Let $\Omega\subset \mathbb{R}^n$ and consider the de Rham cohomology group $H_{dr}^p(\Omega)$ which is the quotient vector space defined by $$H_{dr}^p(\Omega) = \dfrac{\{p\text{-forms on } \Omega \text{ which are smooth and closed}\}}{\{p\text{-forms on } \Omega \text{ which are smooth and exact}\}}$$
Let $\alpha$ and $\beta$ be a closed $p$-form and a closed $q$-form respectively. Denote by $\{\alpha\}$ and $\{\beta\}$ the classes of $\alpha$ and $\beta$ in $H_{dR}^p(\Omega)$ and $H_{dR}^q(\Omega)$. Define the cup product $\smile$ by $\{\alpha\}\smile\{\beta\}= \{\alpha \wedge \beta\}$.
I think it is pretty natural to believe that $\smile$ is not commutative, since $\wedge$ is not commutative. However, I do not know of an explicit example to show this. (I am aware of certain forms that are closed but not exact, such as checking if a 2-form is exact)
I guess what I would like to find are two closed forms $\alpha$ and $\beta$ such that $\alpha \wedge \beta$ is not exact. Unfortunately this seems out of reach. Am I missing something completely obvious?
Actually, the wedge product $\wedge$ is graded-commutative, and therefore $\smile$ is also graded-commutative. Graded-commutative means that if $\alpha$ is a $p$-form and $\beta$ is a $q$-form, then $\alpha \smile \beta = (-1)^{pq} \beta \smile \alpha$.
For an example of failure of commutativity, consider the "torus" $$T = \{(x,y,z,t) \in \mathbb{R}^4 \mid 1 < x^2 + y^2 < 2 \text{ and } 1 < z^2 + t^2 < 2 \}.$$ Let $\alpha = x \, dy - y \, dx$ and $\beta = z \, dt - t \, dz$. Both are closed $1$-forms that are not exact, and you have $\alpha \wedge \beta = - \beta \wedge \alpha$.