Question

The value of the definite integral $\underset{0}{\overset{\pi /2}{\int }}\sqrt{\tan x}dx$ is

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

Answer 1

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

$I=\underset{0}{\overset{\pi /2}{\int }}\sqrt{\tan x}dx\cdots \left(1\right)$
Using property,
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\sqrt{\cot x}dx\cdots \left(2\right)$
Adding (1) and (2),
$2I=\underset{0}{\overset{\pi /2}{\int }}\sqrt{\tan x}+\sqrt{\cot x}dx\Rightarrow I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx\Rightarrow I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x{)}^2}}dx$
Let,
$\sin x-\cos x=t\Rightarrow (\cos x+\sin x)dx=dt$
$I=\frac{1}{\sqrt{2}}\underset{-1}{\overset{1}{\int }}\frac{dt}{\sqrt{1-t^2}}\Rightarrow I=\frac{1}{\sqrt{2}}{\left[{\sin }^{-1}t\right]}_{-1}^1=\frac{\pi }{\sqrt{2}}$