Question

If $\underset{0}{\overset{\pi /4}{\int }}(\sqrt{\tan x}+\sqrt{\cot x})dx$ than value of $I$ is

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

Answer 1

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

$\underset{0}{\overset{\pi /4}{\int }}(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}})dx$
$\underset{0}{\overset{\pi /4}{\int }}\left(\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\right)dx$

Putting,
$\sin x-\cos x=u\Rightarrow du=(\sin x+\cos x)dx(\sin x-\cos x{)}^2=u^2\Rightarrow 1-2\sin x\cos x=u^2\Rightarrow \frac{1-u^2}{2}=\sin x\cos x$
So the integration will be,
$I=\sqrt{2}\underset{-1}{\overset{0}{\int }}\frac{1}{\sqrt{1-u^2}}du$
$\Rightarrow I={\left[\sqrt{2}{\sin }^{-1}u\right]}_{-1}^0\Rightarrow I=\sqrt{2}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{2}}$