Question

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

(a) $2\left(\sin x+x\cos \theta \right)+C$
(b) $2\left(\sin x-x\cos \theta \right)+C$
(c) $2\left(\sin x+2x\cos \theta \right)+C$
(d) $2\left(\sin x-2x\cos \theta \right)+C$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\cos 2x-\cos 2\theta }{\cos x-\cos \theta }\mathrm{d}x \\ =\int \frac{\left(2{\cos }^{2}x-1\right)-\left(2{\cos }^2\theta -1\right)}{\cos x-\cos \theta }\mathrm{d}x \\ =\int \frac{2{\cos }^{2}x-1-2{\cos }^2\theta +1}{\cos x-\cos \theta }\mathrm{d}x \\ =\int \frac{2\left(\cos x-\cos \theta \right)\left(\cos x+\cos \theta \right)}{\cos x-\cos \theta }\mathrm{d}x \\ =\int 2\left(\cos x+\cos \theta \right)\mathrm{d}x \\ =2\left(\sin x+x\cos \theta \right)+C \\ \mathrm{Therefore},\int \frac{\cos 2x-\cos 2\theta }{\cos x-\cos \theta }\mathrm{d}x=2\left(\sin x+x\cos \theta \right)+C \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{a}\right). \end{gathered}$$