Question

Prove that $I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sec x}}{\sqrt{\text{cosec}x}+\sqrt{\sec x}}dx=\frac{\pi }{4}$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sec x}}{\sqrt{\text{cosec}x}\sqrt{\sec x}}dx....\left(i\right)$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sec (\frac{\pi }{2}-x)}}{\sqrt{\text{cosec}(\frac{\pi }{2}-x)}\sqrt{\sec (\frac{\pi }{2}-x)}}$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\text{cosec}x}}{\sqrt{\sec x}+\sqrt{\text{cosec}x}}dx....\left(ii\right)$

On adding Eqs. $\left(i\right)$ and $\left(ii\right)$,we get
$2I={\int }_0^{\frac{\pi }{2}}dx={\left[x\right]}_0^{\frac{\pi }{2}}$

$I=\frac{\pi }{4}$