Question

\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]

Answer 1

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

\[\text{ Putting x }= a \tan^2 \theta\]

\[ \Rightarrow \sqrt{\frac{x}{a}} = \tan \theta\]

\[ \Rightarrow dx = a\left( 2 \tan \theta \right) \sec^2 \text{ θ   dθ  }\]

\[ \therefore I = \int \sin^{- 1} \sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} \left( 2a \tan \theta \right) \sec^2 \text{ θ   dθ  }\]

\[ = \int \sin^{- 1} \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} \left( 2a \tan \theta \sec^2 \theta \right) d\theta\]

\[ = 2a \int \left[ \sin^{- 1} \left( \sin \theta \right)\tan \theta \sec^2 \theta \right] d\theta\]

\[= 2a \int \theta_I \tan \theta_{II} \sec^2 \text{ θ   dθ  }\]

\[ = 2a \left[ \theta\frac{\tan^2 \theta}{2} - \int1\frac{\tan^2 \theta}{2}d\theta \right]\]

\[ = 2a \left[ \frac{\theta . \tan^2 \theta}{2} - \frac{1}{2}\int\left( se c^2 \theta - 1 \right)d\theta \right]\]

\[ = \text{ a }\theta \tan^2 \theta - a \tan \theta + a\theta + C\]

\[ = a\left( \frac{x}{a} \right) \tan^{- 1} \left( \frac{\sqrt{x}}{\sqrt{a}} \right) - a\sqrt{\frac{x}{a}} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C\]

\[ = x \tan^{- 1} \sqrt{\frac{x}{a}} - \sqrt{ax} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C\]