Integration of $\sqrt {\tan x}$ I have tried many ways to integrate $\sqrt {\tan x}$ including integration by parts but didn't get to any final result.
I also assumed,
$$ \tan x = t^2 $$
$$ \int  \sqrt {\tan x} \,dx $$
$$⇒\int \frac{2t^2}{1+t^4}dt$$
but it's getting a bit complicated further, kindly help. Also, are there any simpler ways to integrate this.
Answer,
$$ \frac{1}{\sqrt 2} \tan^{-1}\left[\frac {\sqrt {\tan x}-\sqrt {\cot x}}{\sqrt{2}}\right] +\frac{1}{2\sqrt 2}\ln\left[\frac {\sqrt {\tan x}+\sqrt {\cot x}-\sqrt {2}}{\sqrt {\tan x}+\sqrt {\cot x}+\sqrt {2}}\right] +C $$

You can factor the denominator as the product of two quadratics (finding the [necessarily complex] roots can help there), then use partial fractions. It gets ugly, but Wofram Alpha gives the same ugly answer as you get by doing this, so I assume there isn't a simpler way.
BTW, are you sure about the numerator? I have vague memories of it being $\frac{1}{1+t^4}$, but it was quite a while ago that I did it.
