Answer
\[ = \,\tan \theta - \frac{1}{{\cos \theta }} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{d\theta }}{{1 + \sin \theta }}} \hfill \\
\hfill \\
Multiplying\,numerator\,and\,denominator\, \hfill \\
of\,\,integrand\,\,by\,\,\,\left( {1 - \sin \theta } \right) \hfill \\
\hfill \\
\,integral\,becomes\, \hfill \\
\hfill \\
\int_{}^{} {\frac{{d\theta }}{{1 + \sin \theta }}} = \int_{}^{} {\frac{{1 - \,\sin \,\theta }}{{1 - {{\sin }^2}\theta }}d\theta } \hfill \\
\hfill \\
or \hfill \\
\hfill \\
= \int_{}^{} {\frac{{1 - \sin \theta }}{{{{\cos }^2}\theta }}d\theta } \hfill \\
\hfill \\
split\,\,the\,\,numerator \hfill \\
\hfill \\
= \int_{}^{} {\left( {\frac{1}{{{{\cos }^2}\theta }}} \right) - d\theta } \int_{}^{} {\frac{{\sin \theta }}{{{{\cos }^2}\theta }}d\theta } \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{{\sec }^2}\theta } \right)d\theta } - \int {\frac{{\sin \theta }}{{{{\cos }^2}\theta }}} d\theta \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{{\sec }^2}\theta } \right)d\theta } - \int {{{\left( {\cos \theta } \right)}^{ - 2}}\sin \theta } d\theta \hfill \\
\hfill \\
\,integrate \hfill \\
\hfill \\
= \,\tan \theta - \frac{1}{{\cos \theta }} + C \hfill \\
\end{gathered} \]
