Answer
\[ = - \frac{2}{{1 + \tan \,\left( {\frac{x}{2}} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{1 + \sin x}}} \hfill \\
\hfill \\
se\,t = 2{\tan ^{ - 1}}\,\left( u \right)\,\,\,\,\,then\,\,dx = 2\left( {\frac{{du}}{{1 + {u^2}}}} \right) \hfill \\
and\,\,\sin x = \frac{{2u}}{{1 + {u^2}}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{1 + \sin x}}} = \int_{}^{} {\frac{{\frac{{2du}}{{1 + {u^2}}}}}{{1 + \frac{{2u}}{{1 + {u^2}}}}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \int_{}^{} {\frac{2}{{\,{{\left( {1 + u} \right)}^2}}}du} \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
= - \frac{2}{{1 + u}} + C \hfill \\
\hfill \\
substitute\,\,for\,\,\,u \hfill \\
\hfill \\
= - \frac{2}{{1 + \tan \,\left( {\frac{x}{2}} \right)}} + C \hfill \\
\end{gathered} \]
