∫(cos 2x-cos 2θ)/(cos x-cosθ) dx is equal to

The correct answer is Option (4) → \(2\left(\sin x+x\cos\theta\right)+C\)

Here \(\theta\) is a constant. Writing both terms of the numerator with the identity \(\cos 2A = 2\cos^2A-1\):

    \( \cos 2x-\cos 2\theta = 2\cos^2x-2\cos^2\theta \)

    \( = 2\left(\cos x-\cos\theta\right)\left(\cos x+\cos\theta\right) \)

The factor \(\cos x-\cos\theta\) cancels:

    \( \displaystyle \int\frac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\,dx = \int 2\left(\cos x+\cos\theta\right)dx \)

Since \(\cos\theta\) is a constant, \(\int\cos\theta\,dx = x\cos\theta\):

    \( = 2\left(\sin x+x\cos\theta\right)+C \)