Question

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

Answer 1

Let $I=\int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }\cdot dx$

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

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

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

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

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

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

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

$=2\left[\int \cos x+\int \cos \alpha \cdot dc\right]$

$=2\left[\int \cos x\cdot dx+\cos \alpha \int 1.dx\right]$

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