Integrability of Symplectic structures A symplectic structure on even dimensional manifold is a non-degenerate closed two form and I understood integrability of symplectic structure is closedness as a differential 2-form which comes from involutivity of symplectic vector fields by Frobenius theorem. However, in my calculation of Lie derivative of semi symplectic form $ \omega$ which means merely non-degenerate 2-form with Lie bracket $[X, Y]$ of two symplectic vector fields $X$ and $Y$ is zero without d closed condition.
My question is that did I misunderstand of the notion of integrability of symplectic structures in the sense of Frobenius, or mistake in the following calculation?
Assume that $0=\mathcal{L}_X\omega, \ 0= \mathcal{L}_Y\omega$, since $X$ and $Y$ are symplectic.
We now compute $\mathcal{L}_{[X, Y]}\omega$ using a formula $\mathcal{L}_{[X, Y]}=\mathcal{L}_X \mathcal{L}_Y -\mathcal{L}_Y\mathcal{L}_X$.
\begin{align}
\mathcal{L}_{[X, Y]}\omega & = (\mathcal{L}_X \mathcal{L}_Y -\mathcal{L}_Y\mathcal{L}_X)\omega \\
& = 0. \\
\end{align}
Now $\mathcal{L}_{[X, Y]}\omega$ vanished, it implies that $[X, Y]$ is also symplectic without using d-closed condition.
How should I use d-closed condition to confirm integrability of symplectic structures?

So the notion of Integrability on a Symplectic manifold is by definition the following:
A symplectic structure on a manifold M is a differential $2$-form $\omega$ satisfying two conditions:

*

*$\omega$ is non-degenerate, i.e. for each $p \in M$ and tangent vector $\tilde{u}$ based at $p$, if $\omega_p(\tilde{u},\tilde{v}) = 0$ for all tangent vectors $\tilde{v}$ based at $p$, then $\tilde{u}$ is the zero vector;

*$\omega$ is closed, i.e. the exterior derivative of $\omega$ is zero, i.e. $\mathrm{d}\omega = 0$.

Condition (2) is often noted as the integrability condition.
