Question

$\underset{0}{\overset{a}{\int }}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I=\underset{0}{\overset{a}{\int }}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx \\ \mathrm{Consider},x^2=a^2\cos 2\theta \\ \Rightarrow 2xdx=-2a^2\sin 2\theta d\theta \\ \Rightarrow xdx=-a^2\sin 2\theta d\theta \\ \mathrm{When},x\to 0;\theta \to \frac{\mathrm{\pi }}{4}\mathrm{and}x\to a;\theta \to 0 \\ \mathrm{Now},\mathrm{integral}\mathrm{becomes}, \\ I={\int }_{\frac{\mathrm{\pi }}{4}}^0-a^2\sin 2\theta \sqrt{\frac{a^2-a^2\cos 2\theta }{a^2+a^2\cos 2\theta }}d\theta \\ ={\int }_{\frac{\mathrm{\pi }}{4}}^0-a^2\sin 2\theta \tan \theta d\theta \\ =a^2{\int }_0^{\frac{\mathrm{\pi }}{4}}2\sin \theta \cos \theta \frac{\sin \theta }{\cos \theta }d\theta \\ =a^2{\int }_0^{\frac{\mathrm{\pi }}{4}}2{\sin }^2\theta d\theta \\ =a^2{\int }_0^{\frac{\mathrm{\pi }}{4}}\left[1-\cos 2\theta \right]d\theta \\ =a^2{\left[\theta -\frac{\sin 2\theta }{2}\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =a^2\left[\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right] \end{gathered}$$