Question

Integrate with respect to x : $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$

Answer 1

Given, $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$

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

$=\int \frac{(2{\cos }^{2}x-1)-(2{\cos }^2\alpha -1)}{(\cos x-\cos \alpha )\cdot }dx$

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

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

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

$=2\sin x+2x\cdot \cos \alpha +c$.