Calculus: Early Transcendentals (2nd Edition)

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