${\tan ^{ - 1}}\sqrt {\cfrac{{1 - x}}{{1 + x}}}$

Let $I = \int {{{\tan }^{ - 1}}\sqrt {\cfrac{{1 - x}}{{1 + x}}} dx}$

Let $x = \cos \theta$ $\Rightarrow$ $dx = - \sin \theta d\theta$

$\therefore$ $I = \int {{{\tan }^{ - 1}}\sqrt {\cfrac{{1 - x}}{{1 + x}}} dx = - \int {{{\tan }^{ - 1}}\left( {\tan \cfrac{\theta }{2}} \right)\left( {\sin \theta } \right)d\theta } }$

$= - \int {\cfrac{\theta }{2}\sin \theta d\theta = - \cfrac{1}{2}\left[ {\theta \int {\sin \theta d\theta } - \int {\left( {\cfrac{d}{{d\theta }}\left( \theta \right)\int {\sin \theta d\theta } } \right)d\theta } } \right]}$

$= - \cfrac{1}{2}\left[ {\theta \left( { - \cos \theta } \right) - \int {1\left( { - \cos \theta } \right)d\theta } } \right]$

$= \cfrac{1}{2}\theta \cos \theta - \cfrac{1}{2}\int {\cos \theta d\theta = \cfrac{1}{2}\theta \cos \theta - \cfrac{1}{2}\sin \theta + C}$

$= \cfrac{1}{2}\theta \cos \theta - \cfrac{1}{2}\sqrt {1 - {{\cos }^2}\theta } + C = \cfrac{1}{2}\left[ {x{{\cos }^{ - 1}}x - \sqrt {1 - {x^2}} } \right] + C$