Question

Solve $\int \frac{\cot 2x-\cos 2\alpha }{\cot x-\cos \alpha }dx$

Answer 1

Integrate $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx$

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

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

$(\because \cos 2a=2{\cos }^{2}a-1)$

$\Rightarrow \int \frac{2{\cos }^{2}x-1-2{\cos }^2\alpha +1}{\cos x-\cos \alpha }dx$

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

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

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

$\Rightarrow 2\int (\cos x+\cos \alpha )dx$

$\Rightarrow 2\left[\int \cos xdx+\int \cos \alpha dx\right]$

$\Rightarrow 2\sin x+C_1+2x\cos \alpha +C_2$

$\Rightarrow 2\sin x+2x\cos \alpha +C$ $(\because C=C_1+C_2)$

$\Rightarrow 2(\sin x+x\cos \alpha )+C$