Finding $\int^{\pi}_{0}\sin(8x+8\sin 3x)dx$ 
Finding $\displaystyle \int^{\pi}_{0}\cos^4(x+\sin 3x)dx$

Try: From $\displaystyle \cos^4(x)=(\cos^2(x))^2=\frac{1}{4}\bigg[1+\cos 2x\bigg]^2$
$$=\frac{1}{4}+\frac{1}{8}\bigg(1+\cos^2(4x)+2\cos 4x\bigg)+\frac{\cos 2x}{2}$$
$$=\frac{3}{8}+\frac{1}{16}+\frac{1}{16}\cos(8x)+\frac{1}{4}\cos(4x)+\frac{1}{2}\cos(2x)$$
So $$\cos^4(x+\sin 3x)=\frac{7}{16}+\frac{1}{16}\cos(8x+8\sin 3x)+\frac{1}{4}\cos(4x+4\sin 3x)+\frac{1}{2}\cos(2x+3\sin 3x)$$
How can i solve $$\int^{\pi}_{0}\sin(8x+8\sin 3x)dx$$
$$\int^{\pi}_{0}\sin(4x+4\sin 3x)dx$$  Type integrals 
I have seems that it is not possible in elementry way
could some help me to solve it . thanks
 A: $\cos^4\theta$ can be written in terms of $1,\cos(2\theta),\cos(4\theta)$. Additionally the value of
$$ \int_{0}^{\pi}\cos\left(2kx+2k\sin(3x)\right)\,dx=\frac{1}{2}\text{Re}\int_{-\pi}^{\pi}e^{2k i x}\cdot e^{2ki\sin(3x)}\,dx $$
can be derived from the Jacobi-Anger expansion
$$ e^{iz\sin\theta} = \sum_{n\in\mathbb{Z}} J_n(z) e^{in\theta} $$
leading to
$$ e^{2ki\sin(3x)}=\sum_{n\in\mathbb{Z}}J_n(2k) e^{3nix}. $$
Since for any $a,b\in\mathbb{Z}$ we have $\int_{-\pi}^{\pi}e^{aix}e^{bix}\,dx = 2\pi\delta(a+b)$, for $k\in\mathbb{N}$ the integral $\int_{0}^{\pi}\cos\left(2kx+2k\sin(3x)\right)\,dx$ differs from zero only if $2k$ is a multiple of $3$ (i.e. if $k$ is a multiple of $3$), and in such a case it equals $\pi J_{2k/3}(2k)$ (with $J_n$ being a Bessel function of the first kind). Since neither $2$ or $4$ are multiples of three and $\cos^4(\theta)=\frac{3}{8}+\frac{1}{2}\cos(2\theta)+\frac{1}{8}\cos(4\theta)$, it follows that
$$ \int_{0}^{\pi}\cos^4(x+\sin(3x))\,dx = \color{red}{\frac{3\pi}{8}}.$$
