Question

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

Answer 1

To find : $\int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx$

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

$[\because \cos 2x=2{\cos }^{2}x-1]$

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

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

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

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

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

$=2\int (\cos x.dx+\cos \alpha \int 1.dx)$

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

Where $C$ is constant of integration.