Question

please evaluate this question from integration I'll mark brainliest

Answer 1

You know that, tan(c-b) = (tan c – tan b)/(1+tan c * tan b),
So, (1+tan c * tan b) = (tan c – tan b) / tan (c-b),
Put c= x+a, b=x,
(1+tanx * tan(x+a)) = (tan(x+a) – tan x) / tan(x+a-x) = (tan(x+a) – tan x) / tan a,
Now,seperate the two terms in 2 diff integrals and solve it...
Hope it helps ☺️

Answer 2

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


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


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