Question

Evaluate ${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\sqrt{\tan x}+\sqrt{\cot x})dx=$

• A. $\frac{\pi }{2\sqrt{2}}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{\sqrt{2}}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is C $\frac{\pi }{\sqrt{2}}$

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx$

$={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right)dx$

$={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\left(\frac{1+\tan x}{\sqrt{\tan x}}\right)dx$

putting $\tan x=t^2$

$\Rightarrow {\sec }^{2}xdx=2tdt$

$\Rightarrow \left(1+{\tan }^{2}x\right)dx=2tdt$

$\Rightarrow \left(1+t^4\right)dx=2tdt$

$\Rightarrow dx=\frac{2tdt}{1+t^4}$

$={\int }_1^{\infty }\frac{\left(1+t^2\right)}{t\left(1+t^4\right)}\times 2tdt$

$=2{\int }_1^{\infty }\frac{1+t^2}{1+t^4}dt$

$=2{\int }_1^{\infty }\frac{\left(1+\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)}dt$

$=2{\int }_1^{\infty }\frac{\left(1+\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}-2+2\right)}dt$

$=2{\int }_1^{\infty }\frac{\left(1+\frac{1}{t^2}\right)}{{\left(t-\frac{1}{t}\right)}^2+{\left(\sqrt{2}\right)}^2}dt$

putting $t-\frac{1}{t}=y$

$\Rightarrow \left(1+\frac{1}{t^2}\right)dt=dy$

$=2{\int }_0^{\infty }\frac{dy}{y^2+{\left(\sqrt{2}\right)}^2}$

$=2\times \frac{1}{\sqrt{2}}{\left[{\tan }^{-1}\left(\frac{y}{\sqrt{2}}\right)\right]}_0^{\infty }$

$=\sqrt{2}\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(0\right)\right]$

$=\sqrt{2}\left(\frac{\pi }{2}\right)$

$=\frac{\pi }{\sqrt{2}}$