Question

$\underset{0}{\overset{a}{\int }}{\sin }^{-1}\sqrt{\frac{x}{a+x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I=\underset{0}{\overset{a}{\int }}{\sin }^{-1}\sqrt{\frac{x}{a+x}}dx \\ \mathrm{Let},x=a{\tan }^2\theta \Rightarrow \theta ={\tan }^{-1}\sqrt{\frac{x}{a}} \\ \mathrm{When},x\to x;\theta \to 0andx\to a;\theta \to \frac{\mathrm{\pi }}{4} \\ \mathrm{and}dx=2a\tan \theta se{c}^2\theta d\theta \\ \mathrm{Then}, \\ I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}{\sin }^{-1}\sqrt{\frac{a{\tan }^2\theta }{a+a{\tan }^2\theta }}2a\tan \theta se{c}^2\theta d\theta \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{2a\int }}{\sin }^{-1}\left(\sin \theta \right)\tan \theta se{c}^2\theta d\theta \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{2a\int }}\theta \tan \theta se{c}^2\theta d\theta \\ \mathrm{Let},\tan \theta =t\Rightarrow \theta ={\tan }^{-1}t \\ \Rightarrow se{c}^2\theta d\theta =dt \\ \mathrm{when},\theta \to 0;t\to 0and\theta \to \frac{\mathrm{\pi }}{4};t\to 1 \\ \mathrm{Then},I=2a{\int }_0^1{\tan }^{-1}ttdt \\ =2a{\int }_0^1{\tan }^{-1}ttdt \\ =2a{\left[{\tan }^{-1}t\frac{t^2}{2}\right]}_0^1-\frac{2a}{2}{\int }_0^1\frac{t^2}{1+t^2}dt \\ =2a\left[\frac{\mathrm{\pi }}{4}\times \frac{1}{2}-0\right]-a{\int }_0^1\left[1-\frac{1}{1+t^2}\right]dt \\ =2a\left[\frac{\mathrm{\pi }}{8}\right]-a{\left[t-{\tan }^{-1}t\right]}_0^1 \\ =\frac{\mathrm{\pi a}}{4}-a\left[1-\frac{\pi }{4}\right] \\ =\frac{\mathrm{\pi a}}{4}-a+\frac{\pi a}{4} \\ =\frac{\mathrm{\pi a}}{2}-a \\ =a\left(\frac{\pi }{2}-1\right) \end{gathered}$$