$\displaystyle \int \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta}\, dx$ is equal to:

$\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$ $\Rightarrow \cos 2x - \cos 2\theta = -2\sin(x+\theta)\sin(x-\theta)$ and $\cos x - \cos\theta = -2\sin\dfrac{x+\theta}{2}\sin\dfrac{x-\theta}{2}$ So, $\displaystyle \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} = \dfrac{\sin(x+\theta)\sin(x-\theta)}{\sin\dfrac{x+\theta}{2}\sin\dfrac{x-\theta}{2}} = 4\cos\dfrac{x+\theta}{2}\cos\dfrac{x-\theta}{2} = 2(\cos x + \cos\theta)$ Hence, $\displaystyle \int \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta},dx = \int 2(\cos x + \cos\theta),dx = 2\sin x + 2x\cos\theta + C$ $\boxed{\text{Answer: (A) }2(\sin x + x\cos\theta) + C}$