If $I =\int \frac{\cos 2 x-\cos 2 \alpha}{\sin x-\sin \alpha} d x$, then I equals

$2 \sin x-x \cos \alpha+C$ $2 \cos x-2 x \sin \alpha+C$ $2 \cos x+2 x \sin \alpha+C$ $2 \sin x+x \cos \alpha+C$

$2 \cos x-2 x \sin \alpha+C$

We have, $I=\int \frac{\cos 2 x-\cos 2 \alpha}{\sin x-\sin \alpha} d x$ $\Rightarrow I=\int \frac{\left(1-2 \sin ^2 x\right)-\left(1-2 \sin ^2 \alpha\right)}{\sin x-\sin \alpha} d x$ $\Rightarrow I=-2 \int(\sin x+\sin \alpha) d x$ $\Rightarrow I=2(\cos x-x \sin \alpha)+C$