dx$

Put $x=cos2\theta$

$dx=-sin2\theta .2d\theta$

$=2\int ta{n}^{-1}\sqrt{\frac{1-cos2\theta }{1+cos2\theta }}(-sin2\theta ).d\theta$

$1-cos\left(2\theta \right)=2si{n}^2\left(\theta \right)$

$1+cos\left(2\theta \right)=2co{s}^2\theta$

$=-2\int ta{n}^{-1}\sqrt{ta{n}^2\theta }.sin\left(2\theta \right).d\theta$

$=-2\int \theta .sin\left(2\theta \right).d\theta$ $\left(ta{n}^{-1}\right(tanx)=x)$

Integration by parts;

$=-2[\theta .sin2\theta d\theta -\int (\frac{d\theta }{d\theta }\int sin2\theta .d\theta \left)d\theta \right]$

$=-2[\frac{\theta (-cos2\theta )}{2}-\int -\frac{cos\left(2\theta \right)d\theta }{2}]$

$=\theta .cos\left(2\theta \right)-\frac{sin\left(2\theta \right)}{2}+c$

$=\frac{1}{2}co{s}^{-1}\left(x\right).\left(x\right)-\frac{\left(\sqrt{1-x^2}\right)}{2}+c$

Since, $cos2\theta =x$

$\theta =\frac{1}{2}co{s}^{-1}\left(x\right)$

$sin\left(2\theta \right)=\sqrt{1-co{s}^{2}2\theta }$

$=\sqrt{1-x^2}$