$\int\left(\frac{\cos 2x -\cos 2α}{\cos x -cos α}\right)dx=$
(Given that c is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $2 \sin x + 2x \cos α + c$
\[ \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx = \int \frac{2\cos^2 x - 1 - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx = \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \, dx \]
\[ = \int 2(\cos x + \cos \alpha) \, dx = 2\int \cos x \, dx + 2\cos \alpha \int dx = 2\sin x + 2x\cos \alpha + C \]