Question

\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]

Answer 1

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