Question

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

Answer 1

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

\[\text{ Putting x }= \sin \theta\]

\[ \Rightarrow \theta = \sin^{- 1} x\]

\[ \text{and}\ dx = \cos \text{ θ  dθ }\]

\[ \therefore I = \int \frac{\theta . \cos \theta}{\sin^2 \theta}d\theta\]

\[ = \int \theta . \left( \frac{\cos \theta}{\sin \theta} \right) \times \frac{1}{\sin \theta} d\theta\]

\[ = \int \theta_I . \text{ cosec} _{II}  θ  \cot \text{ θ  dθ }\]

\[ = \theta\int cosec \theta \cot \text{ θ  dθ } - \int\left\{ \frac{d}{d\theta}\left( \theta \right)\int cosec \theta \cot \text{ θ  dθ }\right\}d\theta\]

\[ = \theta \left( - \text{ cosec }\theta \right) - \int1 . \left( - cosec \theta \right) d\theta\]

\[ = - \theta \text{ cosec }\theta + \int cosec \text{ θ  dθ }\]

\[ = - \theta \text{ cosec }\theta + \text{ ln }\left| \text{ cosec }\theta - \cot \theta \right| + C\]

\[ = \frac{- \theta}{\sin \theta} + \text{ ln }\left| \frac{1 - \text{ cos }\theta}{\sin \theta} \right| + C\]

\[ = \frac{- \theta}{\sin \theta} + \text{ ln} \left| \frac{1 - \sqrt{1 - \sin^2 \theta}}{\sin \theta} \right| + C\]

\[ = \frac{- \sin^{- 1} x}{x} + \text{ ln } \left| \frac{1 - \sqrt{1 - x^2}}{x} \right| + C \left[ \because \theta = \sin^{- 1} x \right]\]