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$\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x$ is equal to
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Verified Answer
The correct answer is:
$2(\sin x+x \cos \theta)+C$
$2(\sin x+x \cos \theta)+C$
Suppose
$$
\begin{aligned}
&I=\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x \\
&=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \theta-1\right)}{\cos x-\cos \theta} d x \\
&=2 \int \frac{(\cos x+\cos \theta)(\cos x-\cos \theta)}{(\cos x-\cos \theta)} d x \\
&=2 \int(\cos x+\cos \theta) d x=2(\sin x+x \cos \theta)+C
\end{aligned}
$$
$$
\begin{aligned}
&I=\int \frac{\cos 2 x-\cos 2 \theta}{\cos x-\cos \theta} d x \\
&=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \theta-1\right)}{\cos x-\cos \theta} d x \\
&=2 \int \frac{(\cos x+\cos \theta)(\cos x-\cos \theta)}{(\cos x-\cos \theta)} d x \\
&=2 \int(\cos x+\cos \theta) d x=2(\sin x+x \cos \theta)+C
\end{aligned}
$$
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