Answer
$\frac 1 2 ln|cos(\frac 2 t)|+C$
Work Step by Step
$Let u=cos(\frac 2 t), du=\frac {2sin(\frac 2 t)}{t^2}dt$
$\int \frac{tan\frac2 t}{t^2}dt$
$\frac1 2 \int \frac 1{cos(\frac 2 t)}[\frac {2sin(\frac 2 t)}{t^2}]dt$
$\frac 1 2 ln|cos(\frac 2 t)|+C$