Question

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

Answer 1

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

Let $t^2=\tan x$

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

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

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

=Divide by $t^2$ both integrals

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

$t-\frac{1}{t}=u1+\frac{1}{t^2}dt=du$

$t+\frac{1}{t}=v1-\frac{1}{t^2}dt=dv$

$\int \frac{du}{2+4^2}+\frac{dv}{v^2-2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{4}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \frac{v-\sqrt{2}}{v+\sqrt{2}}+c$

Putting values of u,v then t.

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