Question

\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\frac{\left( 1 + \sin x \right)}{\sin x \left( 1 + \cos x \right)}dx\]
\[\text{ Putting   sin x } = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\text{ and }\text{ cos x }= \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \therefore I = \int\frac{\left( 1 + \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)}{\frac{\left( 2 \tan \frac{x}{2} \right)}{\left( 1 + \tan^2 \frac{x}{2} \right)} \left( 1 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right)}dx\]
\[ = \int\frac{\left( 1 + \tan^2 \frac{x}{2} + 2 \tan \frac{x}{2} \right) \left( 1 + \tan^2 \frac{x}{2} \right)}{\left( 2 \tan \frac{x}{2} \right) \left( 1 + \tan^2 \frac{x}{2} + 1 - \tan^2 \frac{x}{2} \right)}dx\]
\[ = \frac{1}{4}\int\frac{\left( 1 + \tan^2 \frac{x}{2} + 2 \tan \frac{x}{2} \right) \sec^2 \frac{x}{2}}{\tan \frac{x}{2}} \text{ dx}\]
\[\text{ Putting tan} \frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) \text{ dx} = dt\]
\[ \Rightarrow \sec^2 \left( \frac{x}{2} \right) \text{ dx } = 2dt\]
\[ \therefore I = \frac{1}{4}\int\frac{\left( 1 + t^2 + 2t \right) \cdot \left( \text{ 2  dt} \right)}{t}\]
\[ = \frac{1}{2}\int\left( \frac{1}{t} + t + 2 \right) dt\]
\[ = \frac{1}{2} \left[ \text{ ln  }\left| t \right| + \frac{t^2}{2} + 2t \right] + C\]
\[ = \frac{1}{2} \left[ \text{ ln } \left| \text{ tan} \frac{x}{2} \right| + \frac{\tan^2 \left( \frac{x}{2} \right)}{2} + 2 \tan \left( \frac{x}{2} \right) \right] + C....... \left[ \because t = \tan \frac{x}{2} \right]\]
\[ = \frac{1}{2} \text{ ln } \left| \text{ tan  }\frac{x}{2} \right| + \frac{1}{4} \tan^2 \frac{x}{2} + \tan\frac{x}{2} + C\]