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?

  • $\begingroup$ 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? $\endgroup$ – imranfat Oct 13 '16 at 18:46
  • $\begingroup$ @imranfat no, it is a tan(x) function. PS. question already edited :) $\endgroup$ – Roman Nazarkin Oct 13 '16 at 18:47
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    $\begingroup$ @imran, $\mathrm{tg}\,x$ is the notation that was once used in the Soviet Union and Eastern Europe. $\endgroup$ – J. M. is a poor mathematician Oct 13 '16 at 18:55
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    $\begingroup$ 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 $$ u^4+1 = (u^2-u\sqrt{2}+1)(u^2+u\sqrt{2}+1).$$ $\endgroup$ – Jack D'Aurizio Oct 13 '16 at 18:56
  • $\begingroup$ @RomanNazarkin Ok, I see... $\endgroup$ – imranfat Oct 13 '16 at 20:31

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

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


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

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


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