Question

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

Answer 1

We have,

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

Divide and multiply by $\sin (a-b)$

Then,

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

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

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

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

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

Using formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$

So,

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

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

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

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

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

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

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

Hence, this is the answer.