Answer
\[ = \frac{1}{{\cos \,t}} + \frac{{{{\tan }^2}t}}{2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{\sin t + \tan t}}{{{{\cos }^2}t}}} dt \hfill \\
\hfill \\
split\,\;the{\text{ integrand}} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{{\sin t}}{{{{\cos }^2}t}} + \frac{{\tan \,t}}{{{{\cos }^2}t}}} \right)dt} \hfill \\
\hfill \\
for = \int_{}^{} {\frac{{\sin t}}{{{{\cos }^2}t}}dt} \hfill \\
\hfill \\
set\,\,\cos t = u\,\,\, \to \,\, - \sin t\,dt = du \hfill \\
\hfill \\
\int_{}^{} {\frac{{\sin t}}{{{{\cos }^2}t}}dt} = - \int_{}^{} {\frac{{dt}}{{{t^2}}} = \frac{1}{t} + {c_1} = \frac{1}{{\cos x}} + {c_1}} \hfill \\
\hfill \\
for = \int {\frac{{\tan \,t}}{{{{\cos }^2}t}} = } \int_{}^{} {\tan t{{\sec }^2}\,tdt} \hfill \\
\hfill \\
set\,\,\tan \,t = v\,\, \to \,\,{\sec ^2}tdt = dv \hfill \\
\hfill \\
\int {\frac{{\tan \,t}}{{{{\cos }^2}t}} = } \int_{}^{} {vdv} = \frac{{{v^2}}}{2} + {c_2} = \frac{{{{\tan }^2}t}}{2} + {c_2} \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{{\sin t}}{{{{\cos }^2}t}} + \frac{{\tan \,t}}{{{{\cos }^2}t}}} \right)dt} = \frac{1}{{\cos \,t}} + \frac{{{{\tan }^2}t}}{2} + C \hfill \\
\end{gathered} \]
