Question

Evaluate the following: $\int \frac{\sqrt{\tan x}}{\sin x\cdot \cos x}dx$

• A. $\sqrt{\tan x}+C$
• B. $2\sqrt{\tan x}+C$
• C. $2\sqrt{\sec x}+C$
• D. None of these

Answer 1

Answer: (B)

$2\sqrt{\tan x}+C$

Given

$\int \frac{\sqrt{tanx}}{\cos x\sin x}dx$

substitute $\sin x=\frac{\tan x}{\sec x}$, $\cos =\frac{1}{\sec x}$

$=\int {\sec }^{2}x.\frac{1}{\sqrt{tanx}}dx$

substitute $u=\tan x$, $dx=\frac{1}{{\sec }^{2}x}du$

$=\int \frac{1}{\sqrt{u}}du$

$=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$

$=\frac{u^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}$

$=\frac{u^{\frac{1}{2}}}{\frac{-1}{2}}$

$=2u^{\frac{1}{2}}$

$=2\sqrt{u}$

$=2\sqrt{\tan x}+C$