#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise Multiple Choice Questions Question 31 Maths Textbook Solution.

$2(\sin x+x \cos \theta)+C$

Given:

$\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$

Hint:

You must know about the integration of cos x.

Explanation:

Let $\mathrm{I}=\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$

$=\int \frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \theta-1\right)}{\cos x-\cos \theta} d x$                                                    $\left[\because \cos 2 \theta=2 \cos ^{2} \theta-1\right]$

\begin{aligned} &=\int \frac{2 \cos ^{2} x-2 \cos ^{2} \theta+1-1}{\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 \end{aligned}                                                    $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

\begin{aligned} &=2 \int \cos x d x+2 \int \cos \theta d x \\ &=2 \int \cos x d x+2 \cos \theta \int 1 d x \\ &=2 \sin x+2 \cos \theta \cdot x+C \\ &=2(\sin x+x \cos \theta)+C \end{aligned}

Hence,$\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x=2(\sin x+x \cos \theta)+C .$