Calculating the cohomology $H^*(\mathfrak{so}_n(\mathbb{R}); \mathbb{R})$ of $\mathfrak{so}_n(\mathbb{R})$ directly.

Question

I want to be able to calculate the Lie algebra cohomology of $\mathfrak{so}_n(\mathbb{R})$ without appealing to the cohomology of Lie groups. Before I ask this, I do want to make sure this is a reasonable thing to do. My understanding is just as working with Lie algebras often simplifies the calculations of Lie groups, this would be the case for cohomologies as well. But despite that, I struggle to find much online material. Is my idea accurate? Also, as a convention, I will drop the $(-; \mathbb{R})$ as that is implied, and similarly for $\mathfrak{so}_n(\mathbb{R}) = \mathfrak{so}_n$

Given that it is, I've tried to calculate the cohomologies, but I've been struggling with anything beyond the first one. So by definition, we have the cochains are defined as

$$C^k(\mathfrak{so}_n) = \mathrm{Hom}(\wedge^k \mathfrak{so}_n) = \left(\wedge^k \mathfrak{so}_n\right)^*.$$

So by definition, we have $C^0(\mathfrak{so}_n) = \mathbb{R}$, $C^1(\mathfrak{so}_n) = (\mathfrak{so}_n)^*$, and so on. Now, in the case of the trivial module, our map $d: C^k(\mathfrak{so}_n) \to C^{k + 1}(\mathfrak{so}_n)$ is given by

$$d\varphi(x_0, \dots, x_k) =\sum_{0 \leq i < j \leq k} (-1)^{i + j} c([x_i, x_j], x_0, \dots, \hat{x_i}, \dots, \hat{x_j}, \dots, x_k).$$

So for $k = 0$, we have the empty sum and so

$$Z^0(\mathfrak{so}_n) = \mathrm{\ker}(d: C^0(\mathfrak{so}_n) \to C^1(\mathfrak{so}_n)) = \mathbb{R}$$

and

$$B^0(\mathfrak{so}_n) = \mathrm{im}(d: C^{-1}(\mathfrak{so}_n) \to C^0(\mathfrak{so}_n)) = 0,$$

giving $H^0(\mathfrak{so}_n) = \mathbb{R}$. Now, for $H^1(\mathfrak{so}_n)$, it is immediate that $B^1(\mathfrak{so}_n) = 0$. Now, if $\varphi \in C^1(\mathfrak{so}_n)$, we have

$$d \varphi(x_0, x_1) = -\varphi([x_0, x_1]) = \varphi(x_1 x_0) - \varphi(x_0 x_1).$$

I don't really know when I can conclude $\varphi(x_0 x_1) = \varphi(x_1 x_0)$, so I'm not sure what the kernel is. Similarly, I run into problems calculating any higher cohomology groups. I assume it can't be that hard, but I can't find any resources on it. If I could be either given an answer on this, or pointed in the direction of a resource that explicitly calculates $\mathfrak{so}_n$, that would be great. Even some general theorems (based in Lie algebras and not Lie groups) would be helpful. Thanks!

EDIT: I can see that using Whitehead's lemma that $H^1(\mathfrak{so}_n) = H^2(\mathfrak{so}_n) = 0$, but computing anything higher I'm still confused on.