If a symplectic manifold $(M,\omega)$ is given,

is there any efficient method to compute $[\omega]$, the cohomology class of the symplectic form?

I do know that efficient is not a good word to use, but for instance: if one knows for sure that the symplectic form is integral -- the class of $\omega$ belongs to the image of $\check{H}^2(M;\mathbb{Z})\to H_{DR}^2(M;\mathbb{R})$ -- one could choose a set of orientable smooth surfaces $N$ representing the generators of $H_2(M)$ and compute $\int_N\omega$. The value of this integral determines $[\omega]$. However, not knowing ahead this information (but conjecturing that it is the case) implies that one must prove that all the periods have the same value (is that right?) and this seems a hard task.

When the symplectic manifold is a cotangent bundle with the canonical form, it is trivial to compute its cohomology class. This is also the case for a exact symplectic manifold, when one knows a potential 1-form for the symplectic form.

Another "easy" case is $(\mathbb{C}P^{n}, \omega_{FS})$. The only problem is that I do not know how to prove that the Fubini-Study form come from $\mathcal{O}(1)$. I have seen this stated everywhere but no proof was provided.

Bonus question: can someone give me references for that result about the Fubini-Study form?

There are at least two, related, motivations for this question:

1) Given a symplectic manifold, does it admit a hermitian line bundle with connection such that its curvature is proportional to the symplectic form?

2) Given a closed $2n$-dimensional symplectic manifold, is it possible to find a symplectic embedding of it in $(\mathbb{C}P^{2n+1}, \omega_{FS})$ ?

They are both related because if the symplectic form is integral, then there exists an affirmative answer for them (Kostant, Weil and Tischler).

It would be also interesting to know if

is there any kind of classification (or attempts to classify) integral symplectic manifolds?

This is maybe too much, since to my knowledge not even exact symplectic manifolds are classified.

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