Evaluate $I_{n}=\int\frac{\cos nx}{5-4\cos x}=$ I stumbled across the following integral:-
$$I_{n}=\int\frac{\cos nx}{5-4\cos x}dx$$
where $n$ is a positive integer.
I have no idea how to proceed....I tried integration by parts and even writing 
$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$
I couldn't make much headway.....
Any ideas on how to proceed would be appreciated.   
EDIT:
This question is different from @amWhy has marked..I want to evaluate indefinite integral ..Not the definite one.

Hint:
It is just a piece of cake of creating the reduction formula:
$\int\dfrac{\cos nx}{5-4\cos x}~dx$
$=\int\dfrac{2\cos x\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$ (according to http://mathworld.wolfram.com/Multiple-AngleFormulas.html)
$=\dfrac{1}{2}\int\dfrac{4\cos x\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
$=\dfrac{1}{2}\int\dfrac{(4\cos x-5+5)\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
$=-\dfrac{1}{2}\int\cos((n-1)x)~dx+\dfrac{5}{2}\int\dfrac{\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
$=-\dfrac{\sin((n-1)x)}{2(n-1)}+\dfrac{5}{2}\int\dfrac{\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
