Question

If $\underset{-a}{\overset{a}{\int }}\sqrt{\frac{a-x}{a+x}}dx=k\mathrm{\pi },$ then k = ________________.

Answer 1

$$\begin{gathered} \mathrm{Given}:\underset{-a}{\overset{a}{\int }}\sqrt{\frac{a-x}{a+x}}dx=k\mathrm{\pi }...\left(1\right) \\ \mathrm{Let}I=\underset{-a}{\overset{a}{\int }}\sqrt{\frac{a-x}{a+x}}dx \\ \mathrm{We}\mathrm{know}, \\ \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}\left(f\left(x\right)+f\left(-x\right)\right)dx \\ \mathrm{Thus}, \\ I=\underset{0}{\overset{a}{\int }}\left[\sqrt{\frac{a-x}{a+x}}+\sqrt{\frac{a-\left(-x\right)}{a+\left(-x\right)}}\right]dx \\ =\underset{0}{\overset{a}{\int }}\left[\sqrt{\frac{a-x}{a+x}}+\sqrt{\frac{a+x}{a-x}}\right]dx \\ =\underset{0}{\overset{a}{\int }}\left[\frac{\sqrt{a-x}}{\sqrt{a+x}}+\frac{\sqrt{a+x}}{\sqrt{a-x}}\right]dx \\ =\underset{0}{\overset{a}{\int }}\left[\frac{{\left(\sqrt{a-x}\right)}^2+{\left(\sqrt{a+x}\right)}^2}{\sqrt{a+x}\sqrt{a-x}}\right]dx \\ =\underset{0}{\overset{a}{\int }}\left[\frac{a-x+a+x}{\sqrt{a^2-x^2}}\right]dx \\ =\underset{0}{\overset{a}{\int }}\left[\frac{2a}{\sqrt{a^2-x^2}}\right]dx \\ =2a\underset{0}{\overset{a}{\int }}\frac{1}{\sqrt{a^2-x^2}}dx \\ =2a{\left[{\sin }^{-1}\frac{x}{a}\right]}_0^a \\ =2a\left[{\sin }^{-1}\frac{a}{a}-{\sin }^{-1}0\right] \\ =2a\left[{\sin }^{-1}1-0\right] \\ =2a\left[\frac{\mathrm{\pi }}{2}\right] \\ =a\mathrm{\pi }...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right), \\ a\mathrm{\pi }=k\mathrm{\pi } \\ \Rightarrow k=a \end{gathered}$$​

​Hence, k = a.