Answer
$$\frac{1}{2\pi}\sin^2(\pi t)+c$$
Work Step by Step
Given
$$\int \sin( \pi t)\cos( \pi t) dt$$
Let $x= \pi t\ \Rightarrow \ dx= \pi dt $ then
\begin{align*}
\int \sin( \pi t)\cos( \pi t) dt&=\frac{1}{ \pi}\int \sin(x)\cos(x) dx\\
&=\frac{1}{2\pi}\sin^2 x+c\\
&=\frac{1}{2\pi}\sin^2(\pi t)+c
\end{align*}
