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Please help me to evaluate this integral, without using csc function, because we don't use it on class, so must be some easier way to do it.

$$\int{}\frac{\sqrt[7]{\operatorname{ctg}^3(x)}}{1-\cos^2 x}\mathrm dx$$

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    For the unaccustomed: $\mathrm{ctg}=\cot$.2011-08-30
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    Thus far, I've only seen Russians use $\mathrm{tg}$ and $\mathrm{ctg}$... may I ask where you're learning this, if it's fine to tell?2011-08-30
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    This is one exam in Croatian university. Professor is "old skul" obvieously :D2011-08-30

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You mean $\displaystyle \int \frac{\cot^{3/7}(x)}{1-\cos^2(x)}\ dx$? Consider the substitution $u = \cot(x),\ du = \frac{-dx}{\sin^2(x)}$. Then $\displaystyle \int \frac{\cot^{3/7}(x)}{1-\cos^2(x)}\ dx = \int \frac{\cot^{3/7}(x)}{\sin^2(x)}\ dx = \int -u^{3/7}\, du = \frac{-7}{10} \cot^{10/7}(x) + C$.

Also, you can't really do this without (at least implicitly) using the $\csc(x)$ function. The cosecant function is defined as $\csc(x) = \frac{1}{\sin(x)}$, so as long as you have the definition of $\sin(x)$, you also have the definition of $\csc(x)$.

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    WolframAlpha give me a very complicated method, using the csc(x) function in the middle, but it doesn't fit to me, because we don't use it on class. Thank you very much for answer.2011-08-30
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    You're welcome. Could you share the WolframAlpha link? It worked fine for me: `http://www.wolframalpha.com/input/?i=integrate+cot^%283%2F7%29%2Fsin^2+dx`.2011-08-30
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    i was using this: [http://www.wolframalpha.com/input/?i=integrate+%28%28ctg^3+x%29^%281%2F7%29%29%2F%281-cos^2x%29+dx](http://www.wolframalpha.com/input/?i=integrate+%28%28ctg^3+x%29^%281%2F7%29%29%2F%281-cos^2x%29+dx)2011-08-30
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    Hmm, when I view your link, I don't see any exponent carats (^). Perhaps the markdown is eating them up. Edit: Okay now I see them but the link isn't working.2011-08-30
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    Sorry, please use this: [http://goo.gl/smBmx](http://goo.gl/smBmx)2011-08-30
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    (can't edit previous comment) Ah okay. The method that WolframAlpha uses is exactly the same as what I wrote above (perhaps with more words). The only difference is where it writes $du = \csc^2(x)$, which is the same as writing $du = \frac{1}{\sin^2(x)}$.2011-08-30
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    @jbennet: If I click "Show Steps", I see that it did essentially the same thing as azjps... on another note, it's neat the W|A knows $\mathrm{ctg}$...2011-08-30
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    csc is a little strange to me, and i didn't tried to find similarity to other functions. my mistake :)2011-08-30