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

To evaluate the integral $\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx$, simplify the numerator using trigonometric identities.

The difference $\cos 2x - \cos 2\alpha$ becomes $2\cos^2 x - 1 - (2\cos^2 \alpha - 1) = 2(\cos^2 x - \cos^2 \alpha)$.

Factoring the difference of squares gives $2(\cos x - \cos \alpha)(\cos x + \cos \alpha)$, yielding the integrand $\frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha}$.

For $\cos x \neq \cos \alpha$, this reduces to $2(\cos x + \cos \alpha)$.

Integrating term by term, $\int 2\cos x dx = 2\sin x$ and $\int 2\cos \alpha dx = 2\cos \alpha \int dx = 2x\cos \alpha$, since $\cos \alpha$ is constant.

The antiderivative is $2\sin x + 2x\cos \alpha + c$, which corresponds to option C.