Question

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

Answer 1

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

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

Put 

\[t = 1 - x^2\] and \[dt = - 2x dx\] .

So,

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

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

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

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

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