For $a\in\mathbb{Z}$, we have $ \int_0^{2\pi}e^{i\zeta\cos(ax+\phi)}\,\sin^2(ax)\,\mathrm{d}x=\int_0^{2\pi}e^{i\zeta\cos(x+\phi)}\,\sin^2(x)\,\mathrm{d}x\tag{1} $ Mathematica says that this is $ \frac{2\pi}{\zeta}\operatorname{BesselJ}(1,\zeta)=\frac{2\pi}{\zeta}\operatorname{J}_1(\zeta)\tag{2} $ for $\zeta\in\mathbb{R}^+$.
It appears that Mathematica is not correct; i.e. The integral in
$(1)$ is not independent of
$\phi$ as
$(2)$ would indicate. For now, we will keep the same assumptions (
$a\in\mathbb{Z}$ and
$\zeta\in\mathbb{R}^+$). We will also use
$ \begin{align} \int_0^{2\pi}e^{i\zeta\cos(x)}\cos(nx)\,\mathrm{d}x&=2\pi\,i^nJ_n(\zeta)\\ \int_0^{2\pi}e^{i\zeta\cos(x)}\sin(nx)\,\mathrm{d}x&=0 \end{align}\tag{3} $ for
$n\in\mathbb{Z}$.
$ \begin{align} &\int_0^{2\pi}e^{i\zeta\cos(x+\phi)}\,\sin^2(x)\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\,\sin^2(x-\phi)\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2(x-\phi)))\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2x)\cos(2\phi)-\sin(2x)\sin(2\phi))\,\mathrm{d}x\\ &=\pi(J_0(\zeta)+\cos(2\phi)J_2(\zeta))\tag{4} \end{align} $