I = $\int\limits_{-a}^{a} \sqrt{\frac{a-x}{a+x}}$ dx = $\int\limits_{-a}^{a}$ $\frac{a-x}{\sqrt{a^2-x^2}}$ dx $\int\limits_{-a}^{a}$ $\frac{a}{\sqrt{a^2-x^2}} dx$ - $\int\limits_{-a}^{a}$ $\frac{x}{\sqrt{a^2-x^2}}$ dx = $I_1+I_2$ Where $I_1$ = $\int\limits_{-a}^{a} \frac{a}{\sqrt{a^2-x^2}}$ dx , which is the integral of an even function And $I_2$ = $\int\limits_{-a}^{a}$ $\frac{x}{\sqrt{a^2-x^2}}$ , which is the integral of an odd function, and so $I_2$ = 0 Now, I = $I_1$ = $\int\limits_{-a}^{a} \frac{a}{\sqrt{a^2-x^2}}$ dx = 2 $\int\limits_{0}^{a} \frac{a}{\sqrt{a^2-x^2}}$ dx 2a $\int\limits_{0}^{a} \frac{1}{\sqrt{a^2-x^2}}$ dx = 2a $\left[ \sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$ = 2a $\left| \sin^{-1}1 - \sin^{-1}0 \right|$ = 2a $\left( \frac{\pi}{2} \right)$ = πa