Evaluate the following integrals:

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

$\therefore \tan (x-(x+\theta))=\frac{\tan x-\tan (x+\theta)}{1+\tan x \tan (x+\theta)}$

$\therefore \tan (\theta)=\frac{\tan x-\tan (x+\theta)}{1+\tan x \tan (x+\theta)}$

$\Rightarrow \tan (\theta)(1+\tan x \cdot \tan (x+\theta))=\tan (x)-\tan (x+\theta)$

$\Rightarrow(1+\tan x \cdot \tan (x+\theta))=\frac{1}{\tan \theta}(\tan x-\tan (x+\theta)$

$\Rightarrow \int \frac{1}{\tan \theta}(\tan x-\tan (x+\theta)) \cdot d x$

$\Rightarrow \frac{1}{\tan \theta} \int \tan x d x-\int \tan (x+\theta) d x$

$\Rightarrow \frac{1}{\tan \theta}(-\ln |\cos x|-(-\ln |\cos (x+\theta)|+c .$

$\Rightarrow \frac{1}{\tan \theta}(-\ln |\cos x|+\ln |\cos (x+\theta)|+c .$