Question

Integrate ∫ (cos2x-cos2α)/ (cosx-cosα) dx, where α is independent of x.

Answer 1

\[\cos{2x}=\cos^2{x}-\sin^2{x}=\cos^2{x}-(1-\cos^2{x})=2\cos^2{x}-1\]\[\cos{2\alpha}=2\cos^2{\alpha}-1\]\[\int\frac{\cos{2x}-\cos{2\alpha}}{\cos{x}-\cos{\alpha}}dx=\int\frac{2\cos^2{x}-1-(2\cos^2{\alpha}-1)}{\cos{x}-\cos{\alpha}}dx=\]\[=2\int\frac{\cos^2{x}-\cos^2{\alpha}}{\cos{x}-\cos{\alpha}}dx=2\int(\cos{x}+\cos{\alpha})dx=\]\[=2(\sin{x}+x\cos{\alpha})+C\]