Question

# ${\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{\mathrm{tan}x}}dx=$

A

$1$

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B

$-1$

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C

$\frac{\mathrm{\pi }}{2}$

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D

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

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Solution

## The correct option is D $\frac{\mathrm{\pi }}{4}$Explanation for Correct answer:Finding the value of the given integral,Let $I={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{\mathrm{tan}x}}dx\to \left(i\right)$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]$$\begin{array}{rcl}I& =& {\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cot\left(0+\frac{\mathrm{\pi }}{2}-x\right)}}{\sqrt{cot\left(0+\frac{\mathrm{\pi }}{2}-x\right)}+\sqrt{\mathrm{tan}\left(0+\frac{\mathrm{\pi }}{2}-x\right)}}dx\\ & =& {\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{tan}x}}{\sqrt{\mathrm{tan}x}+\sqrt{cotx}}\to \left(ii\right)\left[\because cot\left(\frac{\mathrm{\pi }}{2}-x\right)=\mathrm{tan}x,\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}-x\right)=cotx\right]\\ & & \end{array}$Adding equation $\left(i\right)$ and $\left(ii\right)$ $2I={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{\mathrm{tan}x}}+{\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{tan}x}}{\sqrt{\mathrm{tan}x}+\sqrt{cotx}}\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}1.dx\phantom{\rule{0ex}{0ex}}={\left[x\right]}_{0}^{\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$$\begin{array}{rcl}2I& =& \frac{\mathrm{\pi }}{2}\\ & ⇒& I=\frac{\mathrm{\pi }}{4}\end{array}$Hence, option (D) is the correct answer.

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