Question

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

Answer 1

\[\text{Let I} = \int1 + \tan x \tan \left( x + \theta \right)dx\]
\[ = \int1 + \ tanx\left( \frac{\tan x + \tan \theta}{1 - \tan x \tan \theta} \right)dx\]
\[ = \int\frac{1 + \tan^2 x}{1 - \tan x \tan \theta}dx\]
\[ = \int\frac{\sec^2 x dx}{1 - \tan x \tan \theta}\]
\[Putting\ \ tan\ x = t\]
\[ \Rightarrow \text{sec}^2    x = \frac{dt}{dx} \]
\[ \Rightarrow dx = \frac{dt}{\sec^2 x}\]
\[ \therefore I = \int\frac{1}{1 - t \tan\theta}dt\]
\[ = \frac{- 1}{\tan \theta} \ln \left| 1 - t \tan \theta \right| + C \left[ \because \int\frac{1}{ax + b}dx = \frac{1}{a}\ln \left| ax + b \right| + C \right]\]
\[ = - \cot \theta \ln \left| 1 - \tan\ x \tan \theta \right| + C\]
\[ = \cot \theta \ln \left| \frac{1}{1 - \tan x \tan \theta} \right| + C\]
\[ = \cot \theta \ln \left| \frac{\ cosx \cos\theta}{\cos x \cos \theta - \sin x \sin \theta} \right| + C\]
\[ = \cot \theta \ln \left| \frac{\cos x}{\cos \left( x + \theta \right)} \right| + C' \left[ Let C' = C + \cot \theta \ln \cos\theta \right]\]