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How to integrate $$\int\dfrac{1}{\sqrt[3]{\tan(x)}}\, \mathrm dx?$$

Harry Peter
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user92354
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    Welcome to MSE. What have you tried yet? –  Aug 30 '13 at 11:27
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    Really....why close this? – Ron Gordon Aug 30 '13 at 11:45
  • @RonGordon To few words? – user1729 Aug 30 '13 at 12:28
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    @user1729: it's a fine problem. Yes, spur the OP for what he is stuck on, but why close it? – Ron Gordon Aug 30 '13 at 12:31
  • @RonGordon (I haven't voted to close - my above comment was just an educated guess. Sorry for biting when your comment wasn't directed at me (in a fish way, not a nasty way).) – user1729 Aug 30 '13 at 12:33
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    Some of our users may vote to close homework-type problems until the OP provides their own thoughts. – GEdgar Aug 30 '13 at 17:36
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    @GEdgar: Maybe this is HW, maybe this isn't. It certainly is quite challenging. We have had conversations about this in meta and many of us are split on the topic. That said, if we keep closing problems like this that are inherently interesting because it might be HW, then this site loses something. – Ron Gordon Aug 30 '13 at 19:10
  • Now that there is an answer, there is little chance that user92345 will tell us where it came from. And little chance to teach user92345 how to approach problems like this. Since this is his first question, he may need to be encouraged to say more than a mere statement. (P.S. I did not vote to close, I just guess whey some did.) – GEdgar Aug 30 '13 at 21:42

1 Answers1

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Sub $u=\tan{x}$ and get

$$\int du \frac{u^{-1/3}}{1+u^2}$$

Then sub $u=y^3$ to get

$$3 \int dy \frac{y}{1+y^6} = \frac{3}{2} \int \frac{dv}{1+v^3} = \frac12 \int \frac{dv}{1+v} + \int \frac{dv}{1-v+v^2} - \frac12 \int dv \frac{v}{1-v+v^2} $$

Each of these integrals may be evaluated in turn.

$$\int \frac{dv}{1+v} = \log{(1+v)}$$

$$\int \frac{dv}{1-v+v^2} = \int \frac{dv}{(v-1/2)^2+3/4} = \frac{2}{\sqrt{3}} \arctan{\frac{2 v-1}{\sqrt{3}}}$$

$$\int dv \frac{v}{1-v+v^2} = \int dv \frac{v-1/2}{(v-1/2)^2+3/4} +\frac12 \int \frac{dv}{(v-1/2)^2+3/4} = \\\frac12 \log{[(v-1/2)^2+3/4]} + \frac{1}{\sqrt{3}} \arctan{\frac{2 v-1}{\sqrt{3}}}$$

I get, putting this all together,

$$\int dx \, (\tan{x} )^{-1/3} = \frac12 \log{\frac{1+v}{1-v+v^2}} + \frac{\sqrt{3}}{2} \arctan{\frac{2 v-1}{\sqrt{3}}} +C$$

where $v=(\tan{x})^{-2/3}$.

Ron Gordon
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