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Integrate the function
$\int \frac{1}{\cos (x+a) \cos (x+b)} d x$
$\int \frac{1}{\cos (x+a) \cos (x+b)} d x$
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$\begin{aligned} & \frac{1}{\sin (a-b)} \int \frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)} d x \\=& \frac{1}{\sin (a-b)} \int[\tan (x+a)-\tan (x+b)] d x \\=& \frac{-1}{\sin (a-b)}[\log |\cos (x+a)|-\log \mid \cos (x+b)]+c \\=& \frac{1}{\sin (a-b)} \log \frac{\cos (x+b)}{\cos (x+a)}+c \end{aligned}$
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