Yes. We have
\begin{align}\int {\sin(\pi t)}^2 \,dt
&=\int \underbrace{\sin(\pi t)}_f\cdot \underbrace{\sin(\pi t)}_{g'} \,dt\\
&= \underbrace{\sin(\pi t)}_f\cdot\underbrace{\left(\frac{-\cos(\pi t)}{\pi}\right)}_g-
\int \underbrace{\pi\cos(\pi t)}_{f'}\cdot \underbrace{\left(\frac{-\cos(\pi t)}{\pi}\right)}_g \,dt\\
&= -\frac1\pi\sin(\pi t)\cdot\cos(\pi t)
+\int {\cos(\pi t)}^2 \,dt\\
&= -\frac1\pi\sin(\pi t)\cdot\cos(\pi t)
+\int 1-{\sin(\pi t)}^2 \,dt\\
&= -\frac1\pi\sin(\pi t)\cdot\cos(\pi t)
+t-\int {\sin(\pi t)}^2 \,dt
\end{align}
From which we find that
$$\int {\sin(\pi t)}^2 \,dt=\frac{t}2-\frac{\sin(\pi t)\cdot\cos(\pi t)}{2\pi},$$
up to a constant.