How to integrate the inverse of sin How does one integrate $\int\ {\sin^{-1}(x)}$, using integration by parts, where:
$$ u={\sin^{-1}}, du=\frac{1}{\sqrt{1-x^2}},dv=dx,  v=x ? $$
This is a partial solution, and I do not quite understand how to move forward from here$$\int\ {\sin^{-1}(x)}= x\sin^{-1} - \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx$$
How do I integrate: $$ \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx?$$

Let $$ J= \int\left( \frac{x}{\sqrt{1-x^2}} \right) dx.$$
Then $$ J=\int\left( \color{green}{x} \cdot \color{red}{\frac{1}{\sqrt{1-x^2}}} \right) \color{green}{dx}.$$
Let $t=1-x^2 \implies \frac{dt}{dx}=-2x \iff \color{green}{xdx}= -\frac{1}{2}dt$.
Then $J=-\frac{1}{2}\int \color{red}{\frac{1}{\sqrt{t}}}dt=-\frac{1}{2} \int t^{-1/2}dt=-t^{1/2}+C=\underbrace{-\sqrt{t}+C=\boxed{-\sqrt{1-x^2}+C}}_{\text{since} \ t:=1-x^2}$
