Prove

$\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}=\frac{-2 \sin \frac{2 x+2 \alpha}{2} \sin \frac{2 x-2 \alpha}{2}}{-2 \sin \frac{x+\alpha}{2} \sin \frac{x-\alpha}{2}} \quad\left[\cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}\right]$

$=\frac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)}$

$=\frac{\left[2 \sin \left(\frac{x+\alpha}{2}\right) \cos \left(\frac{x+\alpha}{2}\right)\right]\left[2 \sin \left(\frac{x-\alpha}{2}\right) \cos \left(\frac{x-\alpha}{2}\right)\right]}{\sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)}$

$=\frac{\left[2 \sin \left(\frac{x+\alpha}{2}\right) \cos \left(\frac{x+\alpha}{2}\right)\right]\left[2 \sin \left(\frac{x-\alpha}{2}\right) \cos \left(\frac{x-\alpha}{2}\right)\right]}{\sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)}$

$=4 \cos \left(\frac{x+\alpha}{2}\right) \cos \left(\frac{x-\alpha}{2}\right)$

$=2\left[\cos \left(\frac{x+\alpha}{2}+\frac{x-\alpha}{2}\right)+\cos \frac{x+\alpha}{2}-\frac{x-\alpha}{2}\right]$

$=2[\cos (x)+\cos \alpha]$

$=2 \cos x+2 \cos \alpha$

$\therefore \int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x=\int 2 \cos x+2 \cos \alpha$

$=2[\sin x+x \cos \alpha]+\mathrm{C}$