Question

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

Answer 1

$$LetI=\underset{-a}{\overset{a}{\int }}\sqrt{\frac{a-x}{a+x}}dx$$

$$Considerx=a\cos 2yTheny=\frac{1}{2}{\cos }^{-1}\left(\frac{x}{a}\right)$$

$$\Rightarrow dx=-2a\sin 2ydy$$

$$Whenx\to -a;y\to \frac{\pi }{2}andx\to a;y\to 0$$

$$\text{Now, integral becomes},$$

$$I={\int }_{\frac{\pi }{2}}^0-2a\sin 2y\sqrt{\frac{a-a\cos 2y}{a+a\cos 2y}}dy$$

$$={\int }_0^{\frac{\pi }{2}}2a\sin 2y\tan ydy$$

$$=2a{\int }_0^{\frac{\pi }{2}}2\sin y\cos y\frac{\sin y}{\cos y}dy$$

$$=2a{\int }_0^{\frac{\pi }{2}}2{\sin }^{2}ydy$$

$$=2a{\int }_0^{\frac{\pi }{2}}(1-\cos 2y)dy$$

$$=2a{[y-\frac{\sin 2y}{2}]}_0^{\frac{\pi }{2}}$$

$$=2a{[\frac{\pi }{2}-\frac{\sin 2y}{2}]}_0^{\frac{\pi }{2}}$$

$$=\pi a$$