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

Answer: (B)

$I=\underset{0}{\overset{\pi /2}{\int }}\sqrt{\tan x}dx\dots (1);I=\underset{0}{\overset{\pi /2}{\int }}\sqrt{\cot x}dx$
adding (1) and (2), we get
$2I=\underset{0}{\overset{\pi /2}{\int }}(\sqrt{\tan x}+\sqrt{\cot x})dx=\sqrt{2}\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx=\sqrt{2}\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x{)}^2}}dx$
$=\sqrt{2}\underset{-1}{\overset{1}{\int }}\frac{dt}{\sqrt{1-t^2}}=2\sqrt{2}\underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt{1-t^2}}=\sqrt{2}\pi ($ where $\sin x-\cos x=t)$
$\therefore I=\frac{\pi }{\sqrt{2}}$