Question

evaluate evamurteanswer anaswer ​

Answer 1

Given :-

$\displaystyle \int (1 + \tan x\tan (x+\theta)) \, \, dx\\ \\ \\ \left[ Using \tan (A+B) = \frac{\tan A +\tan B}{1-\tan A\tan B} \right] \\ \\ \\ = \int \left( 1 + \tan x \left( \frac{\tan x + \tan \theta}{1-\tan x \tan \theta} \right) \right) \, \, dx \\ \\ \\ = \int \left( \frac{1\cancel{-\tan x\tan \theta}+\tan^2x+\cancel{\tan x\tan \theta}}{1-\tan x\tan \theta}\right) dx \\ \\ \\ = \int \left( \frac{\sec^2x}{1-\tan x\tan \theta} \right) dx$

$\left[ \begin{array}{l}Let \tan x = t \\ \\ \implies \sec^2x \, dx = dt \\ \\ \\ Also, \, \, Let \, \, \tan \theta = a \end{array}\right]$

$= \displaystyle \int \frac{dt}{1-at} \\ \\ \\ = \frac{\ln (1-at)}{-a} + c \\ \\ \\ = -\frac{1}{\tan \theta} \ln (1-\tan \theta \tan x) +c \\ \\ \\ \\ \implies \boxed{\displaystyle \int (1 + \tan x\tan (x+\theta)) \, \, dx = -\frac{1}{\tan \theta} \ln (1-\tan \theta \tan x) +c}$

Therefore, the required answer.