3
$\begingroup$

Please show me the steps of the following integration. I got an answer in Wolfram, but I need steps..

$$\int \frac{\mathrm dx}{\sqrt[3]{\tan\,x}}$$

  • 0
    @draks [here it is](http://www.wolframalpha.com/input/?i=Integrate%5B1%2F%28ArcTan%5Bx%5D%29%5E%281%2F3%29%2C+x%5D). And WA do not know closed form for this integral2012-07-31
  • 0
    Well, I'm sorry its not arctan, its just tan2012-07-31
  • 0
    This is the [link](http://www.wolframalpha.com/input/?i=Integrate[1%2F%28Tan[x]%29^%281%2F3%29%2C+x])2012-07-31
  • 0
    @draks sorry, I've typed it wrong. It's now corrected. Can you evaluate now?2012-07-31
  • 0
    Aha, so please edit your question accordingly!!! and click on "show steps" in your linked W|A page...2012-07-31

1 Answers 1

10

We try the substitution $t^3 = \tan^2 x$. Therefore, $3t^2 dt = 2 \tan x \sec^2 x dx$, giving us $\frac{dx}{\sqrt{t}} = \frac{3 dt}{2(1+t^3)}$.

Thus, we will only evaluate $\int \frac{3 dt}{1+t^3} $, divide by $2$ and substitute back. Note that $3 = (1-t+t^2) + (2-t)(1+t)$, reducing our integral to $$ \int \frac{dt}{1+t} + \int \frac{(2-t)dt}{1-t + t^2} $$ I won't elaborate further, since our integrals are already in standard forms.

  • 0
    Wow..a cleaver approach. You are a genius..!2012-08-01