$\begin{equation} \int\limits u\; \mathrm{d} v\end{equation}$ = $uv-\begin{equation} \int\limits v\; \mathrm{d} u\end{equation}$
Let $u = \sin^{-1}\; \; \; \; \; \; \; \; \; \; \; dv =dx$
$\; \; \; du = \frac{1}{\sqrt{1-x^2}} dx \; \; \; \; \; \; \; \; \; \; v = x$
$\therefore \begin{equation} \int\limits_0^1 \sin^{-1} x\ \mathrm{d} x \end{equation} = (x \sin^{-1} x)^1_0$ -$\begin{equation}\int\limits_0^1 \frac{x \;dx}{\sqrt{1-x^2}} \end{equation}$
put $1-x^2 = t $
$\implies -2 x \;dx = dt \; \; \; \; \therefore x\;dx = \frac{dt}{2}$
where x = 0 and x = 1
$ \; \; \; \; t = 1\;and \;t=0$
$\therefore \frac{-1}{2} \begin{equation} \int\limits_0^1 \frac{dt}{\sqrt t} \end{equation}$ = $-\frac{1}{2} [\frac{\sqrt t}{ {\frac {1}{2}} } ]_1^0 = +1$
$\therefore I = [x \; \sin^{-1} x]_0^1 - 1$
$ =[1 \; \; \sin^{-1} (1) \;\;0 \;\; -1]$
$=\frac{\pi}{2} - 1$