Solution
= Solution
A transformation of <phase space> is <canonical transformation>[canonical] when it preserves the <symplectic form>, or equivalently all <Poisson brackets>. For one degree of freedom this is
$$
dQ\wedge dP=dq\wedge dp,
$$
equivalently $\{Q,P\}_{q,p}=1$.
Suppose a <type-two generating function for a canonical transformation> $F(q,P)$ defines
$$
p=F_q,
\qquad Q=F_P.
$$
Then
$$
\begin{aligned}
dq\wedge dp&=F_{qP}\,dq\wedge dP,\\
dQ\wedge dP&=F_{Pq}\,dq\wedge dP.
\end{aligned}
$$
Equality of the mixed partial derivatives gives $dq\wedge dp=dQ\wedge dP$, so the transformation is canonical wherever it is locally invertible.
For
$$
F_0(q,P)=\int_0^q\sqrt{2P-u^2}\,du,
$$
the <fundamental theorem of calculus> and differentiation under the integral sign give
$$
p=\sqrt{2P-q^2},
\qquad
Q=\int_0^q\frac{du}{\sqrt{2P-u^2}}
=\arcsin\frac{q}{\sqrt{2P}}.
$$
Thus
$$
q=\sqrt{2P}\sin Q,
\qquad
p=\sqrt{2P}\cos Q,
$$
and in particular $P=(p^2+q^2)/2$.
For the unit-frequency <simple harmonic motion>, the transformed <Hamiltonian> is simply $H=P$. <Hamilton's equations> become
$$
\dot Q=\frac{\partial H}{\partial P}=1,
\qquad
\dot P=-\frac{\partial H}{\partial Q}=0.
$$
Therefore
$$
P(t)=E,
\qquad Q(t)=t-t_0,
$$
and transformation back gives the familiar solution
$$
q(t)=\sqrt{2E}\sin(t-t_0),
\qquad
p(t)=\sqrt{2E}\cos(t-t_0).
$$
Now consider the weak <quartic oscillator>
$$
H(q,p)=\frac12(p^2+q^2)+\epsilon q^4.
$$
Choose the modified generating function
$$
F_\epsilon(q,P)
=\int_0^q\sqrt{2P-u^2-2\epsilon u^4}\,du.
$$
It gives
$$
p=\sqrt{2P-q^2-2\epsilon q^4},
$$
so the transformed Hamiltonian is again $H=P$. Hence $P=E$ and $Q=t-t_0$. Expanding the square root with the <Taylor series> gives
$$
F_\epsilon(q,P)
=F_0(q,P)
-\epsilon\int_0^q\frac{u^4}{\sqrt{2P-u^2}}\,du
+O(\epsilon^2).
$$
Differentiation with respect to $P$ therefore yields
$$
Q
=\arcsin\frac{q}{\sqrt{2P}}
+\epsilon I(q,P)+O(\epsilon^2),
$$
where
$$
I(x,y)=\int_0^x\frac{u^4}{(2y-u^2)^{3/2}}\,du.
$$
Set $\theta=t-t_0$ and
$$
q_0=\sqrt{2E}\sin\theta,
\qquad
p_0=\sqrt{2E}\cos\theta.
$$
Writing $q=q_0+\epsilon q_1+O(\epsilon^2)$ and expanding the <inverse sine> relation at $q_0$ gives
$$
0=\frac{q_1}{p_0}+I(q_0,E),
\qquad
q_1=-p_0I(q_0,E).
$$
Expanding $p^2=2E-q^2-2\epsilon q^4$ similarly gives
$$
p=p_0+\epsilon p_1+O(\epsilon^2),
\qquad
p_1=q_0I(q_0,E)-\frac{q_0^4}{p_0}.
$$
Consequently
$$
\begin{aligned}
\frac qp
&=\frac{q_0}{p_0}
+\epsilon\left(\frac{q_1}{p_0}
-\frac{q_0p_1}{p_0^2}\right)
+O(\epsilon^2)\\
&=\tan\theta
-\epsilon I(q_0,E)(1+\tan^2\theta)
+\epsilon q_0^2\tan^3\theta
+O(\epsilon^2).
\end{aligned}
$$
Since $\theta=t-t_0$, this is the required expression. The formula is understood away from turning points where $p_0=0$ and the ratio $q/p$ itself is singular.