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 $n\in\mathbb{Z}$.
$$
\begin{align}
&\int_0^{2\pi}e^{i\zeta\cos(x+\phi)}\,\sin^2(x)\,\mathrm{d}x\\
&=\int_0^{2\pi}e^{i\zeta\cos(x)}\,\sin^2(x-\phi)\,\mathrm{d}x\\
&=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2(x-\phi)))\,\mathrm{d}x\\
&=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2x)\cos(2\phi)-\sin(2x)\sin(2\phi))\,\mathrm{d}x\\
&=\pi(J_0(\zeta)+\cos(2\phi)J_2(\zeta))\tag{4}
\end{align}
$$