Integrals of forms are equal implies they differ by $d\mu$

The problem is about the proof of the following result.

If $\omega_1,\omega_2 \in \Omega^n_c(X)$ (where $X$ is smooth manifold) are such that $\int_X\omega_1=\int_X \omega_2$ then there is $\mu\in\Omega^{n-1}_c(X)$ such that $\omega_1-\omega_2=d\mu$.

I am using text at link. Part where I am stuck is on numerical page 202 which is page 30 in the file. It says that using partition of unity argument in Step 4 we can assume $\omega_1,\omega_2$ are supported in some parametrizable open sets.

Can someone explain to me why is this true?

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Note that $\omega\in \Omega_c^n(X)$ means $\omega$ is a $\mathcal{C}^\infty$ $n$-form, compactly supported on the manifold $X$. The notation is introduced on page 106 of the document. Partitions of unity are discussed starting on page 188.
Regarding the beginning of step 4: By assumption, $\int_X \omega_i = \int_{U_i}\omega_i = 0$. Step 1 tells us that $\omega_i = d\mu_i$ for $\mu_i \in \Omega_c^{n-1}(U_i)$. This implies $\omega_1\sim \omega_2$.
Another way to look at this proof is to ignore the argument regarding the special case $c=0$. Assume $\int_X\omega_1 = \int_X\omega_2 = c$ where $c$ can be zero. Don't divide by $c$. Go through step 5 but instead choose $\int_X \alpha_i = c$. The rest of step 5 goes through as written, giving $(a)\Rightarrow (b)$.
I understand the notation, the problem is if we have $\int_X \omega_1=\int_X \omega_2$ and $\rho$ is any function from the partition of the unity we do not know that $\int_X \rho \omega_1=\int_X\rho\omega_2$? I feel they used this in the proof. Is there any explanation for this? – dmm Jul 29 '12 at 8:27