= Solution
Put every <formal pseudodifferential operator> in normal order, with all <coefficient> <functions> on the left. Its <formal pseudodifferential residue> is the <coefficient> of $\partial^{-1}$. The <Adler trace> is
$$
\boxed{\operatorname{Tr}P=\int\operatorname{res}P\,dx.}
$$
Here and below, use periodic <coefficients> and integrate over a period, or impose convergence and vanishing boundary terms on the real line. Equivalently, the algebraic integration <functional> must annihilate total <derivatives>. This convention is needed for the requested trace property; arbitrary elements of $C^\infty(\mathbb R)$ do not supply it automatically.
It suffices to compute with monomials $A=a\partial^p$, $B=b\partial^q$. Write $r=p+q+1$. If $r<0$, no term in either product has power $\partial^{-1}$, so both residues are zero. If $r\geq0$, the <formal pseudodifferential composition rule> gives
$$
\operatorname{res}(AB)=\binom pr a b^{(r)},\qquad
\operatorname{res}(BA)=\binom qr b a^{(r)}.
$$
Since $q=r-1-p$, reversal of the factors in the numerator proves
$$
\binom qr=\frac{(r-1-p)\cdots(-p)}{r!}=(-1)^r\binom pr.
$$
After $r$ integrations by parts, $\int ab^{(r)}dx=(-1)^r\int ba^{(r)}dx$. Hence the two product traces agree. More explicitly, for $r\geq1$ their residue difference is the total <derivative>
$$
\operatorname{res}[A,B]
=\binom pr\frac{d}{dx}\left(\sum_{j=0}^{r-1}(-1)^j a^{(j)}b^{(r-1-j)}\right).
$$
For $r=0$ the residue difference is $ab-ba=0$. In products of operators bounded above in order, only finitely many monomial pairs can contribute to a residue: the required $r$ is nonnegative, so the two downward <coefficient> indices have a bounded sum. Summing the monomial result therefore proves
$$
\boxed{\operatorname{Tr}[P_1,P_2]=0,\qquad
\operatorname{Tr}(P_1P_2)=\operatorname{Tr}(P_2P_1).}
$$
The boundary convention is substantive. Without it, take $A=\partial^2$, $B=(\tanh x)\partial^{-2}$. Both individual residues vanish, but $\operatorname{res}[A,B]=2(\tanh x)'=2\operatorname{sech}^2x$ and its integral over $\mathbb R$ is $4$, not zero. The primitive has unequal limits at the two ends. Also a generic smooth residue need not have a convergent integral at all. Thus the proof supplies the intended cyclic trace under its necessary analytic or formal integration convention.
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