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प्रश्न
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उत्तर
\[\text{We have}, \]
\[I = \int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^\frac{3}{2}} dx\]
\[\text{ Putting sin}^{- 1} x = \theta\]
\[ \Rightarrow x = \sin\theta\]
\[ \Rightarrow dx = \cos\text{ θ dθ}\]
\[ \therefore I = \int\frac{\text{ sin θ θ cosθ dθ }}{\left( 1 - \sin^2 \theta \right)^\frac{3}{2}}\]
\[ = \int\frac{\theta \sin\theta \cos\text{ θ dθ}}{\left( \cos^2 \theta \right)^\frac{3}{2}}\]
\[ = \int\theta\frac{\sin\theta}{\cos^2 \theta} d\theta\]
\[ = \int \theta_I \sec \theta_{II} \tan \text{ θ dθ}\]
\[ = \theta \times \sec\theta - \int1 \times \sec\text{ θ dθ}\]
\[ = \theta \times \sec\theta - \int\sec \text{ θ dθ}\]
\[ = \theta \times \sec\theta - \text{ log }\left| \sec\theta + \tan\theta \right| + C\]
\[ = \frac{\theta}{\cos\theta} - \text{ log }\left| \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \right| + C\]
\[ = \frac{\theta}{\sqrt{1 - \sin^2 \theta}} - \text{ log }\left| \frac{1 + \sin\theta}{\cos\theta} \right| + C\]
\[ = \frac{\theta}{\sqrt{1 - \sin^2 \theta}} - \text{ log }\left| \frac{1 + \sin\theta}{\sqrt{1 - \sin^2 \theta}} \right| + C\]
\[ = \frac{\theta}{\sqrt{1 - \sin^2 \theta}} - \text{ log} \left| \frac{\sqrt{1 + \sin\theta}}{\sqrt{1 - \sin\theta}} \right| + C\]
\[ = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}} - \text{ log }\left| \frac{\sqrt{1 + x}}{\sqrt{1 - x}} \right| + C\]
\[ = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}} - \frac{1}{2} \text{ log} \left| \frac{1 + x}{1 - x} \right| + C\]
