Evaluate $I_{n}=\int\frac{\cos nx}{5-4\cos x}=$

Question

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.

Answer 1

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)

$=\dfrac{1}{2}\int\dfrac{4\cos x\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$

$=\dfrac{1}{2}\int\dfrac{(4\cos x-5+5)\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$

$=-\dfrac{1}{2}\int\cos((n-1)x)~dx+\dfrac{5}{2}\int\dfrac{\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$

$=-\dfrac{\sin((n-1)x)}{2(n-1)}+\dfrac{5}{2}\int\dfrac{\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$