Question

Integrate wrt x
$\int \sqrt{\tan x}dx$

Answer 1

$\int \sqrt{\tan x}dx$

Let $t^2=\tan x$

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

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

$\Rightarrow dx=\frac{2tdt}{1+t^4}$

$=\int \frac{2t^2}{1+t^4}dt$

$=\int \frac{(1+t^2-(1-t^2)}{1+t^4}dt$

$=\int \frac{1+t^2}{1+t^4}dt-\int \frac{1-t^2}{1+t^4}dt$

$=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt+\int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt$

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

Let $z=t-\frac{1}{t}$

$dz=(1+\frac{1}{t^2})dt$

and

let $u=t+\frac{1}{t}$

$du=(1-\frac{1}{t^2})dt$

$=\int \frac{dz}{z^2+(\sqrt{2}{)}^2}+\int \frac{du}{u^2-(\sqrt{2}{)}^2}$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{z}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right|+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+C$

$\therefore \int \sqrt{\tan x}dx$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+C$