Evaluate $\displaystyle \int \limits_{0}^{1} \sqrt{\dfrac{1 - x}{1 + x}} \; dx$
Let $I = \displaystyle \int \limits_{0}^{1} \sqrt{\dfrac{1 - x}{1 + x}} \; dx$ $\;\;\; \cdots \; (1)$
Let $1 + x = u$ $\;\;\; \cdots \; (2)$
Differentiating equation $(2)$ gives
$dx = du$ $\;\;\; \cdots \; (2a)$
$\text{When } \begin{cases}
x = 0, & u = 1 + 0 = 1 \\
x = 1, & u = 1 + 1 = 2
\end{cases}$ $\;\;\; \cdots \; (2b)$
Also, from equation $(2)$, $x = u - 1$
$\therefore$ $\;$ $1 - x = 2 - u$ $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can be written as
$I = \displaystyle \int \limits_{1}^{2} \dfrac{\sqrt{2 - u}}{\sqrt{u}} \; du$ $\;\;\; \cdots \; (3)$
Let $\sqrt{u} = v$ $\;\;\; \cdots \; (4)$
Differentiating equation $(4)$ gives
$\dfrac{1}{2 \sqrt{u}} \; du = dv$
$\implies$ $\dfrac{du}{\sqrt{u}} = 2 \; dv$ $\;\;\; \cdots \; (4a)$
$\text{When } \begin{cases}
u = 1, & v = \sqrt{1} = 1 \\
u = 2, & v = \sqrt{2}
\end{cases}$ $\;\;\; \cdots \; (4b)$
Also, from equation $(4)$, $u = v^2$ $\;\;\; \cdots \; (4c)$
$\therefore$ $\;$ In view of equations $(4)$, $(4a)$, $(4b)$ and $(4c)$, equation $(3)$ can be written as
$I = 2 \displaystyle \int \limits_{1}^{\sqrt{2}} \sqrt{2 - v^2} \; dv$ $\;\;\; \cdots \; (5)$
Let $v = \sqrt{2} \sin \theta$ $\;\;\; \cdots \; (6)$
Differentiating equation $(6)$ gives
$dv = \sqrt{2} \cos \theta \; d \theta$ $\;\;\; \cdots \; (6a)$
$\text{When } \begin{cases}
v = 1, & \theta = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4} \\
v = \sqrt{2}, & \theta = \sin^{-1} \left(\dfrac{\sqrt{2}}{\sqrt{2}}\right) = \dfrac{\pi}{2}
\end{cases}$ $\;\;\; \cdots \; (6b)$
Also, from equation $(6)$, $\theta = \sin^{-1} \left(\dfrac{v}{\sqrt{2}}\right)$ $\;\;\; \cdots \; (6c)$
$\therefore$ $\;$ In view of equations $(6)$, $(6a)$, $(6b)$ and $(6c)$, equation $(5)$ can be written as
$\begin{aligned}
I & = 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{2 - 2 \sin^2 \theta} \times \sqrt{2} \cos \theta \; d \theta \\\\
& = 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{2} \left(\sqrt{1 - \sin^2 \theta}\right) \times \sqrt{2} \cos \theta \; d \theta \\\\
& = 4 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d \theta \\\\
& = 4 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2 \theta \; d \theta \\\\
& = 4 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d \theta \\\\
& = 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta + 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2 \theta \; d \theta \\\\
& = 2 \left[\theta\right]_{\pi / 4}^{\pi / 2} + 2 \left[\dfrac{\sin 2 \theta}{2}\right]_{\pi / 4}^{\pi / 2} \\\\
& = 2 \left[\dfrac{\pi}{2} - \dfrac{\pi}{4}\right] + \left[\sin \left(\pi\right) - \sin \left(\dfrac{\pi}{2}\right)\right] \\\\
& = \dfrac{\pi}{2} + \left(0 - 1\right) \\\\
& = \dfrac{\pi}{2} - 1
\end{aligned}$