$W(t)$ is Brownian a motion. I proved the title using Itô lemma for the product of two Itô processes.
Let $Y(t) = \int_0^t W(s)ds$
$W(t)Y(t) = \int_0^t W(s)dY(s) + \int_0^t Y(s) dW(s) + \int_0^t dW(t)dY(t) = \int_0^t W^2(s)ds + \int_0^t Y(s) dW(s) + 0$
Since the second term is a martingale, taking expectation on both side I obtain :
$\mathbb{E}[W(t)\int_0^t W(s)ds] = \mathbb{E}[\int_0^t W^2(s)ds] = \int_0^t\mathbb{E}[W^2(s)]ds = \frac{t^2}{2}$
Is everything ok in my algebra ? Do you see a less brutal way to compute $\mathbb{E}[W(t)\int_0^t W(s)ds]$ ?