The induced metric on a sphere may be given by,
$$ds^2 = d\theta^2 + \sin^2\theta\, d\phi^2$$
By using Cartan's method of moving frames, one can compute the curvature 2-form in an orthonormal basis ($e^\theta = d\theta$ and $e^\phi = \sin\theta \, d\phi$), that is,$^\dagger$
$$\Omega = \left( \begin{array}{cc} 0 & \sin\theta \\ -\sin\theta & 0 \end{array} \right) (d\theta \wedge d\phi)$$
Now, the second Chern class of the tangent bundle is given by,
$$c_2 = \frac{1}{8\pi^2} (\mathrm{Tr} \Omega^2 - \mathrm{Tr}^2 \Omega) = -\frac{\sin^2 \theta}{4\pi^2} (d\theta \wedge d\phi)^2$$
What disturbs me is the factor of $(d\theta \wedge d\phi)^2$ which arises from the matrix multiplication to obtain $\Omega^2$. How does it make sense to integrate,
$$\int_{S^2}c_2 \sim \int_{S^2} \sin^2 \theta \, (d\theta \wedge d\phi)^2$$
Are we suppose to read this as $(\dots)^2 = (\dots) \wedge (\dots)$? Also, isn't $c_2$ a $4$-form then? I feel this may be a silly question, and I'm just misinterpreting something.
$\dagger$ From Cartan's second equation, $\Omega^a_b = d\omega^a_b + \omega^a_c \wedge \omega^c_b$, with $\omega^a_b$ the spin connection. I double checked the curvature form to make sure by plugging it into $\int_{S^2} \mathrm{Pf}[\Omega]$ to get the right Euler characteristic for a sphere.
The $n$th Chern class $c_n$ is a $2n$-form, in general, so yes, $c_2$ is a $4$-form. And yes, $x^2 = x \wedge x$ when $x$ is a differential form. Here it's simple, there is no nonzero $4$-form on $S^2$, so $c_2$ is simply zero.