# Solving trigonometric indefinite integral $\int \frac{dx}{\sqrt{\tan x}}$

Continuing the series of horrible integrals, my instructor gave me exercise to solve next indefinite integral:

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

Seems simple and short, but wolframalpha gives me totally horrible answer.

Is there any way to simplify this integral or any hints on solving it? Maybe some trigonometric formulas?

• I am not getting it, $t$ and $g$ are just plain numbers? and so we are anti deriving with respect to $x$? That's straight forward, isn't it? – imranfat Oct 13 '16 at 18:46
• @imranfat no, it is a tan(x) function. PS. question already edited :) – Roman Nazarkin Oct 13 '16 at 18:47
• @imran, $\mathrm{tg}\,x$ is the notation that was once used in the Soviet Union and Eastern Europe. – J. M. is a poor mathematician Oct 13 '16 at 18:55
• Through the substitutions $x=\arctan t$ and $t=u^2$ the problem boils down to computing $\int\frac{du}{1+u^4}$ that is doable through partial fraction decomposition, since $$u^4+1 = (u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1).$$ – Jack D'Aurizio Oct 13 '16 at 18:56
• @RomanNazarkin Ok, I see... – imranfat Oct 13 '16 at 20:31

Let $u=\sqrt{\tan x}$ ,

Then $x=\tan^{-1}u^2$

$dx=\dfrac{2u}{u^4+1}~du$

$\therefore\int\dfrac{dx}{\sqrt{\tan x}}=\int\dfrac{2}{u^4+1}~du$

The only key point is how to evaluate $\int\dfrac{du}{u^4+1}$ .

You can factorize $u^4+1$ and partial fraction decomposition as usual (as foolish as WolframAlpha), or getting the smarter approach e.g. in Evaluating $\int \frac{1}{{x^4+1}} dx$.