Answer
$$\int \sin ^{-1} y dy=y\sin ^{-1} y+ (1-y^2)^{\frac{1}{2}}
+C $$
Work Step by Step
Given $$\int \sin ^{-1} y dy $$ Using integration by parts, we get: $$ u=\sin ^{-1} y \Rightarrow du= \frac{1}{\sqrt{1-y^2}}dy $$ $$ dv= dy \Rightarrow v=y $$ So, we get \begin{aligned}
I&=\int \sin ^{-1} y dy\\
&=uv- \int vdu\\
&=y\sin ^{-1} y- \int \frac{y}{\sqrt{1-y^2}}dy\\
&=y\sin ^{-1} y- \frac{1}{2}\int 2y(1-y^2)^{-\frac{1}{2}}dy\\
&=y\sin ^{-1} y+ (1-y^2)^{\frac{1}{2}}
+C \end{aligned}
