Question

The value of the definite integral ${\int }_0^{\frac{\pi }{2}}\sqrt{\tan x}dx$ 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 $\frac{\pi }{\sqrt{2}}$

$I={\int }_0^{\frac{\pi }{2}}\sqrt{\tan x}dx$..........(1)
$I={\int }_0^{\frac{\pi }{2}}\sqrt{\cot x}dx$.......(2)
Adding equations(1) and (2), we get
$2I={\int }_0^{\frac{\pi }{2}}(\sqrt{\tan x}+\sqrt{\cot x})dx$
$=-\sqrt{2}{\int }_0^{\frac{\pi }{2}}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$
$-\sqrt{2}{\int }_0^{\frac{\pi }{2}}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x{)}^2}}dx$
$\sqrt{2}{\int }_{-1}^1\frac{dt}{\sqrt{1-t^2}}$ (where $\sin x-\cos x=t$)
$=2\sqrt{2}{\int }_0^1\frac{dt}{\sqrt{1-t^2}}=\sqrt{2}\pi$
or
$I=\frac{\pi }{\sqrt{2}}$