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