Simple or challenging integral?, can't find table anywhere.

Dec 2013
31
4
United States
I'm looking for solutions to the form \(\displaystyle \int \frac{cos(a/x)}{1-cos(a/x)}dx\), where a is a constant.

I've looked for tables that include that form but have found none. I've tried some guess and check functions but have found none yet. I don't have access to mathematica, maple, etc., atm to see if they can crunch it, wolfram alpha couldn't. Solutions, or info and/or techniques on this form are appreciated.
 
Dec 2013
2,000
757
Colombia
For rational functions of trigonometric functions, the best bet is usually the substitution \(\displaystyle t=\tan \tfrac{x}{2}\), although given the unusual arguments for the cosines, you might try \(\displaystyle t = \tan \tfrac{a}{2x}\) instead.
 
Dec 2013
31
4
United States
Thanks for the suggestion Archie, but I still haven't cracked it.

After working with the project more, it turns out I'm looking for the solutions to 3 specific integrals related to the form above. They are: \(\displaystyle \int_{a}^{b}\frac{\cos \frac{2\pi x}{k}}{2*\left ( 1-\cos \frac{1}{k} \right )}dk\), \(\displaystyle \int_{a}^{b}\frac{\cos \frac{1}{k}}{2*\left ( 1-\cos \frac{1}{k} \right )}dk\), and \(\displaystyle \int_{a}^{b}\frac{\left | \cos \frac{2\pi x}{k}-\cos \frac{1}{k} \right |}{2*\left ( 1-\cos \frac{1}{k} \right )}dk\).

And yes, x is a constant, k is the variable, and the 3rd one numerator is an absolute value . Any help is appreciated.