Question

${\int }_0^{\pi /4}\frac{\sqrt{\tan x}}{\sin x\cos x}dx$ equals-

• A. $1$
• B. $2$
• C. $0$
• D. $4$

Answer 1

The correct option is B $2$

${\int }_0^{\pi /4}\frac{\sqrt{\tan x}}{\sin x\cos x}dx$

${\int }_0^{\pi /4}\frac{\sqrt{\tan x}/{\cos }^{2}x}{\sin x\cos x/{\cos }^{2}x}dx$

$={\int }_0^{\pi /4}\frac{\sqrt{\tan x}.{\sec }^{2}x}{\frac{\sin x}{\cos x}}dx$

$={\int }_0^{\pi /4}\frac{\sqrt{\tan x}{\sec }^{2}x}{\tan x}dx$

$={\int }_0^{\pi /4}\frac{{\sec }^{2}x}{\sqrt{\tan x}}dt$

Let $\tan x=t^2$

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

when $x=0,t=0$

$x=\pi /4,t=1$

$={\int }_0^1\frac{2tdt}{t}$

$=2{\int }_0^{1}1dt=2[t{]}_0^1$

$=2[1-0]=2$