Question

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

Answer 1

$\int \frac{cos2x-cos2\alpha }{cosx-cos\alpha }dx$

$cos2x=2co{s}^{2}x-1$, use in above Integral

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

$\int \frac{2co{s}^{2}x-2co{s}^2\alpha -1+1}{cosx-cosx}dx$

$\int \frac{2\left[\right(cosx+cos\alpha \left)\right(cosx-cos\alpha \left)\right]}{cosx-cos\alpha }dx$

$\int \frac{2(co{s}^{2}x-co{s}^2\alpha )}{cosx-cos\alpha }dx$

$2\int cosx+cos\alpha dx$

$2sinx+2xcos\alpha +c$ $[cos\alpha$ a constant, Since use are Integrating wrt x]

$=2sinx+2xcos\alpha +c$