Answer
\[ = \frac{1}{4}\ln \left| {\frac{{1 - \sin t}}{{1 + \sin t}}} \right| - \frac{1}{2}\frac{1}{{\sin t + 1}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{\sec t}}{{1 + \sin t}}\,dt} \hfill \\
\hfill \\
Given\,\,funcion \hfill \\
\hfill \\
\frac{{\sec t}}{{1 + \sin t}} \hfill \\
\hfill \\
{\text{using}}\,\,{\text{trigonometric identities}} \hfill \\
\hfill \\
\frac{{\sec t}}{{1 + \sin t}} = \frac{1}{{\cos t\,\left( {1 + \sin t} \right)}} \hfill \\
\hfill \\
multiply\,\,the\,\,numerator\,\,and\,\,{\text{denominator}}\,\,by\,\,\cos \,\,t \hfill \\
\hfill \\
= \frac{{\cos t}}{{{{\cos }^2}t\,\left( {1 + \sin t} \right)}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \frac{{\cos t}}{{\,\left( {1 - \sin t} \right)\,{{\left( {1 + \sin t} \right)}^2}}} \hfill \\
\hfill \\
\int_{}^{} {\frac{{\sec t}}{{1 + \sin t}}\,dt} \, = \int_{}^{} {\,\left( {\frac{{\cos t}}{{\,\left( {1 - \sin t} \right)\,{{\left( {1 + \sin t} \right)}^2}}}} \right)} \,dt \hfill \\
\hfill \\
{\text{using}}\,\,\,partial\,\,fractions \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{1}{4}\frac{1}{{1 + \sin t}} + \frac{1}{2}\frac{1}{{\,{{\left( {1 + \sin t} \right)}^2}}} - \frac{1}{4}\frac{1}{{\sin t - 1}}} \right)\,\cos tdt} \hfill \\
\hfill \\
Upon\,\,\,integrating \hfill \\
\hfill \\
\,Let,\,\,\,\,\,u = \sin t \hfill \\
\hfill \\
\int_{}^{} {\frac{{\sec t}}{{1 + \sin t}}\,dt} = \int_{}^{} {\left( {\frac{1}{4}\frac{1}{{u + 1}} + \frac{1}{2}\frac{1}{{\,{{\left( {u + 1} \right)}^2}}} - \frac{1}{4}\frac{1}{{u - 1}}} \right)} \cos tdt \hfill \\
\hfill \\
{\text{integrating}} \hfill \\
\hfill \\
= \frac{1}{4}\ln \left| {u + 1} \right| - \frac{1}{2}\frac{1}{{u + 1}} - \frac{1}{4}\ln \left| {u - 1} \right| - C \hfill \\
\hfill \\
= \frac{1}{4}\ln \left| {\frac{{u + 1}}{{u - 1}}} \right| - \frac{1}{2}\frac{1}{{u + 1}} + C \hfill \\
\hfill \\
Substitute\,\,back\,\,u = \sin t \hfill \\
\hfill \\
= \frac{1}{4}\ln \left| {\frac{{1 - \sin t}}{{1 + \sin t}}} \right| - \frac{1}{2}\frac{1}{{\sin t + 1}} + C \hfill \\
\end{gathered} \]
