Question

$\int (si{n}^{-1}{)}^{2}dx$

Answer 1

Let $I=\int {\left({\sin }^{-1}x\right)}^{2}dx$

$\int {\left({\sin }^{-1}x\right)}^{2}dx=\int 1\times {\left({\sin }^{-1}x\right)}^{2}dx$

Applying integration by parts, we have

$I=x{\left({\sin }^{-1}x\right)}^2-\int x\left(\frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}\right)dx$

$I=x{\left({\sin }^{-1}x\right)}^2-2\int {\sin }^{-1}x\left(\frac{x}{\sqrt{1-x^2}}\right)dx$

Now solving for $\int {\sin }^{-1}x\left(\frac{x}{\sqrt{1-x^2}}\right)dx$,

Let

$u={\sin }^{-1}x$

$v=\frac{x}{\sqrt{1-x^2}}\Rightarrow \int vdx=-\sqrt{1-x^2}$

$\therefore \int {\sin }^{-1}x\left(\frac{x}{\sqrt{1-x^2}}\right)dx={\sin }^{-1}x(-\sqrt{1-x^2})-\int \frac{-\sqrt{1-x^2}}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}{\sin }^{-1}x+x+C$

$\therefore I=x{\left({\sin }^{-1}x\right)}^2-2(-\sqrt{1-x^2}{\sin }^{-1}x+x)+C$

$\Rightarrow I=x{\left({\sin }^{-1}x\right)}^2+2\sqrt{1-x^2}{\sin }^{-1}x-2x+C$

Hence $\int {\left({\sin }^{-1}x\right)}^{2}dx=x{\left({\sin }^{-1}x\right)}^2+2\sqrt{1-x^2}{\sin }^{-1}x-2x+C$.