Evaluate $\displaystyle \int \limits_{0}^{1} \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right) \; dx$
$\begin{aligned}
\text{Let } I & = \int \limits_{0}^{1} \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right) \; dx \\\\
& \left[\begin{aligned}
\text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\
& \text{Here } u = \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right); \; v = 1
\end{aligned}\right] \\\\
& = \left[\sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right) \int dx\right]_{0}^{1} \\
& \hspace{5em} - \int \limits_{0}^{1} \left\{\int dx \times \dfrac{d}{dx} \left[\sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right)\right] \right\} \; dx \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
\text{Now, } \dfrac{d}{dx} \left[\sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right)\right] & = \dfrac{1}{\sqrt{1 - \dfrac{x}{x + 1}}} \times \dfrac{d}{dx} \left(\sqrt{\dfrac{x}{x + 1}}\right) \\\\
& = \sqrt{x + 1} \times \left(\dfrac{\sqrt{x + 1} \times \dfrac{1}{2 \sqrt{x}} - \sqrt{x} \times \dfrac{1}{2 \sqrt{x + 1}}}{x + 1}\right) \\\\
& = \sqrt{x + 1} \times \left(\dfrac{x + 1 - x}{2 \left(x + 1\right) \sqrt{x} \sqrt{x + 1}}\right) \\\\
& = \dfrac{1}{2 \left(x + 1\right) \sqrt{x}} \;\;\; \cdots \; (2)
\end{aligned}$
$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ becomes
$\begin{aligned}
I & = \left[x \; \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right)\right]_{0}^{1} - \dfrac{1}{2} \int \limits_{0}^{1} \dfrac{x \; dx}{\left(x + 1\right) \sqrt{x}} \\\\
& = 1 \times \sin^{-1} \left(\sqrt{\dfrac{1}{2}}\right) - 0 - \dfrac{1}{2} \int \limits_{0}^{1} \dfrac{x \; dx}{\left(x + 1\right) \sqrt{x}} \;\;\; \cdots \; (3)
\end{aligned}$
Consider $\displaystyle \int \limits_{0}^{1} \dfrac{\sqrt{x}}{x + 1} \; dx$ $\;\;\; \cdots \; (4)$
Put $\sqrt{x} = u$ $\;\;\; \cdots \; (5)$
Differentiating equation $(5)$ gives
$\dfrac{1}{2 \sqrt{x}} \; dx = du$ $\implies$ $dx = 2 \sqrt{x} \; du = 2 u \; du$ $\;\;\; \cdots \; (5a)$
$\text{When } \begin{cases}
x = 0, & u = \sqrt{0} = 0 \\
x = 1, & u = \sqrt{1} = 1
\end{cases}$ $\;\;\; \cdots \; (5b)$
$\therefore$ $\;$ In view of equations $(5)$, $(5a)$ and $(5b)$, equation $(4)$ becomes
$\begin{aligned}
\int \limits_{0}{1} \dfrac{\sqrt{x}}{x + 1} \; dx & = \int \limits_{0}^{1} \dfrac{u \times 2u}{u^2 + 1} \; du \\\\
& = 2 \int \limits_{0}^{1} \dfrac{u^2}{u^2 + 1} \; du \\\\
& = 2 \left[\int \limits_{0}^{1} \dfrac{u^2 + 1}{u^2 + 1} \; du - \int \limits_{0}^{1} \dfrac{du}{u^2 + 1}\right] \\\\
& = 2 \left[\int \limits_{0}^{1} du - \int \limits_{0}^{1} \dfrac{du}{u^2 + 1}\right] \\\\
& = 2 \left\{\left[u\right]_{0}^{1} - \left[\tan^{-1} u\right]_{0}^{1} \right\} \\\\
& = 2 \left[1 - 0 - \tan^{-1} \left(1\right) + \tan^{-1} \left(0\right)\right] \\\\
& = 2 \left[1 - \dfrac{\pi}{4}\right] \\\\
& = 2 - \dfrac{\pi}{2} \;\;\; \cdots \; (6)
\end{aligned}$
$\therefore$ $\;$ Substituting equation $(6)$ in equation $(3)$ gives
$\begin{aligned}
I & = \dfrac{\pi}{4} - \dfrac{1}{2} \left(2 - \dfrac{\pi}{2}\right) \\\\
& = \dfrac{\pi}{4} - 1 + \dfrac{\pi}{4} = \dfrac{\pi}{2} - 1
\end{aligned}$