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प्रश्न
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उत्तर
\[\text{ Let I }= \int \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]
\[\text{ Putting x }= a \tan^2 \theta\]
\[ \Rightarrow \sqrt{\frac{x}{a}} = \tan \theta\]
\[ \Rightarrow dx = a\left( 2 \tan \theta \right) \sec^2 \text{ θ dθ }\]
\[ \therefore I = \int \sin^{- 1} \sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} \left( 2a \tan \theta \right) \sec^2 \text{ θ dθ }\]
\[ = \int \sin^{- 1} \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} \left( 2a \tan \theta \sec^2 \theta \right) d\theta\]
\[ = 2a \int \left[ \sin^{- 1} \left( \sin \theta \right)\tan \theta \sec^2 \theta \right] d\theta\]
\[= 2a \int \theta_I \tan \theta_{II} \sec^2 \text{ θ dθ }\]
\[ = 2a \left[ \theta\frac{\tan^2 \theta}{2} - \int1\frac{\tan^2 \theta}{2}d\theta \right]\]
\[ = 2a \left[ \frac{\theta . \tan^2 \theta}{2} - \frac{1}{2}\int\left( se c^2 \theta - 1 \right)d\theta \right]\]
\[ = \text{ a }\theta \tan^2 \theta - a \tan \theta + a\theta + C\]
\[ = a\left( \frac{x}{a} \right) \tan^{- 1} \left( \frac{\sqrt{x}}{\sqrt{a}} \right) - a\sqrt{\frac{x}{a}} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C\]
\[ = x \tan^{- 1} \sqrt{\frac{x}{a}} - \sqrt{ax} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C\]
