Question

Find the integrals of the function :
$\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$

Answer 1

Given that ,

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

We know that $\cos 2A=2{\cos }^{2}A-1$

Therefore,

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

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

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

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

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

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

Hence, this is the answer.