Question

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

• A. $2[\cos x+(\sin \alpha \left)x\right]+c$
• B. $2[\cos x-x\sin \alpha ]+c$
• C. $-2[\cos x-x\sin \alpha ]+c$
• D. $-2$[$\cos x-\sin \alpha$] $+c$

Answer 1

The correct option is B $2[\cos x-x\sin \alpha ]+c$

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

We know, $\cos 2x=1-2{\sin }^{2}x;\cos 2\alpha =1-2{\sin }^2\alpha$

Where $\alpha$ is const.

Substituting the values, we get

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

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

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

$=-2\int (\sin x+\sin \alpha ).dx$

$=-2\left[\right(-\cos x)+x\sin \alpha +c].$

$=2[\cos x-x\sin \alpha ]+c$