Question

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

Answer 1

\[\int \frac{x . \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]
\[\text{ Let} \sin^{- 1} x = \theta\]
\[x = \sin \theta\]
\[dx =  \text{ cos   θ  dθ }\]
\[ \therefore \int \frac{x . \sin^{- 1} x}{\sqrt{1 - x^2}}dx = \int \frac{\left( \sin \theta \right) . \theta}{\sqrt{1 - \sin^2 \theta}} . \text{ cos   θ  dθ }\]
\[ = \int \frac{\left( \sin \theta \right) . \theta}{\cos \theta} . \text{ cos   θ  dθ }\]
\[ = \int \theta_I . \sin_{II} \text{   θ  dθ }\]
\[ = \theta\int\sin \text{    θ  dθ }- \int\left\{ \frac{d}{d\theta}\left( \theta \right)\int\sin \text{    θ  dθ }\right\}d\theta\]
\[ = \theta\left( - \cos \theta \right) - \int 1 . \left( - \cos \theta \right) d\theta\]
\[ = - \theta \cos \theta + \sin \theta + C\]
\[ = - \theta \sqrt{1 - \sin^2 \theta} + \sin \theta + C\]
\[ = - \sin^{- 1} x \sqrt{1 - x^2} + x + C \left( \because \sin^{- 1} x = \theta \right)\]