Answer
$$\int\sin^{-1}xdx=x\sin^{-1}x+\cos(\sin^{-1}x)+C$$
Work Step by Step
$$A=\int\sin^{-1}xdx$$
Take $y=f^{-1}(x)=\sin^{-1}x$
So $x=f(y)=\sin y$
Applying the formula $\int f^{-1}(x)dx=xf^{-1}x-\int f(y)dy$, we have
$$A=x\sin^{-1}x-\int\sin ydy$$ $$A=x\sin^{-1}x-(-\cos y)+C$$ $$A=x\sin^{-1}x+\cos y+C$$ $$A=x\sin^{-1}x+\cos(\sin^{-1}x)+C$$
