Question

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

Answer 1

$\frac{1}{\cos (x+a)\cos (x+b)}$

Multiplying and dividing by $\sin (a-b)$, we obtain

$\frac{1}{\sin (a-b)}\left[\frac{\sin (a-b)}{\cos (x+a)\cos (x+b)}\right]$
$=\frac{1}{\sin (a-b)}\left[\frac{\sin \left[\right(x+a)-(x+b\left)\right]}{\cos (x+a)\cos (x+b)}\right]$
$=\frac{1}{\sin (a-b)}\left[\frac{\sin (x+a)\cdot \cos (x+b)-\cos (x+a)\sin (x+b)}{\cos (x+a)\cos (x+b)}\right]$

$=\frac{1}{\sin (a-b)}[\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x+b)}]$
$=\frac{1}{\sin (a-b)}\left[\tan \right(x+a)-\tan (x+b\left)\right]$
$\therefore \int \frac{1}{\cos (x+a)\cos (x+b)}dx=\frac{1}{\sin (a-b)}\int \left[\tan \right(x+a)-\tan (x+b\left)\right]dx$

$=\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|+\log |\cos (x+b\left)|\right]+C$

$=\frac{1}{\sin (a-b)}\log \left|\frac{\cos (x+b)}{\cos (x+a)}\right|+C$