Let's have Mathieu equation $$ \tag 1 y''(t) + (A - 2qcos(t))y(t) = 0 $$ Let's assume case $q << 1$. Then Eq. $(1)$ can be rewritten in a form $$ \tag 2 \frac{1}{2\epsilon }y''(t) + (b - cos(t))y(t) = 0, \quad \frac{1}{2\epsilon} \equiv \frac{1}{2q}, \quad b \equiv \frac{A}{2q} >> 1 $$ It seems that it is possible to construct the solution of Eq. $(2)$ as WKB series: $$ y = \text{Exp}\left[i \int \limits_{t_{\text{in}}}^{t}\left( \epsilon p_{0}{'}(\tau ) + p_{1}{'}(\tau ) + \frac{1}{\epsilon} p_{2}{'}(\tau ) + ... \right)d\tau\right] $$ Zeroth and first orders in $\epsilon$ give $$ p_{0} = \sqrt{2(b - cos(2t))}, \quad p_{1} = \frac{i}{2}\frac{p_{0}{'}}{p_{0}}, $$ where " ' " denotes derivative. Thus $$ \int \limits_{0}^{t}\left(\frac{1}{\epsilon}p_{0}{'} + p_{1}{'}\right)dt = \frac{1}{\epsilon}p_{0}\bigg|_0^t + p_{1}\bigg|_0^t $$ The second summand gives $$ \tag 2 \frac{i}{2} \frac{sin(t)}{b - cos(t)}\bigg|_{0}^{t} \approx \frac{i}{2b}sin(t)(1 + cos(t)) $$ I know that for small values of $q$ parameter the Floquet index $\mu$ (exponential index of solution, $y(t) \sim e^{\mu t}$ ) for some values of $A$, specifically $A \approx \frac{P}{Q}$, where $P, Q$ are integer numbers, is proportional to $\frac{q}{2}$. But I don't see how WKB method imply this value, see $(2)$.
The question: how to get expression for Floquet index by using WKB method?