Question

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

Answer 1

\[\text{We have}, \]

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

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

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

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

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

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

\[ = 2a \int \theta \tan \theta \sec^2 \text{  θ  dθ}\]

\[\text{Considering  θ as first function and  tan   θ  sec}^2 \text{  θ  as second function}\]

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

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

\[ = a\left[ \theta \tan^2 \theta - \tan \theta + \theta \right] + C\]

\[ = a\left[ \theta \times \left( 1 + \tan^2 \theta \right) - \tan \theta \right] + C\]

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

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