= Solution
Define the <canonical momentum>[canonical momenta]
$$
p_i=\frac{\partial\mathcal L}{\partial\dot q_i}.
$$
When the velocity Hessian is nonsingular, these equations can be inverted for $\dot q_i(q,p,t)$, and the <Hamiltonian> is the <Legendre transform>
$$
\boxed{H(q,p,t)=\sum_i p_i\dot q_i-\mathcal L(q,\dot q,t).}
$$
Its differential is
$$
dH=\sum_i\dot q_i\,dp_i
-\sum_i\frac{\partial\mathcal L}{\partial q_i}\,dq_i
-\frac{\partial\mathcal L}{\partial t}\,dt,
$$
because the $p_i\,d\dot q_i$ terms cancel. The <Euler-Lagrange equation> identifies $\partial\mathcal L/\partial q_i=\dot p_i$, so coefficient comparison gives <Hamilton's equations>
$$
\boxed{\dot q_i=\frac{\partial H}{\partial p_i},
\qquad
\dot p_i=-\frac{\partial H}{\partial q_i}.}
$$
For the <symmetric top>, the momenta are
$$
p_\theta=A\dot\theta,
\qquad
p_\psi=B(\dot\psi+\dot\phi\cos\theta),
\qquad
p_\phi=A\dot\phi\sin^2\theta+p_\psi\cos\theta.
$$
Thus
$$
\dot\theta=\frac{p_\theta}{A},
\qquad
\dot\phi=\frac{p_\phi-p_\psi\cos\theta}{A\sin^2\theta},
\qquad
\dot\psi=\frac{p_\psi}{B}-\dot\phi\cos\theta.
$$
The Legendre transform gives
$$
\boxed{H=
\frac{p_\theta^2}{2A}
+\frac{(p_\phi-p_\psi\cos\theta)^2}{2A\sin^2\theta}
+\frac{p_\psi^2}{2B}
+Mgl\cos\theta.}
$$
Both $\phi$ and $\psi$ are <cyclic coordinate>[cyclic coordinates], so $p_\phi$ and $p_\psi$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\phi,\psi$,
$$
\{H,p_\phi\}=\{H,p_\psi\}=\{p_\phi,p_\psi\}=0
$$
under the <Poisson bracket>. Hence three independent <first integrals in involution> are
$$
\boxed{H,\qquad p_\phi,\qquad p_\psi.}
$$
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