Question

\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int \frac{x^2 . \sin^{- 1} x dx}{\left( 1 - x^2 \right)^\frac{3}{2}}\]
\[\text{ Putting  x } = \sin \theta \]
\[ \Rightarrow dx = \cos \text{  θ   dθ } \]
\[\text{and} \theta = \sin^{- 1} x\]
\[ \therefore I = \int \frac{\sin^2 \theta . \theta . \cos \theta \text{ dθ }}{\left( 1 - \sin^2 \theta \right)^\frac{3}{2}}\]
\[ = \int \frac{\sin^2 \theta . \theta . \cos \text{  θ   dθ }}{\left( \cos^2 \theta \right)^\frac{3}{2}}\]
\[ = \int \frac{\sin^2 \theta . \theta . \cos \text{  θ   dθ } }{\cos^3 \theta}\]
\[ = \int \tan^2 \theta . \text{  θ   dθ } \]
\[ = \int \left( \sec^2 \theta - 1 \right)\theta . d\theta\]
\[ = \int \theta_I . \sec^2_{II} \text{  θ   dθ } - \int \theta . d\theta\]
\[ = \theta\int \sec^2 \text{  θ   dθ }  - \int\left\{ \frac{d}{d\theta}\left( \theta \right)\int \sec^2 \text{  θ   dθ } \right\}d\theta - \int \theta . d\theta\]
\[ = \theta \tan \theta - \int 1 . \tan\text{  θ   dθ }  - \frac{\theta^2}{2}\]
\[ = \theta . \tan \theta - \text{ ln }\left| \sec \theta \right| - \frac{\theta^2}{2} + C\]
\[ = \theta . \frac{\sin \theta}{\cos \theta} + \text{ ln }\left| \cos \theta \right| - \frac{\theta^2}{2} + C\]
\[ = \theta . \frac{\sin \theta}{\cos \theta} + \text{ ln }\left| \sqrt{1 - \sin^2 \theta} \right| - \frac{\theta^2}{2} + C\]
\[ = \frac{\theta . \sin \theta}{\sqrt{1 - \sin^2 \theta}} + \frac{1}{2}\text{ ln} \left| 1 - \sin^2 \theta \right| - \frac{\theta^2}{2} + C\]
\[ = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} + \frac{1}{2}\text{  ln }\left( 1 - x^2 \right) - \frac{1}{2} \left( \sin^{- 1} x \right)^2 + C \left[ \because \theta = \sin^{- 1} x \right]\]