Let $X$ be a smooth proper algebraic variety over $\mathbb{C}$.

I know that in the analytic world, there is an isomorphism between the de Rham cohomology and the cohomology of the constant sheaf $\underline{\mathbb{C}}$: $$ H^i_{dR}(X^{an}/\mathbb{C}) \cong H^i(X^{an}, \underline{\mathbb{C}}). $$

The proof of this relies on the fact that the sequence of sheaves on $X^{an}$ $$ 0 \to \underline{\mathbb{C}} \to \Omega^0_{X^{an}} \to \Omega^1_{X^{an}} \to \Omega^2_{X^{an}} \to \cdots $$ is exact.

There is also an isomorphism between $H^i_{dR}(X^{an}/\mathbb{C})$ and $H^i_{dR}(X^{Zar}/\mathbb{C})$. I assume that this also works for $H^i_{dR}(X^{et}/\mathbb{C})$.

However $H^i(X^{Zar}, \underline{\mathbb{C}}) = 0$ because a constant Zariski sheaf is flabby.

What is $H^i(X^{et}, \underline{\mathbb{C}})$?

So far as I can see, neither of the above arguments to calculate this work: a constant étale sheaf is not flabby, because étale open sets need not be connected (actually I am not sure if the definition of flabby sheaves in the Zariski topology makes sense for the étale site).

And the sequence $$ 0 \to \underline{\mathbb{C}} \to \Omega^0_{X^{et}} \to \Omega^1_{X^{et}} \to \Omega^2_{X^{et}} \to \cdots $$ is not exact because, if $X=\mathbb{P}^1$, its stalks at the origin are $$ 0 \to \mathbb{C} \to R \to R \, dX \to 0 $$ where $R$ is the strict Henselization of $\mathbb{C}[X]_{(X)}$. $R$ is algebraic over $\mathbb{C}[X]$ so does not contain $\log(1+X)$ and hence the 1-form $dX/(1+X)$ is not exact.