Question

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

Answer 1

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

$y=\int \frac{\left(2{\cos }^{2}x-1\right)-(2{\cos }^2\alpha -1)}{\cos x-\cos \alpha }dx[\cos 2\theta =2{\cos }^2\theta -1]$

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

$y=\int \frac{2{\cos }^{2}x-2{\cos }^2\alpha }{(\cos x-\cos \alpha )}dx$

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

$y=2\int \frac{(\cos x-\cos \alpha )(\cos x+\cos \alpha )}{(\cos x-\cos \alpha )}dx{a}^2-b^2=(a-b)(a+b)$

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

$y=2\left[\int \cos x+\int \cos \alpha dx\right]$

$y=2[\sin x+x.\cos \alpha +c]$ $\cos \alpha$ is a constant.

$y=2\sin x+2x\cos \alpha +c$