Question

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

Answer 1

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

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

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

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

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