Question

${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{\tan x}}dx=$

• A. $1$
• B. $-1$
• C. $\frac{\mathrm{\pi }}{2}$
• D. $\frac{\mathrm{\pi }}{4}$

Answer 1

The correct option is D

$\frac{\mathrm{\pi }}{4}$

Explanation for Correct answer:

Finding the value of the given integral,

Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{\tan x}}dx\to \left(i\right)$

We know that, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

Replace $x\to 0+\frac{\mathrm{\pi }}{2}-x\left[\text{where}a=0,b=\frac{\mathrm{\pi }}{2}\right]$

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cot\left(0+{\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}}{\sqrt{cot\left(0+\frac{\mathrm{\pi }}{2}-x\right)}+\sqrt{\tan \left(0+\frac{\mathrm{\pi }}{2}-x\right)}}dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{cotx}}\to \left(ii\right)\left[\because cot\left(\frac{\mathrm{\pi }}{2}-x\right)=\tan x,\tan \left(\frac{\mathrm{\pi }}{2}-x\right)=cotx\right]\\ & & \end{array}$

Adding equation $\left(i\right)$ and $\left(ii\right)$

$$\begin{gathered} 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{\tan x}}+{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{cotx}} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1.dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

$\begin{array}{rcl}2I& =& \frac{\mathrm{\pi }}{2}\\ & \Rightarrow & I=\frac{\mathrm{\pi }}{4}\end{array}$

Hence, option (D) is the correct answer.