Answer
$$\frac{1}{\pi }{e^{\sec \pi t}} + C $$
Work Step by Step
$$\eqalign{
& \int {{e^{\sec \pi t}}\sec \pi t} \tan \pi tdt \cr
& {\text{set }}u = \sec \pi t{\text{ then }}\frac{{du}}{{dt}} = \sec \pi t\tan \pi t\left( \pi \right) \to \frac{{du}}{{\pi \sec \pi t\tan \pi t}} = dt \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{e^{\sec \pi t}}\sec \pi t} \tan \pi tdt = \int {{e^{\sec \pi t}}\sec \pi t} \tan \pi t\left( {\frac{{du}}{{\pi \sec \pi t\tan \pi t}}} \right) \cr
& {\text{cancel common terms}} \cr
& = \int {{e^u}\left( {\frac{{du}}{\pi }} \right)} \cr
& = \frac{1}{\pi }\int {{e^u}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{\pi }{e^u} + C \cr
& {\text{replace }}\sec \pi t{\text{ for }}u \cr
& = \frac{1}{\pi }{e^{\sec \pi t}} + C \cr} $$
