Question

Evaluate: $\int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx$.

Answer 1

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

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

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

$I=\int 2\cos \frac{x+\alpha }{2}\times 2\cos \frac{x-\alpha }{2}dx$

$\cos (A+B)\cos (A-B)={\cos }^{2}A-{\sin }^{2}B$

$I=\int (4{\cos }^2\frac{x}{2}-{\sin }^2\frac{a}{2})dx$

$I=4\int {\cos }^2\frac{x}{2}-4\int {\sin }^2\frac{a}{2}dx$

$I=2\int (1+\cos x)dx-4x{\sin }^2\frac{a}{2}$

$I=2x+2\sin x-4{\sin }^2\frac{a}{2}+c$