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

Question

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. ain't a 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

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

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 $\int \frac{1}{{x^4+1}} dx$.

edited Apr 13 '17 at 12:21

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answered Oct 14 '16 at 16:01

Harry Peter

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

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