Answer
$\dfrac{1}{\pi }e^{\sec \pi t} +c$
Work Step by Step
Solve $\int e^{\sec \pi t} \sec \pi t \tan \pi tdt$
Let us $p=\sec \pi t$ and $dp=(\sec \pi t \tan \pi t) \pi dt$
This implies $\int e^{\sec \pi t} \sec \pi t \tan \pi tdt=\dfrac{1}{\pi}\int e^p dp$
$=\dfrac{1}{\pi} e^p +c$
$=\dfrac{1}{\pi }e^{\sec \pi t} +c$
Hence,$\int e^{\sec \pi t} \sec \pi t \tan \pi tdt=\dfrac{1}{\pi }e^{\sec \pi t} +c$
