Question

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

Answer 1

Given equation is,

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

Multiply and divided by $\sin (a-b)$ in the equation, we get

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

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

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

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

We know that ,

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

Therefore,

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

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

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

We know; $\int \tan x=-\log \left|\cos x\right|+c$

Therefore,

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

$\Rightarrow \log m-\log n=\log \frac{m}{n}$

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

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