Evaluate $\displaystyle \int \tan^{-1} \left(\sqrt{\dfrac{1 - x}{1 + x}}\right) \; dx$
$\begin{aligned}
\text{Let } I & = \int \tan^{-1} \left(\sqrt{\dfrac{1 - x}{1 + x}}\right) \; dx \\\\
& = \int \tan^{-1} \left(\dfrac{1 - x}{\sqrt{1 - x^2}}\right) \; dx \;\;\; \cdots \; (1)
\end{aligned}$
Let $x = \cos \theta$ $\;\;\; \cdots \; (2a)$
Differentiating equation $(2a)$ gives
$dx = - \sin \theta \; d\theta$ $\;\;\; \cdots \; (2b)$
$\begin{aligned}
\text{Now, } \dfrac{1 - x}{\sqrt{1 - x^2}} & = \dfrac{1 - \cos \theta}{\sqrt{1 - \cos^2 \theta}} \;\;\; \left[\text{by equation }(2a)\right] \\\\
& = \dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{\sin \theta} \;\;\; \left[\text{Note: }1 - \cos 2 \theta = 2 \sin^2 \theta\right] \\\\
& = \dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)} \;\;\; \left[\text{Note: } \sin 2 \theta = 2 \sin \theta \cos \theta\right] \\\\
& = \dfrac{\sin \left(\dfrac{\theta}{2}\right)}{\cos \left(\dfrac{\theta}{2}\right)}
= \tan \left(\dfrac{\theta}{2}\right) \;\;\; \cdots \; (2c)
\end{aligned}$
$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes
$\begin{aligned}
I & = \int \tan^{-1} \left[\tan \left(\dfrac{\theta}{2}\right)\right] \left(- \sin \theta\right) \; d\theta \\\\
& = - \int \dfrac{\theta}{2} \; \sin \theta \; d\theta \\\\
& \left[\begin{aligned}
\text{Note: } & \int u \; v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \\\\
& \text{Here } u = \theta, \;\; v = \sin \theta
\end{aligned}\right] \\\\
& = \dfrac{-1}{2} \left\{\theta \int \sin \theta \; d \theta - \int \left[\int \sin \theta \; d \theta \times \dfrac{d}{d\theta} \left(\theta\right)\right] \; d\theta \right\} \\\\
& = \dfrac{-1}{2} \left\{- \theta \cos \theta + \int \cos \theta \; d \theta \right\} \\\\
& = \dfrac{\theta \; \cos \theta}{2} - \dfrac{\sin \theta}{2} + c \;\;\; \cdots \; (3)
\end{aligned}$
From equation $(2a)$,
$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$ $\;\;\; \cdots \; (4a)$
and $\theta = \cos^{-1}\left(x\right)$ $\;\;\; \cdots \; (4b)$
$\therefore$ $\;$ In view of equations $(4a)$ and $(4b)$, equation $(3)$ becomes
$I = \dfrac{x \; \cos^{-1}x}{2} - \dfrac{\sqrt{1 - x^2}}{2} + c$