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I'm trying to solve the following integral:

$$\int \frac{1}{1+\cot^3(x)}dx$$

While the solution can be found in Wolfram Alpha, I am not completely sure how to reduce the above integral to get the solution referenced. Pointers would be appreciated.

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On Courant and John first volume of Calculus, there is a section called Integration of some other kind of functions. There, under the subsection "Integration of $R\big(\cos(x),\sin(x)\big)$", you should find the proper way to solve the integral. Looking at the solution given by WA, it might be a good idea to transform the integrand to the form $\frac{du}{u}$. –  Pragabhava Oct 10 '12 at 0:54

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Let $u=\cot(x)$; then $dx = -\frac{1}{u^2+1}du$. Now the integral becomes

$$ I = - \int \frac{1}{(u^2+1)(u^3+1)} du $$

which can be resolved into partial fractions as:

$$ I =- \int \frac{1-2u}{3(u^2-u+1)} + \frac{u+1}{2(u^2+1)} + \frac{1}{6u+6} du $$

each sub-integral of which can be readily evaluated.

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If you make the change of variables $ x=\arctan(t) $ you get

$$ \int \!{\frac {{t}^{3}}{ \left( 1+{t}^{3} \right) \left( 1+{t}^{2} \right) }}{dt}\,. $$

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