find the distribution of $X$ such that $X(\omega) = \int_{0}^{1} B_s^2(\omega) ds$ where $(B_t)_{t \geq0}$ is standard Brownian motion

Question

since for a fixed $\omega$, $B_s(\omega)$ is continuous in $s$ then $B^2_s(\omega)$ is Riemann integrable w.r.t. $s$ :

$X(\omega) = \int_{0}^{1} B_s^2(\omega) ds \implies X = \lim_{n \to \infty} \frac{1}{n}\sum_{k = 1}^{n} B^2_{\frac{k}{n}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k = 1}^{n} Y_k$

I would like to use the strong law of large numbers but I don't have independence of the $Y_i$'s.

I know for one that for $t \geq s$, the increments $B_t - B_s$ are independent of $B_s$, maybe we can somehow use this property here ?