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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: $\mathrm{ctg}=\cot$. – J. M. Aug 30 '11 at 22:09
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? – J. M. Aug 30 '11 at 22:13
This is one exam in Croatian university. Professor is "old skul" obvieously :D – jbennet Aug 30 '11 at 22:21
up vote 4 down vote accepted

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. – jbennet Aug 30 '11 at 22:20
You're welcome. Could you share the WolframAlpha link? It worked fine for me:^%283%2F7%29%2Fsin^2+dx. – azjps Aug 30 '11 at 22:21
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. – azjps Aug 30 '11 at 22:28
Sorry, please use this: – jbennet Aug 30 '11 at 22:33

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