= Solution
For a localized charge and current distribution, take the retarded vector potential
$$
A(x,t)=\frac{\mu_0}{4\pi}
\int\frac{J(x',t-|x-x'|/c)}{|x-x'|}\,d^3x'.
$$
Let $R=|x|$, $n=x/R$, and $t_R=t-R/c$. If the source size is much smaller than both $R$ and the characteristic radiation wavelength, the leading dipole term is
$$
A(x,t)\simeq\frac{\mu_0}{4\pi R}\int J(x',t_R)\,d^3x'
=\frac{\mu_0}{4\pi R}\dot p(t_R),
$$
where the last identity follows from charge conservation.
Keeping only the $1/R$ terms in the <radiation zone> gives the transverse fields
$$
E_{\rm rad}
=\frac{\mu_0}{4\pi R}n\times(n\times\ddot p(t_R)),
\qquad
B_{\rm rad}=\frac1c n\times E_{\rm rad}.
$$
The radial <Poynting vector> therefore yields
$$
\frac{d\mathcal P}{d\Omega}
=\frac{\mu_0}{16\pi^2c}
|n\times\ddot p(t_R)|^2.
$$
Using
$$
\int n_i n_j\,d\Omega=\frac{4\pi}{3}\delta_{ij}
$$
or equivalently $\int\sin^2\theta\,d\Omega=8\pi/3$, we recover the <electric dipole radiation> formula
$$
\boxed{\mathcal P\simeq
\frac{\mu_0}{6\pi c}|\ddot p(t-R/c)|^2}.
$$
This requires a localized, nonrelativistic source of size $d\ll\lambda$, observation distance $R\gg d$ and $R\gg\lambda$, and neglect of higher multipoles and faster-decaying near fields.
For the pulsar, write its magnetic dipole moment as $m(t)$ to distinguish it from the electric dipole above. The <magnetic dipole radiation> formula differs by a factor $c^{-2}$:
$$
\mathcal P=\frac{\mu_0}{6\pi c^3}|\ddot m|^2.
$$
Here
$$
\ddot m(t)=-p_0\Omega^2\sin\alpha
[\cos(\Omega t)\widehat x+\sin(\Omega t)\widehat y],
$$
and hence
$$
\boxed{\mathcal P=C\Omega^4,
\qquad
C=\frac{\mu_0p_0^2\sin^2\alpha}{6\pi c^3}}.
$$
On the slow spin-down timescale, energy conservation gives the <magnetic-dipole spin-down> equation
$$
I\Omega\dot\Omega=-C\Omega^4,
$$
so
$$
\frac1{\Omega(t)^2}=\frac1{\Omega_0^2}+\frac{2C}{I}t.
$$
Half of the initial rotational energy remains when $\Omega=\Omega_0/\sqrt2$. Therefore
$$
t_{1/2}=\frac{I}{2C\Omega_0^2}
=\boxed{\frac{6\pi M\mathcal R^2c^3}
{5\mu_0p_0^2\Omega_0^2\sin^2\alpha}},
$$
where $I=2M\mathcal R^2/5$. Since $E_0=I\Omega_0^2/2$, the equivalent expression is
$$
\boxed{t_{1/2}=\frac{6\pi M^2\mathcal R^4c^3}
{25\mu_0p_0^2E_0\sin^2\alpha}}.
$$
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