I have tried the method that I know about to solve this by proper substitution (contour integration), but in the end it reaches a point where I have to find the residue at $z=0$ by: $\frac{d^{n-2}}{dz^{n-2}}(\frac{z^{2n}}{z^2+2\alpha z+1})$. Now this is pretty difficult to find. Please help me, I'm stuck, even if there is a different method to do this.
Edit: I tried to solve this by first finding out $\int_{0}^{2\pi}\frac{e^{in\theta}}{(\cos(\theta)+\alpha)^2}d\theta$ and got the value as real (lets say $a$). So, is it right to do this?:$$\int_{0}^{2\pi}\frac{e^{in\theta}}{(\cos(\theta)+\alpha)^2}d\theta=\int_{0}^{2\pi} \frac{\cos(n\theta)}{(\cos(\theta)+\alpha)^2}d\theta\space +i\int_{0}^{2\pi} \frac{\sin(n\theta)}{(\cos(\theta)+\alpha)^2}d\theta=a$$ $$\implies\int_{0}^{2\pi} \frac{\sin(n\theta)}{(\cos(\theta)+\alpha)^2}d\theta=0\space ;\int_{0}^{2\pi}\frac{e^{in\theta}}{(\cos(\theta)+\alpha)^2}d\theta=a $$ Edit 2: Yes the above method is correct, and DinosaurEgg's answer sums up the question.
Doing the calculation with residues is indeed the easiest way. Write first
$$J_n(a)=\int_{0}^{2\pi}\frac{\cos n\theta}{a+\cos\theta}d\theta=\text{Re}\int_{0}^{2\pi}\frac{e^{i n\theta}}{a+\cos\theta}d\theta=\text{Re}~~(-2i)\oint\frac{z^n}{z^2+2az+1}dz$$
to express it as an integral on the unit circle. Here you can use simple residue calculus to obtain
$$J_n(a)=(-1)^n\frac{2\pi}{\sqrt{a^2-1}}\left(a-\sqrt{a^2-1}\right)^n$$
However the integral you want is given by $-dJ_n(a)/da$. Take a derivative here to show
$$\int_{0}^{2\pi}\frac{\cos n\theta}{(a+\cos\theta)^2}d\theta=(-1)^n\frac{2\pi}{(a^2-1)^{3/2}}\left(a-\sqrt{a^2-1}\right)^n\left(a+n\sqrt{a^2-1}\right)$$
Interestingly, one can obtain an infinite series in various ways from this integral, which can be seen as a generating function for the following combinatorial coefficients:
$$\sum_{m=0}^{\infty}{2m+n\choose m}a^{2m+n+1}=(-1)^n\frac{J_n(1/a)}{2\pi}=\frac{(1-\sqrt{1-a^2})^n}{a^{n-1}\sqrt{1-a^2}}$$