= Solution
With relativistically normalized one-particle states, <Fermi's golden rule> becomes
$$
\Gamma=\frac1{2M}\int |\mathcal M|^2(2\pi)^4\delta^{(4)}(P-p_1-p_2)
\frac{d^3p_1}{(2\pi)^3 2E_1}\frac{d^3p_2}{(2\pi)^3 2E_2}.
$$
In the parent <rest frame>, the three-momentum <Dirac delta function>[delta function] fixes $\mathbf p_2=-\mathbf p_1$. The <energy> <Dirac delta function>[delta function] fixes the common magnitude $p^*$; its radial <Jacobian determinant>[Jacobian] converts the remaining <Lorentz-invariant phase space>[phase-space] factor into
$$
\boxed{\Gamma=\frac{p^*}{32\pi^2M^2}\int|\mathcal M|^2\,d\Omega.}
$$
Only one <momentum> magnitude remains because the two final <momenta> are constrained, not independent. Explicitly,
$$
p^*=\frac{\sqrt{[M^2-(m_1+m_2)^2][M^2-(m_1-m_2)^2]}}{2M}.
$$
For leptonic <W boson> decay both final <masses> are small compared with $m_W$, including the <Tau lepton>[tau] <mass> at the precision requested, so $p^*\simeq m_W/2$.
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