Evaluate the following integrals:

Let $I=\int \frac{1}{\cos (x+a) \cos (x+b)} d x$

Dividing and multiplying I by $\sin (a-b)$ we get,

$I=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x+a) \cos (x+b)} d x$

$I=\frac{1}{\sin (a-b)} \int \frac{\sin \{(x+a)-(x+b)\}}{\cos (x+a) \cos (x+b)} d x$

$I=\frac{1}{\sin (a-b)} \int \frac{\sin (x+a) \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)} d x$

$I=\frac{1}{\sin (a-b)} \int\{\tan (x+a)-\tan (x+b)\} d x$

$\int \tan x d x=|\log \sec x|+c$

Therefore,

$1=\frac{1}{\sin (a-b)}\left\{\frac{\log (\sec (x+a))}{x+a}-\frac{\log (\sec (x+b))}{x+b}\right\}+c$