Answer
$\int sin~\pi t~cos~\pi t~dt = \frac{1}{2\pi}~(sin~\pi t)^2+C$
Work Step by Step
$\int sin~\pi t~cos~\pi t~dt$
Let $u = sin~\pi t$
$\frac{du}{dt} = \pi~cos~\pi t$
$dt = \frac{du}{\pi~cos~\pi t}$
$\int (u)(cos~\pi t)~(\frac{du}{\pi~cos~\pi t})$
$=\int \frac{1}{\pi}~u~du$
$=\frac{1}{2\pi}~u^2+C$
$=\frac{1}{2\pi}~(sin~\pi t)^2+C$