Clarify derivation: Why is the last term of (5.37) in Modern Cosmology by Dodelson and Schmidt zero? I'm going through Modern Cosmology by Dodelson and Schmidt 2nd edition, and I'm stuck at (5.37) where they present the Boltzmann equation for dark matter and take the integral of all the terms over momentum. The result is,
$$
\begin{gathered}
\frac{\partial}{\partial t} \int \frac{d^{3} p}{(2 \pi)^{3}} f_{\mathrm{c}}+\frac{1}{a} \frac{\partial}{\partial x^{i}} \int \frac{d^{3} p}{(2 \pi)^{3}} f_{\mathrm{c}} \frac{p\, \hat{p}^{i}}{E(p)}-[H+\dot{\Phi}] \int \frac{d^{3} p}{(2 \pi)^{3}} p \frac{\partial f_{\mathrm{c}}}{\partial p} \\
-\frac{1}{a} \frac{\partial \Psi}{\partial x^{i}} \int \frac{d^{3} p}{(2 \pi)^{3}} \frac{\partial f_{\mathrm{c}}}{\partial p} E(p) \hat{p}^{i}=0 .
\end{gathered}
$$
Now they say,

Integration by parts shows that the last term vanishes. The remainder of the terms are all relevant, though.

I don't see how this is true. How can we get rid of the last term using integration by parts?
Attempt
What is commonly done is that we pass to spherical coordinates of the integral with the (momentum) volume element being $d^{3}p = p^{2}\sin\theta\, dp\, d\theta\, d\phi = p^{2}\, dp\, d\Omega$. Then we get
\begin{align*}
\int \frac{d^{3}p}{(2\pi)^{3}} \frac{\partial f_{c}}{\partial p} E(p) \hat{p}^{i} &= \frac{1}{(2\pi)^{3}}\int_{0}^{\infty} dp\, p^{2} E\, \frac{\partial}{\partial p}\int f_{c}\hat{p}^{i} \,d\Omega \\[1.6ex]
&= (\text{boundary term that goes to zero}) \\[1.6ex]
&\qquad - \frac{1}{(2\pi)^{3}}\int_{0}^{\infty} dp\, \left(2pE(p) + p^{2}\frac{\partial E}{\partial p}\right) \int f_{c}\hat{p}^{i} \,d\Omega
\end{align*}
I don't see how this could possibly be zero, nor do I see any other way to perform integration by parts that is fruitful.
 A: Going through Modern Cosmology by Dodelson and Schmidt 1st edition helped tremendously. In Chapter 4, we have an analogous equation (4.69). In the paragraph afterwards the book says,

The last term here can be neglected since the integral over the direction vector is nonzero only for the perturbed part of $f_{dm}$. Thus the integral is first order and it multiplies the first-order term $\partial\Psi/\partial x^{i}$. The rest of the terms are all relevant, though.

To explain this, we can take $f_{c} = f_{\text{zeroth-order}} + f_{\text{perturbation}}$, and the point is that $f_{\text{zeroth-order}}$ is isotropic (as has been assumed in previous sections). Thus, the angular integral $\int f_{c}\hat{p}^{i}\, d\Omega$ is zero for the zeroth-order part and nonzero for the perturbation. This means the integral
\begin{align*}
\int \frac{d^{3}p}{(2\pi)^{3}} \frac{\partial f_{c}}{\partial p} E(p) \hat{p}^{i} &= - \frac{1}{(2\pi)^{3}}\int_{0}^{\infty} dp\, \left(2pE(p) + p^{2}\frac{\partial E}{\partial p}\right) \int f_{c}\hat{p}^{i} \,d\Omega
\end{align*}
is a first-order term. Plugging this back into (5.37), the last term of (5.37) is
$$ -\frac{1}{a} \underbrace{ \frac{\partial \Psi}{\partial x^{i}} }_{\text{first-order}} \underbrace{ \int \frac{d^{3}p}{(2\pi)^{3}} \frac{\partial f_{c}}{\partial p} E(p) \hat{p}^{i} }_{\text{first-order}} . $$
We have a first-order factor times a first-order factor, giving us a second-order term. We are interested in zeroth-order and first-order terms only, so we neglect this term altogether.
Sadly, the second edition of the book does a worse job at explaining/clarifying this than the first edition.
