Question

Solve $\int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx$

Answer 1

$\int \frac{cos2x-cos2\alpha }{cosx-cos\alpha }dx$

$=\int \frac{2{\cos }^{2}x-1-2{\cos }^2\alpha +1}{cosx-cos\alpha }dx$

$=\int \frac{2{\cos }^{2}x-2{\cos }^2\alpha }{cosx-cos\alpha }dx$

$=\int \frac{2({\cos }^{2}x-{\cos }^2\alpha )}{cosx-cos\alpha }dx$

$=\int \frac{2(cosx+cos\alpha )(cosx-cos\alpha )}{(cosx-cos\alpha )}dx$

$=\int 2(cosx+cos\alpha )dx$

$=2\int \cos xdx+2\cos \alpha \int 1dx$

$=2\sin x+2x\cos \alpha +C$

Hence, this is the answer.