Question

Find : $$\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$$ .

Answer 1

$$I=\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx={\sin }^{-1}x\int \frac{x}{\sqrt{1-x^2}}dx-\int [\frac{d}{dx}({\sin }^{-1}x)\int \frac{x}{\sqrt{1-x^2}}dx]dx$$ parts)

Firstly, let us evaluate the integral $$\int \frac{x}{\sqrt{1-x^2}}dx$$ .

Put 

$$t=1-x^2$$ and $$dt=-2xdx$$ .

So,

$$\int \frac{x}{\sqrt{1-x^2}}dx=-\frac{1}{2}\int \frac{dt}{\sqrt{t}}=-\sqrt{t}=-\sqrt{1-x^2}$$

$$\therefore I=\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$$

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

$$=-\sqrt{1-x^2}{\sin }^{-1}x+\int dx$$

$$=-\sqrt{1-x^2}{\sin }^{-1}x+x+C$$