Question

${\int }_0^{\pi /2}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}$dx is equal to

• (A) 1
• (B) -1
• (C) $\frac{\pi }{2}$
• (D) $\frac{\pi }{4}$

Answer 1

The Correct Option is D

Explanation:
$$\text{ Let }I={\int }_0^{\pi /2}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx$$Now$$\begin{array}{rl}I& ={\int }_0^{\pi /2}\frac{\sqrt{\cot (\pi /2-x)}}{\sqrt{\cot (\pi /2-x)}+\sqrt{\tan (\pi /2-x)}}dx\\ \Rightarrow I& ={\int }_0^{\pi /2}\frac{\sqrt{\tan x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx\end{array}$$On adding Equations(i) and (ii), we get$$\begin{array}{rl}2I& ={\int }_0^{\pi /2}1dx\\ \Rightarrow 2I& =\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}\end{array}$$