Question

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

Answer 1

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

\[ = 2 \times \frac{1}{2\sqrt{2}}\text{ ln }\left| \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} \right| + C\]
\[ = \frac{1}{\sqrt{2}}\text{ ln }\left| \frac{\sqrt{2} + \tan \frac{x}{2} - 1}{\sqrt{2} - \tan \frac{x}{2} + 1} \right| + C\]