Question

$\int \frac{1}{\cos (x-a)\cos (x-b)}dx$

Answer 1

$\int \frac{1}{\cos (x-a)\cos (x-b)}dx$

$=\int \frac{\sin (a-b)}{\sin (a-b)}\times \frac{1}{\cos (x-a)\cos (x-b)}dx$

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

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

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

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

Using $\sin (A-B)=sinAcosB-cosAsinB$

Replace $A$ by $(x-b)$, & $B$ by $(x-a)$

$\therefore$ $\sin ((x-b)-\sin (x-a))=\sin (x-b)\cos (x-a)-\cos (x-b)\sin (x-a)$

$\therefore$ $\frac{1}{\sin (a-b)}\int \frac{\sin (x-b)-(x-a)}{\cos (x-a)\cos (x-b)}dx$

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

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

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

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

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

[Note : $\int tanxdx=-log\left|cosx\right|+C$]

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

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

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