How to show that $(\Lambda^2(g))^g = H^2(g)$?

Question

Let $g$ be a semisimple Lie algebra and $\Lambda^2(g) = g \wedge g \subset g \otimes g$ the exterior square of $g$. Consider the adjoint action of g on $g \wedge g$ and let $$(\Lambda^2(g))^g = \{x \in g: x.(a \wedge b) = [x,a] \wedge b + a \wedge [x,b] = 0, \forall x \in g\}$$ be the $g$-invariants of $(\Lambda^2(g))^g$ under the adjoint action. How to show that $(\Lambda^2(g))^g = H^2(g)$? Thank you very much.