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