Answer
$$\int \sin ^{-1} t d t=\sqrt{1-t^{2}}+t \sin ^{-1} t+C$$
Work Step by Step
Given $$ \int \sin ^{-1} t d t $$
Let
\begin{align*}
u&=\sin^{-1}t \ \ \ \ \ \ \ \ dv=dt\\
du&=\frac{dt}{\sqrt{1-t^2}}\ \ \ \ \ \ \ \ v= t
\end{align*}
Then
\begin{align*}
\int \sin ^{-1} t d t &= t\sin^{-1}t -\int \frac{tdt}{\sqrt{1-t^2}}\\
&=t \sin ^{-1} t+\sqrt{1-t^{2}} +C
\end{align*}
