# 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

$$\cos^4\theta$$ 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 $$\int_{0}^{\pi}\cos^4(x+\sin(3x))\,dx = \color{red}{\frac{3\pi}{8}}.$$