Question

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

Answer 1

\[\text{We have}, \]

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

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

\[ \Rightarrow x = \sin\theta\]

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

\[ \therefore I = \int\frac{\text{ sin θ   θ  cosθ  dθ }}{\left( 1 - \sin^2 \theta \right)^\frac{3}{2}}\]

\[ = \int\frac{\theta \sin\theta \cos\text{ θ    dθ}}{\left( \cos^2 \theta \right)^\frac{3}{2}}\]

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

\[ = \int \theta_I \sec \theta_{II}  \tan   \text{ θ    dθ}\]

\[ = \theta \times \sec\theta - \int1 \times \sec\text{ θ    dθ}\]

\[ = \theta \times \sec\theta - \int\sec \text{ θ    dθ}\]

\[ = \theta \times \sec\theta - \text{ log }\left| \sec\theta + \tan\theta \right| + C\]

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

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

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

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

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

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