Is it possible to evalute $\int_{-\pi} ^\pi \cos( \frac{\sqrt 3 \cos x}{2} )\cos(\frac{\sin(x)}{2}) dx $? I want to know how to integrate this function.
$$
\int_{-\pi}^{\pi}
\cos\left(\,\sqrt{\, 3\, }\cos\left(x\right) \over 2\right)
\cos\left(\sin\left(x\right) \over 2\right)\,\mathrm{d}x
$$
Is it possible to use Bessel function to integrate this function ??.
I want to know the answer analytically.
 A: OP ask for generalization in comment. Answer updated to address that.

For any $r > 0$, notice
$$\begin{align}
f(r,x) & \stackrel{def}{=}
\cos\left(\frac{\sqrt{3}}{2}r\cos(x)\right)\cos\left(\frac12r\sin(x)\right)\\
& = \frac12\left[
\cos\left(r\left(\frac{\sqrt{3}}{2}\cos(x)-\frac12\sin(x)\right)\right)
+\cos\left(r\left(\frac{\sqrt{3}}{2}\cos(x)+\frac12\sin(x)\right)\right)
\right]\\
& =  \frac12\left[
\cos\left(r\cos\left(x + \frac{\pi}{6}\right)\right) +
\cos\left(r\cos\left(x - \frac{\pi}{6}\right)\right)
\right]
\end{align}
$$
and everything is periodic in $x$ with period $2\pi$. We have
$$\int_{-\pi}^\pi f(r,x) dx = \int_{-\pi}^\pi \cos(r\cos(x)) dx
= 4\int_0^{\pi/2}\cos(r\cos(x)) dx
= 4\int_0^{\pi/2}\cos(r\sin(x)) dx
$$
Compare RHS with following integral representation of Bessel function of first kind,
$$J_n(z) = \frac{1}{\pi}\int_0^\pi \cos(nx - z\sin(x)) dx$$
We find $$\int_{-\pi}^\pi f(r,x) dx = 2\pi J_0(r)$$
For the original integral where $r = 1$, it evaluates to
$$2\pi J_0(1) \approx 4.807878861268825996546649785183...$$
A: Using the Taylor series of $\cos
 $ we have $$I=\int_{-\pi}^{\pi}\cos\left(\frac{\sqrt{3}\cos\left(x\right)}{2}\right)\cos\left(\frac{\sin\left(x\right)}{2}\right)dx=2\int_{0}^{\pi}\cos\left(\frac{\sqrt{3}\cos\left(x\right)}{2}\right)\cos\left(\frac{\sin\left(x\right)}{2}\right)dx
 $$ $$=2\sum_{n\geq0}\frac{\left(-1\right)^{n}2^{-2n}}{\left(2n\right)!}\int_{0}^{\pi}\cos\left(\frac{\sqrt{3}\cos\left(x\right)}{2}\right)\sin^{2n}\left(x\right)dx
 $$ and now using this identity involving the Bessel function of the first kind $$J_{v}\left(z\right)=\frac{\left(\frac{z}{2}\right)^{v}}{\Gamma\left(v+\frac{1}{2}\right)\sqrt{\pi}}\int_{0}^{\pi}\cos\left(z\cos\left(x\right)\right)\sin^{2v}\left(x\right)dx,\,\textrm{Re}\left(v\right)>-\frac{1}{2}
 $$ we get, using the formula for the Gamma function with half integers argument and the properties of the double factorial, $$I=2\pi\sum_{n\geq0}\frac{\left(-1\right)^{n}2^{-2n}}{\left(2n\right)!}\left(\frac{4}{\sqrt{3}}\right)^{n}\Gamma\left(n+\frac{1}{2}\right)J_{n}\left(\frac{\sqrt{3}}{2}\right)
 =2\pi\sum_{n\geq0}\frac{\left(-1\right)^{n}}{n!}\left(\frac{1}{4\sqrt{3}}\right)^{n}J_{n}\left(\frac{\sqrt{3}}{2}\right)
 $$ and, as tired suggest, using the multiplication theorem of the Bessel functions, we have $$I=\color{red}{2\pi J_{0}\left(1\right)}.$$
