Question

$\int \frac{\cos 2x - \cos 2\theta}{\cos x -\cos \theta} dx$ is equal to

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

Answer 1

Answer: (A)

Let $I =\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$
$=\int \frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \theta-1\right)}{\cos x-\cos \theta} d x $
$=2 \int \frac{\cos ^{2} x-\cos ^{2} \theta}{\cos x-\cos \theta} d x $
$=2 \int \frac{(\cos x-\cos \theta)(\cos x+\cos \theta)}{\cos x-\cos \theta} d x $
$=2 \int(\cos x+\cos \theta) d x$
$=2[\sin x+x \cos \theta]+C $