Find the integral of the function $\frac{1}{\cos (x-a) \cos (x-b)}$.

(N/A) To evaluate the integral $I = \int \frac{1}{\cos (x-a) \cos (x-b)} dx$,we multiply and divide by $\sin(a-b)$:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x-a) \cos (x-b)} dx$
Since $a-b = (x-b) - (x-a)$,we can rewrite the numerator:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin [(x-b) - (x-a)]}{\cos (x-a) \cos (x-b)} dx$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin (x-b) \cos (x-a) - \cos (x-b) \sin (x-a)}{\cos (x-a) \cos (x-b)} dx$
$I = \frac{1}{\sin (a-b)} \int [\tan (x-b) - \tan (x-a)] dx$
Integrating $\tan(x)$ gives $\ln|\sec(x)|$ or $-\ln|\cos(x)|$:
$I = \frac{1}{\sin (a-b)} [-\ln|\cos (x-b)| + \ln|\cos (x-a)|] + C$
$I = \frac{1}{\sin (a-b)} \ln \left| \frac{\cos (x-a)}{\cos (x-b)} \right| + C$,where $C$ is an arbitrary constant.