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I can't compute this integral, $$\int{\frac{\mathrm{d}x}{\sqrt[3]{\tan(x)}}}$$

I tried making the substitution $\tan(x) = t$, but it does not lead anywhere.

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Try instead $\tan(x)=t^3$... I cannot guarantee that this works, but certainly it is a better alternative. –  Matemáticos Chibchas Feb 24 '13 at 19:41
    
Clearly wrong but a question: What if we make the substitution t=tan(x/2)? Then we will get that in terms of t; tanx=2t/1-t^2 and a new integral. Wolfram alpha suggest that solution to this does not exist in elementary functions(hypergeometric functions). Can we from here conclude that our first integral does not have a solution in elementary functions as well? –  EricAm Feb 24 '13 at 21:06
    
Since the differential element $\frac{dx}{\sqrt[3]{\tan x}}$ do not modify if we switch $x$ by $-x$ so the substitution $t=\cos x$ works as well. –  Sami Ben Romdhane Feb 24 '13 at 22:18
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2 Answers

You were on a good track.

If you let $(\tan x)^{1/3} = t$, we have $t^3 = \tan x$.

Substituting, yields:

$$\int\frac{3t}{t^6+1} dt$$

Using the partial fraction expansion of this, yields:

$$\frac{3t}{t^6+1} = \frac{t}{t^2+1} + \frac{\sqrt{3}t-3}{2\sqrt{3}(-t^2+\sqrt{3}t -1)} + \frac{-\sqrt{3}t-3}{2\sqrt{3}(t^2+\sqrt{3}t+1)}$$

A little nasty, but solvable.

Make sure to pay attention to all steps!

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I like to point out, the "positives" in an OP's work, as well as the mistakes! +1 –  amWhy Apr 30 '13 at 0:42
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If $t^3=\tan(x)$ then $3t^2dt=\sec^2(x)dx=(1+t^6)dx$, so your integral becomes

$$\int\frac{3t^2\,dt}{t(1+t^6)}=\int\frac{3t\,dt}{1+t^6}\,,$$

which can be solved via partial fractions.

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This website may also help you finish the job: mathwarehouse.com/algebra/polynomial/… –  Alex Zorn Feb 24 '13 at 19:47
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