Find the integral of the function $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$.

Consider the expression $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$.
Using the formula $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$,we have:
$\frac{-2 \sin(x+\alpha) \sin(x-\alpha)}{-2 \sin(\frac{x+\alpha}{2}) \sin(\frac{x-\alpha}{2})} = \frac{\sin(x+\alpha) \sin(x-\alpha)}{\sin(\frac{x+\alpha}{2}) \sin(\frac{x-\alpha}{2})}$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$= \frac{[2 \sin(\frac{x+\alpha}{2}) \cos(\frac{x+\alpha}{2})] [2 \sin(\frac{x-\alpha}{2}) \cos(\frac{x-\alpha}{2})]}{\sin(\frac{x+\alpha}{2}) \sin(\frac{x-\alpha}{2})}$
$= 4 \cos(\frac{x+\alpha}{2}) \cos(\frac{x-\alpha}{2})$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$= 2[\cos(\frac{x+\alpha}{2} + \frac{x-\alpha}{2}) + \cos(\frac{x+\alpha}{2} - \frac{x-\alpha}{2})]$
$= 2[\cos x + \cos \alpha] = 2 \cos x + 2 \cos \alpha$
Now,integrating with respect to $x$:
$\int (2 \cos x + 2 \cos \alpha) dx = 2 \sin x + 2x \cos \alpha + C$,where $C$ is an arbitrary constant.