Question

Integrate $\int \frac{\cos 2x-\cos 2a}{\cos x-\cos a}dx=$

• A. $2\left(\sin x-x\cos a\right)+C$
• B. $2\left(-\sin x+x\cos a\right)+C$
• C. $2\left(\sin x+x\cos a\right)+C$
• D. $2\left(-\sin x-x\cos a\right)+C$

Answer 1

The correct option is C $2\left(\sin x+x\cos a\right)+C$

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

$=\int \frac{2{\cos }^{2}x-1-2{\cos }^2\alpha +1}{\cos x-\cos \alpha }dx$ $\because \cos 2x={\cos }^{2}x-{\sin }^{2}x=2{\cos }^{2}x-1$

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

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

$=2\int \cos xdx+2\int \cos \alpha dx$

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

Hence, the answer is $2\sin x+2x\cos \alpha +c.$