Integrate the master piece

Let $I(m,n) \; = \; \int_0^\pi \cos m\big(x + \sin nx\big)\,dx \hspace{2cm} m,n \in \mathbb{N}$ Then, using some symmetry and the complex substitution $z = e^{ix}$ , $\begin{aligned} I(m,n) & = \; \tfrac12 \int_{-\pi}^\pi \cos m\big(x + \sin nx\big)\,dx \; = \; \tfrac12 \int_{-\pi}^\pi e^{im(x + \sin nx)}\,dx \\ & = \; \tfrac12 \int_{|z|=1} z^m e^{\frac12m(z^n-z^{-n})}\,\frac{dz}{iz} \; = \; \tfrac{1}{2i} \int_{|z|=1} F_{m,n}(z)\,\frac{dz}{z} \end{aligned}$ where $F_{m,n}(z) \; = \; z^m e^{\frac12m(z^n-z^{-n})}$ is holomorphic on $\mathbb{C} \backslash \{0\}$ , and has an isolated essential singularity at $0$ . Thus $I(m,n) \; =\; \pi \mathrm{Res}_{z=0} z^{-1}F_{m,n}(z)$ is equal to $\pi$ times the constant term in the Laurent expansion of $F_{m,n}(z)$ about $0$ . But since $F_{m,n}(z) \; = \; \sum_{u,v \ge 0} \frac{(\frac12m)^{u+v}(-1)^v}{u!v!}z^{m+n(u-v)} \hspace{2cm} z \neq 0$ we see that $I(m,n) = 0$ except when $n$ divides $m$ . This is all we need for this question (the integrals $I(kn,n)$ can be evaluated in terms of Bessel functions, however).

Since $\sin^4\theta \; \equiv \; \tfrac18\cos4\theta - \tfrac12\cos2\theta + \tfrac38$ we deduce that $\int_0^\pi \sin^4\big(x + \sin 3x\big)\,dx \; = \; \tfrac18I(4,3) - \tfrac12I(2,3) + \tfrac38\pi \; = \; \tfrac38\pi$ making the required answer $\tfrac{3\times8}{6} = \boxed{4}$ .

I guessed. Too hard for me.