Question

$\int {\sin }^{-1}\sqrt{\frac{x}{a+x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\sin }^{-1}\sqrt{\frac{x}{a+x}}dx \\ \mathrm{Putting}x=a{\tan }^2\theta \\ \Rightarrow \sqrt{\frac{x}{a}}=\tan \theta \\ \Rightarrow dx=a\left(2\tan \theta \right){\sec }^2\mathit{}\theta d\theta \\ \therefore I=\int {\sin }^{-1}\sqrt{\frac{a{\tan }^2\theta }{a+a{\tan }^2\mathit{}\theta }}\left(2a\tan \mathit{}\theta \right){\sec }^2\mathit{}\theta d\theta \\ =\int {\sin }^{-1}\sqrt{\frac{{\tan }^2\theta }{{\sec }^2\theta }}\left(2a\tan \theta {\sec }^2\theta \right)d\theta \\ =2a\int \left[{\sin }^{-1}\left(\sin \mathit{}\theta \right)\tan \theta {\sec }^2\theta \right]d\theta \end{gathered}$$

$$\begin{gathered} =2a\int \underset{I}{\theta }\underset{\mathrm{II}}{\tan \theta {\sec }^2\theta }d\theta \\ =2a\left[\theta \frac{{\tan }^2\theta }{2}-\int 1\frac{{\tan }^2\theta }{2}d\theta \right] \\ =2a\left[\frac{\mathrm{\theta }.{\tan }^2\mathrm{\theta }}{2}-\frac{1}{2}\int \left(se{c}^2\theta -1\right)d\theta \right] \\ =a\mathrm{\theta }{\tan }^2\mathrm{\theta }-a\tan \mathrm{\theta }+a\mathrm{\theta }+\mathrm{C} \\ =a\left(\frac{x}{a}\right){\tan }^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)-a\sqrt{\frac{x}{a}}+a{\tan }^{-1}\sqrt{\frac{x}{a}}+\mathrm{C} \\ =x{\tan }^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+a{\tan }^{-1}\sqrt{\frac{x}{a}}+\mathrm{C} \end{gathered}$$