Question

If $I=\int \frac{\cos 2x-\cos 2\alpha }{\sin x-\sin \alpha }dx$, then $I$ equals

• A. $2\sin x-x\cos \alpha +C$
• B. $2\cos x-2x\sin \alpha +C$
• C. $2\cos x+2x\sin \alpha +C$
• D. $2\sin x+x\cos \alpha +C$

Answer 1

Answer: (C)

$2\cos x+2x\sin \alpha +C$

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

We know that ${\cos }^{2}x=1-{\sin }^{2}x$

$I=\int \frac{((1-2{\sin }^{2}x)-(1-2{\sin }^2\alpha ))}{\sin x-\sin \alpha }dx$

$I=\int \frac{(-2{\sin }^{2}x+2{\sin }^2\alpha )}{\sin x-\sin \alpha }dx$

We know that, $a^2-b^2=(a-b)(a+b)$

$I=-2\int \left(\sin x+\sin \alpha \right)dx$

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