$\int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx$ is equal to

The integral to evaluate is $\int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx$, where α is a constant.

Using the double angle formula $\cos 2θ = 2\cos^2 θ - 1$:

$\cos 2x = 2\cos^2 x - 1$

$\cos 2α = 2\cos^2 α - 1$

The numerator becomes:

$\cos 2x - \cos 2α = (2\cos^2 x - 1) - (2\cos^2 α - 1)$

$= 2\cos^2 x - 1 - 2\cos^2 α + 1$

$= 2\cos^2 x - 2\cos^2 α$

$= 2(\cos^2 x - \cos^2 α)$

Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:

$2(\cos^2 x - \cos^2 α) = 2(\cos x - \cos α)(\cos x + \cos α)$

The integral becomes:

$\int \frac{2(\cos x - \cos α)(\cos x + \cos α)}{\cos x - \cos α} dx$

$= \int 2(\cos x + \cos α) dx$

$= \int 2\cos x \, dx + \int 2\cos α \, dx$

Evaluating each term:

$\int 2\cos x \, dx = 2\sin x$

$\int 2\cos α \, dx = 2x\cos α$ (since α is a constant)

Therefore:

$\int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx = 2\sin x + 2x\cos α + C$