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

Answer 1

Answer: (A)

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