Question

Evaluate: $\int \sqrt{\tan x}dx,(0<x<\frac{\pi }{2})$

Answer 1

Let $\tan x=t^2\Rightarrow \frac{d}{dx}\left(\tan x\right)=\frac{d}{dt}\left(t^2\right)\Rightarrow {\sec }^{2}x.dx=2t.dt\Rightarrow dx=\frac{2t}{1+t^2}dt$

$\int \sqrt{t^2}\frac{2t}{1+t^2}dt\Rightarrow \int \frac{2t^2}{1+t^2}dt$

$=\int 2(1-\frac{1}{1+t^2})dt=\int 2dt-2\int \frac{1}{1+t^2}dt=2t-2{\tan }^{-1}t+c$

We know that $t^2=\tan x\Rightarrow t=\sqrt{\tan x}$

Substituting the value of t , we will get

$2{\tan }^{-1}x-2{\tan }^{-1}\left(\sqrt{\tan x}\right)+c$