Question

Integrate $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$

Answer 1

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

$=\int \frac{-2\sin (x+\alpha ).\sin (x-\alpha )}{-2.\sin \left(\frac{x+\alpha }{2}\right).\sin \left(\frac{x-\alpha }{2}\right)}dx$

$=\int \frac{\sin (x+\alpha ).\sin (x-\alpha )}{\sin \left(\frac{x+\alpha }{2}\right).\sin \left(\frac{x-\alpha }{2}\right)}dx$

$=\int \frac{2\sin \left(\frac{x+\alpha }{2}\right).\cos \left(\frac{x+\alpha }{2}\right).2.\sin \left(\frac{x-\alpha }{2}\right)\cos \left(\frac{x-\alpha }{2}\right)}{\sin \left(\frac{x+\alpha }{2}\right).\sin \left(\frac{x-\alpha }{2}\right)}dx$

$=4\int \cos \left(\frac{x+\alpha }{2}\right).\cos \left(\frac{x-\alpha }{2}\right)dx$

$=2\int 2\cos \left(\frac{x+\alpha }{2}\right).\cos \left(\frac{x-\alpha }{2}\right)dx$

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

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

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