Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a question NOT on how to actually solve the problem but more about a concern in what one needs to know before making a sort of substitution.

Last day I tried to solve this integral: $\int \frac{1}{\sqrt[3]{\tan x}}dx$. Now this integral can fairly easy be found in terms of elementary functions. However, I tried with the universal substitution: $t=\tan (\frac{x}{2})$ which gives me:

$\sin x=\frac{2t}{1+t^2}$ $\cos x=\frac{1-t^2}{1+t^2}$ and further $\tan x=\frac{2t}{1-t^2}$. This didn't really simplify the problem but I plugged in this into Wolfram Alpha which gives a solution in terms of non-elemtary functions(hypergeometric functions). Totally different from the actual solution. I should probably say that i have no knowledge in non-elementary functions.

So where did I go wrong and how can I avoid falling into this trap?

share|cite|improve this question
Actually, those hypertrigonometric functions can be simplified further because all trigonometric functions(circular or hyperbolic) are combos of exponential functions. WolframAlpha shows this behaviour because it's a computer program after all. Try plugging the function directly. It returns an easy solution – Cheeku Feb 27 '13 at 14:48
So what you mean is that hyper trigonometric functions are composed of elementary functions? – EricAm Feb 27 '13 at 14:54… See this link. The relation between hyperbolic functions and exponentials are given – Cheeku Feb 27 '13 at 14:58
Did you remember to also put in $dx = 2dt/(1+t^2)$ ... ? – GEdgar Feb 27 '13 at 15:04
@Cheeku: Adam got hypergeometric functions, not hyperbolic functions. If you leave out the $dx$ when you do his substitution, that is what happens. – GEdgar Feb 27 '13 at 15:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.