$\int_{0}^{\pi/2}\frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{ an x}}dx$ is equal to

Question

$\int_{0}^{\pi/2}\frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}}dx$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Let $I = \int_{0}^{\pi/2}\frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{	an x}}dx$ …(i)
$\Rightarrow \ I = \int_{0}^{\pi/2}{\frac{\sqrt{	an x}}{\sqrt{\cot x} + \sqrt{	an x}}dx}$ …(ii)
$\left\lbrack put\ x = \left( \frac{\pi}{2} - x ight) ightbrack$
On adding Eqs. (i) and (ii), we get
$2I = \int_{0}^{\pi/2}{1\ dx = \frac{\pi}{2}\  \Rightarrow \ I = \frac{\pi}{4}}$

Q. $\int_{0}^{\pi/2}\frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}}dx$ is equal to

Explanation

Reviewed by Dr. Abubacker

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