Question

The value of $\int \frac{\sqrt{\tan x}}{\sin x\cos x}dx$ is equal to

• A. $2\sqrt{\tan x}+C$
• B. $2\sqrt{\cot x}+C$
• C. $\frac{\sqrt{\tan x}}{2}+C$
• D. $\sqrt{\tan x}+C$

Answer 1

The correct option is A $2\sqrt{\tan x}+C$

Given $\int \frac{\sqrt{\tan x}}{\sin x\cos x}$

simplifying the function

$=\int \frac{\sqrt{\tan x}}{\sin x.\cos x.\frac{\cos x}{\cos x}}$

$=\int \frac{\sqrt{\tan x}}{\sin x.\frac{{\cos }^{2}x}{\cos x}}$

$=\int \frac{\sqrt{\tan x}}{{\cos }^{2}x.\frac{\sin x}{\cos x}}$

$=\int \frac{\sqrt{\tan x}}{{\cos }^{2}x.\tan x}$

$=\int \frac{\sqrt{\tan x}.(\tan x{)}^{-1}}{{\cos }^{2}x}$

$=\int \frac{(\tan x{)}^{1/2^{-1}}}{{\cos }^{2}x}$

$=\int \frac{(\tan x{)}^{-1/2}}{{\cos }^{2}x}$

$=\int (\tan x{)}^{-1/2}.\frac{1}{{\cos }^{2}x}$

$=\int (\tan x{)}^{-1/2}.{\sec }^{2}x$

Let $\tan x=t$

So, ${\sec }^{2}x=\frac{dt}{dx}$

$\Rightarrow dx=\frac{dt}{{\sec }^{2}x}$

$\therefore \int (\tan x{)}^{-1/2}.{\sec }^{2}x.dx$

$=\int (t{)}^{-1/2}.{\sec }^{2}x.\frac{dt}{{\sec }^{2}x}$

$=\int (t{)}^{-1/2}dt$

$=\frac{t^{-1/2}+1}{-1/2+1}+C$ $\left\{as\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right\}$

$=\frac{t^{1/2}}{1/2}+C$

$=2t^{1/2}+C$

$=2\sqrt{t}+C$

Substituting $t=\tan x$

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