Question

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

Answer 1

Given $\int \sqrt{\tan x}dx$
Let $\tan x=t^2$
$\Rightarrow {\sec }^{2}xdx=2tdt$ [$\because {\sec }^{2}x=1+{\tan }^2$]
$\Rightarrow dx=\frac{2t}{(1+t^4)}dt$

Now given expression can be written as $\int \sqrt{\tan x}dx=\int \frac{2t^2}{(1+t^4)}dt$

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

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

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

$\Rightarrow \int \frac{(1+\frac{1}{t^2})}{(t-\frac{1}{t}{)}^2+2}dt+\int \frac{(1-\frac{1}{t^2})}{(t+\frac{1}{t}{)}^2-2}dt$......(i)

Let $t-\frac{1}{t}=u$ for the first integral $\Rightarrow (1+\frac{1}{t^2})dt=du$

and $t+\frac{1}{t}=v$ for the 2nd integral $\Rightarrow (1-\frac{1}{t^2})dt=dv$

Substitute above integrals in equation(i), we get
$=\int \frac{du}{(u^2+2)}+\int \frac{dv}{(v^2-2)}$

$=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{u}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log |\frac{(v-\sqrt{2})}{(v+\sqrt{2})}|+c$

$=\left(\frac{1}{\sqrt{2}}\right)ta{n}^{-1}\left[\frac{(t^2-1)}{t\sqrt{2}}\right]+\left(\frac{1}{2\sqrt{2}}\right)\log \frac{(t^2+1-t\sqrt{2})}{(t^2+1+t\sqrt{2})}+c$

=$\left(\frac{1}{\sqrt{2}}ta{n}^{-1}\right[\frac{(tanx-1)}{\sqrt{2tanx}}]+\frac{1}{2\sqrt{2}}\log \frac{[tanx+1-\sqrt{\left(2tanx\right)}]}{[tanx+1+\sqrt{\left(2tanx\right)}]}+c$