Integrate the function: $\frac{1}{\cos (x+a) \cos (x+b)}$

To integrate $I = \int \frac{1}{\cos (x+a) \cos (x+b)} dx$,we multiply and divide by $\sin (a-b)$:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x+a) \cos (x+b)} dx$
We can rewrite the numerator as $\sin [(x+a) - (x+b)] = \sin (a-b)$:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin [(x+a) - (x+b)]}{\cos (x+a) \cos (x+b)} dx$
Using the identity $\sin (A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin (a-b)} \int \left[ \frac{\sin (x+a) \cos (x+b) - \cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)} \right] dx$
$I = \frac{1}{\sin (a-b)} \int [\tan (x+a) - \tan (x+b)] dx$
Integrating $\tan \theta$ gives $\ln |\sec \theta|$ or $-\ln |\cos \theta|$:
$I = \frac{1}{\sin (a-b)} [-\ln |\cos (x+a)| + \ln |\cos (x+b)|] + C$
Using the property $\ln m - \ln n = \ln (m/n)$:
$I = \frac{1}{\sin (a-b)} \ln \left| \frac{\cos (x+b)}{\cos (x+a)} \right| + C$