Question

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

Answer 1

$\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\frac{-2\sin \frac{2x+2\alpha }{2}\sin \frac{2x-2\alpha }{2}}{-2\sin \frac{x+\alpha }{2}\sin \frac{x-\alpha }{2}},[\because \cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}]$

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

$=\frac{\left[2\sin \left(\frac{x+\alpha }{2}\right)\cos \left(\frac{x+\alpha }{2}\right)\right]\left[2\sin \left(\frac{x-\alpha }{2}\right)\cos \left(\frac{x-\alpha }{2}\right)\right]}{\sin \left(\frac{x+\alpha }{2}\right)\sin \left(\frac{x-\alpha }{2}\right)}$

$=4\cos \left(\frac{x+\alpha }{2}\right)\cos \left(\frac{x-\alpha }{2}\right)$

$=2[\cos (\frac{x+\alpha }{2}+\frac{x-\alpha }{2})+\cos (\frac{x-\alpha }{2}-\frac{x-\alpha }{2})]$

$=2\left[\cos \right(x)+\cos \alpha ]$

$=2\cos x+2\cos \alpha$

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

$=2[\sin x+x\cos \alpha ]+C$