Please show me the steps of the following integration. I got an answer in Wolfram, but I need steps..
$$\int \frac{\mathrm dx}{\sqrt[3]{\tan\,x}}$$
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$\begingroup$ @draks here it is. And WA do not know closed form for this integral $\endgroup$– NorbertJul 31, 2012 at 11:02
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$\begingroup$ Well, I'm sorry its not arctan, its just tan $\endgroup$– AnubisJul 31, 2012 at 11:05
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$\begingroup$ This is the link $\endgroup$– AnubisJul 31, 2012 at 11:08
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$\begingroup$ @draks sorry, I've typed it wrong. It's now corrected. Can you evaluate now? $\endgroup$– AnubisJul 31, 2012 at 11:10
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$\begingroup$ Aha, so please edit your question accordingly!!! and click on "show steps" in your linked W|A page... $\endgroup$– draks ...Jul 31, 2012 at 11:10
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1 Answer
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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.
