Using the property: $\displaystyle\int_0^{\pi} x \cdot f(\sin x) \, dx = \frac{\pi}{2}\int_0^{\pi} f(\sin x) \, dx$
Here $f(\sin x) = \sin^2(x)\cos(x)$. However, $\cos(x)$ changes sign, so we split:
$$I = \frac{\pi}{2}\int_0^{\pi} \sin^2(x)\cos(x) \, dx$$
Let $u = \sin(x)$, $du = \cos(x)\,dx$. From $0$ to $\pi$, $\sin$ goes $0 \to 0$, so $I = 0$.
$$\boxed{I = 0}$$