= Solution
Since $f(0)=0$, the <fundamental theorem of calculus> gives $f=Tf'$, with the same oriented integral on the negative half-interval. Set $h(t)=t^2$. The adjoint calculation gives the real-valued function
$$
\boxed{g(t)=(T^*h)(t)=\begin{cases}(\pi^3-t^3)/3,&0<t\leq\pi,\\-(\pi^3+t^3)/3,&-\pi<t<0.\end{cases}}
$$
Choose any value at zero. This function is bounded and hence belongs to $L^2(-\pi,\pi]$. Since it is real, the <adjoint operator> identity yields
$$
\int_{-\pi}^{\pi}t^2f(t)dt=\langle Tf',h\rangle=\langle f',T^*h\rangle=\int_{-\pi}^{\pi}g(t)f'(t)dt.
$$
One can also check this by <integration by parts> separately on each half-interval: $g'=-t^2$, $g(\pm\pi)=0$, and the two possible jump terms at zero vanish because $f(0)=0$.
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