If $I =\int \frac{\cos 2 x-\cos 2 \alpha}{\sin x-\sin \alpha} d x$, then I equals
Answer & explanation
Correct answer: option 2
We have,
$I=\int \frac{\cos 2 x-\cos 2 \alpha}{\sin x-\sin \alpha} d x$
$\Rightarrow I=\int \frac{\left(1-2 \sin ^2 x\right)-\left(1-2 \sin ^2 \alpha\right)}{\sin x-\sin \alpha} d x$
$\Rightarrow I=-2 \int(\sin x+\sin \alpha) d x$
$\Rightarrow I=2(\cos x-x \sin \alpha)+C$