Question

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

Answer 1

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

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

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

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

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

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

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