The value of integral $\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x$, where $\alpha$ is a constant is :
Answer & explanation
Correct answer: option 4
$I =\int \frac{\cos 2 x-\cos 2 x}{\cos x-\cos x} d x$
$=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 x-1\right)}{\cos x-\cos \alpha} d x$
as $\cos 2 \theta = 2\cos^2 \theta - 1$
$=\int \frac{2 \cos ^2 x-2 \cos ^2 x-1+1}{\cos x-\cos x} d x$
$=2 \int \frac{\cos ^2 x-\cos ^2 x}{\cos x-\cos x} d x$
$=2 \int \cos x+\cos \alpha d x$
$=2 \sin x+2 x \cos k+C$