Question

\[\int\frac{1 + \tan x}{1 - \tan x} dx\]

Answer 1

\[\text{Let I} = \int\left( \frac{1 + \ tanx}{1 - \ tanx} \right)dx\]
\[ = \int\left( \frac{1 + \frac{\sin  x}{\cos x}}{1 - \frac{\sin x}{\cos x}} \right)dx\]
\[ = \int\left( \frac{\cos x + \sin x}{\cos x - \sin x} \right)dx\]
\[Putting\ \cos\ x - \sin x = t\]
\[ \Rightarrow \left( - \sin x - \cos x \right)dx = dt\]
\[ \Rightarrow \left( \sin x + \cos x \right)dx = - dt\]
\[ \therefore I = - \int\frac{1}{t}dt\]
\[ = - \text{ln }\left| t \right| + C\]
\[ = - \text{ln }\left| \cos x - \sin x \right| + C \left[ \because t = \cos x - \sin x \right]\]