Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 10

Evaluate $\displaystyle \int \sin^{-1} x dx$
If we let $u = \sin^{-1} x$ and $dv = dx$, then
$\displaystyle du = \frac{1}{\sqrt{1-x^2}} dx$ and $\displaystyle v = \int dx = x$

So,
$\displaystyle \int \sin^{-1} x dx = uv - \int v du = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx$
To evaluate $\displaystyle \int \frac{x}{\sqrt{1-x^2}} dx$ we let $u_1 = 1 - x^2$, then $du_1 = -2xdx$


Then,

$
\begin{equation}
\begin{aligned}
\int \frac{x}{\sqrt{1-x^2}} dx &= \int \frac{\frac{-du_1}{2}}{\sqrt{u_1}} = \frac{-1}{2} \int u_1^{-\frac{1}{2}} du_1 = -\frac{1}{2} \left[ \frac{u_1^{\frac{1}{2}}}{\frac{1}{2}} \right]\\
\\
&= -(u_1)^{\frac{1}{2}} + c = -(1-x^2)^{\frac{1}{2}} + c
\end{aligned}
\end{equation}
$


Therefore,

$
\begin{equation}
\begin{aligned}
\int \sin^{-1} x dx &= x \sin^{-1} x - \left[ -(1-x^2)^{\frac{1}{2}} \right] + c\\
\\
&= x \sin^{-1} x + (1-x^2)^{\frac{1}{2}} + c
\end{aligned}
\end{equation}
$