Proof of the General Stokes Theorem in Munkres In "Analysis on manifolds" Munkres proves the general Stokes theorem
$ \int_{\partial M}\omega = \int_Md\omega $ in the case where the support of $ \omega$ can be covered by a single coordinate patch and says the general case can be easily proved from that. However, I have found myself having quite a bit of trouble with this.
If  $ \{ \phi_i \} $ is a partition of unity on M, we have $ \int_M\eta = \sum_i\int_M\phi_i\eta $ by definition.
So, for $ \omega $, I compute:
$ \int_Md\omega = \sum_i\int_M\phi_id\omega = \sum_i\int_{U_i}\alpha_i^*(\phi_id\omega) = \sum_i\int_{U_i}(\phi_i \circ \alpha_i)d\alpha_i^*\omega$, where $ \alpha_i:U_i \to V_i $ is a coordinate patch on M with the support of $ \phi_i $ contained in $ V_i $.
On the other hand, $ \int_{\partial M}\omega = \sum_i\int_{\partial M}\phi_i\omega = \sum_i\int_Md(\phi_i\omega) = \sum_i\int_{U_i}d(\alpha_i^*(\phi_i\omega)) = \sum_i\int_{U_i}d((\phi_i \circ \alpha_i)(\alpha_i^*\omega)) $, which looks quite similar to the previous expression, but not really equal to it. Can I get some suggestions about what I'm missing here?
Edit: Ok, now I'm aware of the simple proof
 $ \omega = \phi_1\omega + ... + \phi_k\omega$
$ \int_{M}d\omega = \int_Md(\sum_i\phi_i\omega) = \sum_i\int_Md(\phi_i\omega) = \sum_i\int_{\partial M}\phi_i\omega = \int_{\partial M}\omega $.
But I'd still like to know where I went wrong in my initial attempt.
 A: Your attempt that led to
$$\int_{\partial M}\omega = \sum_i\int_{\partial M}\phi_i\omega = \sum_i\int_Md(\phi_i\omega) = \sum_i\int_{U_i}d(\alpha_i^*(\phi_i\omega)) = \sum_i\int_{U_i}d((\phi_i \circ \alpha_i)(\alpha_i^*\omega))$$
isn't wrong, you just didn't take it far enough.
You have $\alpha_i^\ast\left(d\phi_i\omega\right)$ - it's in my opinion nicer to have the pull-back outside here. Then you differentiate to get
$$\alpha_i^\ast\left(d\phi_i\wedge\omega + \phi_id\omega \right) = \alpha_i^\ast(d\phi_i\wedge\omega) + \alpha_i^\ast \phi_id\omega.$$
Now the second part is exactly what you want, and all that remains is that the first,
$$\alpha_i^\ast (d\phi_i \wedge\omega),$$
sums to $0$ because we have $\sum\limits_i \phi_i \equiv 1$.

The proof usually goes
$$\begin{align}
\int_{\partial M} \omega &= \int_{\partial M} \sum_i \phi_i\omega\\
&= \sum_i \int_{\partial M} \phi_i\omega\\
&= \sum_i \int_M d(\phi_i\omega)\\
&= \sum_i \int_M (d\phi_i \wedge \omega + \phi_id\omega)\\
&= \sum_i \int_M d\phi_i \wedge \omega + \sum_i \int_M \phi_id\omega\\
&= \int_M \left(\sum_i d\phi_i\right)\wedge \omega + \int_M \left(\sum_i\phi_i\right)d\omega\\
&= \int_M d\left(\sum_i \phi_i\right)\wedge\omega + \int_M d\omega\\
&= \int_M 0\wedge\omega + \int_M d\omega\\
&= \int_M d\omega.
\end{align}$$
