Is it generalizable?

If we define $J(m,D) \; =\; \int_0^{2\pi} \cos 2m(x + \sin(Dx))\,dx \hspace{2cm} m,D \in \mathbb{N}$ then $\begin{aligned} J(m,D) & = \; \mathfrak{Re}\left[\int_0^{2\pi} e^{2mi(x + \sin (Dx))}\,dx\right] \; = \; \mathfrak{Re}\left[\int_{|z|=1} z^{2m} e^{m(z^D - z^{-D})}\,\frac{dz}{iz}\right] \\ & = \; 2\pi\mathfrak{Re}\left[\mathrm{Res}_{z=0} z^{2m-1} e^{m(z^D-z^{-D})}\right] \end{aligned}$ if $m \neq 0$ , then the Laurent series of $e^{m(z^D-z^{-D})}$ will only contain terms of the form $z^{Dk}$ , where $k \in \mathbb{Z}$ . Thus $z^{2m-1}e^{m(z^D-z^{-D})}$ will only have a term in $z^{-1}$ if $2m$ is a multiple of $D$ . Thus we can certainly ensure that $J(m,D) \; = \; \left\{ \begin{array}{lll} 2\pi & \hspace{1cm} & m = 0 \\ 0 & & 1 \le m \le n \end{array}\right.$ provided that $D$ is odd and $n < D$ . Since $\sin^{2n}\theta \; \equiv \; \frac{1}{2^{2n-1}}\sum_{j=1}^n (-1)^j \binom{2n}{n+j}\cos2j\theta + \frac{1}{2^{2n}}\binom{2n}{n}$ we deduce that $\begin{aligned} \int_0^\pi \sin^{2n}(x + \sin(Dx))\,dx & = \; \tfrac12\int_0^{2\pi} \sin^{2n}(x + \sin(Dx))\,dx \\ & = \; \frac{1}{2^{2n}}\sum_{j=1}^n (-1)^j\binom{2n}{n+j}J(j,D) + \frac{1}{2^{2n+1}}\binom{2n}{n}J(0,D) \\ & = \; \frac{\pi}{2^{2n}}\binom{2n}{n} \end{aligned}$ provided that $D$ is odd and $n < D$ . In the case $n=10$ , $D=15$ we get an integral of $\tfrac{46189}{262144}\pi$ making the answer $\boxed{308333}$ .

The answer is not as straightforward when $n \ge D$ , since there would be other residues to consider. For example $\int_0^\pi \sin^{2D}\big(x + \sin(Dx)\big)\,dx \; = \; \frac{\pi}{2^{2D}}\binom{2D}{D} - \frac{\pi}{2^{2D-1}}J_2(2D)$ for odd positive integers $D$ . In this case $J(D,D)$ is nonzero. We are using $J_2$ , a Bessel function of the first kind, here.

Here is the generalized version you needed. There might be a mistake,please verify it once.