Answer
$$\int \sin^{-1}ydy=y\sin^{-1}y+\sqrt{1-y^2}+C$$
Work Step by Step
$$A=\int \sin^{-1}ydy$$
Set $u=\sin^{-1}y$ and $dv=dy$
Then we would have $du=\frac{1}{\sqrt{1-y^2}}dy$ and $v=y$
Using the formula $\int udv= uv-\int vdu$:
$$A=y\sin^{-1}y-\int\frac{y}{\sqrt{1-y^2}}dy$$
Set $a=\sqrt{1-y^2}$
Then we would have $$da=\frac{(1-y^2)'}{2\sqrt{1-y^2}}dy=\frac{-2y}{2\sqrt{1-y^2}}dy=-\frac{y}{\sqrt{1-y^2}}dy$$
Therefore, $$\frac{y}{\sqrt{1-y^2}}dy=-da$$
Substitute these back into $A$: $$A=y\sin^{-1}y-\int-da$$ $$A=y\sin^{-1}y+a+C$$ $$A=y\sin^{-1}y+\sqrt{1-y^2}+C$$
