Let \( f(x) \) be a function satisfying \( f(x) + f(\pi - x) = \pi^2, \forall x \in \mathbb{R} \). Then \(\displaystyle \int_{0}^{\pi} f(x)\sin x \, dx \) is equal to:
\( \pi^2 \)
\( \dfrac{\pi^2}{4} \)
\( \dfrac{\pi^2}{2} \)
\( 2\pi^2 \)