Question

Evaluate: ${\int }_{-a}^a\frac{\sqrt{a-x}}{\sqrt{a+x}}dx$

Answer 1

We have to evaluate ${\int }_{-a}^a\frac{\sqrt{a-x}}{\sqrt{a+x}}dx$

Let it be $I={\int }_{-a}^a\frac{\sqrt{a-x}}{\sqrt{a+x}}dx={\int }_{-a}^a\frac{\sqrt{a+x}}{\sqrt{a-x}}dx$

$\Rightarrow 2I={\int }_{-a}^a\frac{\sqrt{a-x}}{\sqrt{a+x}}dx+{\int }_{-a}^a\frac{\sqrt{a+x}}{\sqrt{a-x}}dx={\int }_{-a}^a\frac{2a}{\sqrt{a^2-x^2}}dx$

$\Rightarrow I={\int }_{-a}^a\frac{a}{\sqrt{a^2-x^2}}dx$

Now take $x=asin\theta$ , we get $dx=acos\theta d\theta$

$\Rightarrow I=a{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}d\theta =a\pi$