\(\int\displaystyle\left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right)dx =\)
(Given that c is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2 \sin x + 2x \cos \alpha + c\)
Using \(\cos 2\theta = 2\cos^2\theta - 1\), the numerator becomes a difference of squares:
\(\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1)\)
\(= 2(\cos^2 x - \cos^2 \alpha)\)
\(= 2(\cos x - \cos \alpha)(\cos x + \cos \alpha)\)
Cancelling \(\cos x - \cos \alpha\):
\(\displaystyle \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\,dx = \int 2(\cos x + \cos \alpha)\,dx\)
Here \(\cos \alpha\) is a constant, so its integral with respect to x is \(x\cos \alpha\):
\(= 2\sin x + 2x\cos \alpha + c\)
NCERT (2026-27 edition):
Integration Using Trigonometric Identities
