Question

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

Answer 1

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