Question

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

Answer 1

\[\text{ Let I } = \int\frac{1}{1 - \tan x}dx\]
\[ = \int\frac{1}{1 - \frac{\sin x}{\cos x}}dx\]
\[ = \int\frac{\cos x}{\cos x - \sin x}dx\]
\[ = \frac{1}{2}\int\frac{2 \cos x}{\cos x - \sin x}dx\]
\[ = \frac{1}{2}\int\left( \frac{\cos x + \sin x + \cos x - \sin x}{\cos x - \sin x} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{\cos x + \sin x}{\cos x - \sin x} \right)dx + \frac{1}{2}\int dx\]
\[\text{ Putting cos x }- \sin x = t\]
\[ \Rightarrow \left( - \sin x - \cos x \right)dx = dt\]
\[ \Rightarrow \left( \sin x + \cos x \right)dx = - dt\]
\[ \therefore I = - \frac{1}{2}\int\frac{dt}{t} + \frac{x}{2} + C\]
\[ = - \frac{1}{2} \text{ ln }\left| \cos x - \sin x \right| + \frac{x}{2} + C\]
\[ = \frac{x}{2} - \frac{1}{2} \text{ ln }\left| \cos x - \sin x \right| + C\]