Which integral Steifel-Whitney classes are universally $0$? Let $BO(n)$ denote the classifying space of the orthogonal group $O(n)$. Then there is the well-known ring isomorphism 
$$H^*(BO(n);\mathbb{Z}/2) \cong \mathbb{Z}/2[w_1,\dots,w_n] $$
where $w_i \in H^i(BO(n);\mathbb{Z}/2)$ is the $i$-th universal Steifel-Whitney class.
From the short-exact sequence $\mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to \mathbb{Z}/2$ there is a natural Bockstein homomorphism $\beta\colon H^k(-;\mathbb{Z}/2) \to H^{k+1}(-;\mathbb{Z})$, which in particular has the property that $\beta(c) = 0$ iff $c$ is the mod-$2$ reduction of some integral class. Then we can define the integral Steifel-Whitney classes
$$ W_i = \beta(w_{i-1}) \in H^i(BO(n);\mathbb{Z}).$$
I haven't found much information about these in the usual sources other than the definition. My question is whether these are all non-zero, and if not whether there is a complete description of which ones vanish. In particular I would like to know if the universal $W_4$ is $0$.

This question is (tangentially) related to this other question about the Universal Coefficient Theorem, which led me to wonder if there is $2$-torsion in $H^4(BO(n);\mathbb{Z})$ for $n\geq 4$. I know that $H^*(BO(n);\mathbb{Z})$ consists of a free part generated by Pontryagin classes and a $2$-torsion part given by $im(\beta)$, but I can't determine whether $\beta(w_3)=0$ or not.
Note: I am aware of papers by Brown and Feshbach giving fairly explicit/complete descriptions of the ring $H^*(BO(n);\mathbb{Z})$, but I have only been able to find them on JSTOR and I don't have access.

Edit: An idea I had was to try to use the formula for $Sq^i(w_j)$ (for example here) and then I think it's true that $Sq^1 = (\text{reduction mod-}$2$)\circ\beta$. I computed $Sq^1(w_3) = w_1w_3 + w_4$, which is non-zero in $H^*(BO(n);\mathbb{Z})$ by algebraic independence of SW classes, but then this would mean $W_4 = \beta(w_3) \neq 0$. Is this argument valid?
 A: Let $\rho\colon H^*(-;\mathbb{Z}) \to H^*(-;\mathbb{Z}/2)$ be the natural "reduction mod-$2$" map induced by $\mathbb{Z} \to \mathbb{Z}/2$. Then we can use the fact that the first steenrod square $Sq^1$ is actually the composition $\rho \circ \beta$, in conjuction with the formula (linked in the question)
$$Sq^i(w_j) = \sum_{t=0}^i\binom{j-i-1+t}{t}w_{i-t}w_{j+t} $$
where $i < j$ (the Steenrod square vanishes if $j > i$ for degree reasons, and $Sq^j w_j = w_j^2$). In particular, let $i = 1$ and $j> 1$: then the formula simplifies considerably to 
$$Sq^1(w_j) = w_1w_j + (j-1)w_{j+1}. $$
Since the Steifel-Whitney classes are algebraically independent in $H^*(BO(n);\mathbb{Z}/2)$ it follows from this formula that $Sq^1(w_j)$ and hence $\beta(w_j)$ are non-zero for $j > 1$, therefore $W_j \neq 0$ for $j > 2$; moreover $Sq^1(w_1) = w_1^2 \neq 0$ so $W_2$ is non-zero as well. 
Edit: As Connor Malin observes in a comment the class $W_1$ is actually $0$, since the natural transformation $\beta\colon H^0(-;\mathbb{Z}/2) \to H^1(-;\mathbb{Z})$ is given by a map $\mathbb{Z}/2 \to K(\mathbb{Z},1)\simeq S^1$ which cannot be homotopically non-trivial. In other words it's not just $Sq^1\colon H^0(-;\mathbb{Z}/2) \to H^1(-;\mathbb{Z}/2)$ which is trivial in this degree, but $\beta$ itself.
