The Theorem 1.5 and 1.6 of


Brown, Edgar H., Jr. The cohomology of BSOn and BOn with integer coefficients. Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288. 


give a general answer for $H^d(BSO\_n,Z)$ and $H^d(BO\_n,Z)$, which are very complicated. I wonder what are $H^d(BSO\_\infty,Z)$ and $H^d(BO\_\infty,Z)$ for $d=1,2,3,4,5$?



What are the cohomology groups $H^d(BSO_\infty,Z)$ and $H^d(BO_\infty,Z)$?
The Theorem 1.5 and 1.6 you quote give the answer.


More precisely, for $SO$, in the range $d<6$, the only polynomial generators
are $p\_1$ which has degree 4, $\delta(w\_2)$ with degree 3 and $\delta(w\_4)$ with degree 5. The only relations are $2\delta(w\_{2i})=0$, which gives $$H^d(BSO\_{\infty};\mathbb{Z})\cong 0,\quad (d=1,2),$$
$$H^d(BSO\_{\infty};\mathbb{Z})\cong \mathbb{Z}/2 ,\quad (d=3,5), $$
$$H^4(BSO\_{\infty};\mathbb{Z})\cong \mathbb{Z}.$$


In the case of $BO$, there are more generators $\delta(w\_1)$ and
 $\delta(w\_1w\_2)$ in degrees 2 and 4. Thus in degrees 4 and 5 we also have products $\delta(w\_1)^2$ and $\delta(w\_1)\delta(w\_2)$.


All of these lead to
$$H^1(BO\_{\infty};\mathbb{Z})\cong 0,$$
$$H^i(BO\_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$
$$H^4(BO\_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$
$$H^5(BO\_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$


