Evaluate $\displaystyle \int \dfrac{\sin^{-1} x \; dx}{x^2}$ $\;\;\;$ $x \in \left(0,1\right)$
Let $I = \displaystyle \int \dfrac{\sin^{-1} x \; dx}{x^2}$ $\;\;\; \cdots$ (1)
Let $\sin^{-1} x = p$ $\;\;\; \cdots$ (2)
$\implies$ $x = \sin p$ $\;\;\; \cdots$ (2a)
Differentiating equation (2a) gives
$dx = \cos p \; dp$ $\;\;\; \cdots$ (2b)
In view of equations (2), (2a) and (2b), equation (1) becomes
$\begin{aligned}
I & = \int \dfrac{p \; \cos p \; dp}{\sin^2 p} \\\\
& = \int p \; \text{cosec }p \; \cot p \; dp \;\;\; \cdots (3)\\
\end{aligned}$
$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$
Here $u = p$ and $v = \csc p \cot p$
$\begin{aligned}
\therefore \; I & = p \int \csc p \; \cot p \; dp - \int \left\{\int \csc p \; \cot p \; dp \times \dfrac{d}{dp} \left(p\right) \right\} \; dp \\\\
& = - p \; \csc p + \int \csc p \; dp \\\\
& = - p \; \csc p + \log \left|\csc p - \cot p\right| + c \;\;\; \cdots (3a)
\end{aligned}$
From equation (2a), $\csc p = \dfrac{1}{x}$ $\;\;\; \cdots$ (4a)
and $\cot p = \sqrt{1 - \csc^2 p} = \sqrt{\dfrac{1}{x^2} - 1} = \dfrac{\sqrt{1 - x^2}}{x}$ $\;\;\; \cdots$ (4b)
$\therefore$ In view of equations (2), (4a) and (4b), equation (3a) becomes
$I = \dfrac{- \sin^{-1} x}{x} + \log \left|\dfrac{1 - \sqrt{1 - x^2}}{x}\right| + c$