Question

$\int \left\{1+\tan x\tan \left(x+\mathrm{\theta }\right)\right\}dx$

Answer 1

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