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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 $\frac{du}{u}$.2012-10-10

2 Answers 2

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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}\,. $