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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1$\begingroup$ 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 $\frac{du}{u}$. $\endgroup$– PragabhavaOct 10, 2012 at 0:54
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2 Answers
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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
$$ \int \!{\frac {{t}^{3}}{ \left( 1+{t}^{3} \right) \left( 1+{t}^{2} \right) }}{dt}\,. $$
answered Oct 10, 2012 at 0:58