Anyone recognizing or knowing how to solve this integral? I would like to find closed-form (or, failing that, series) expressions for the following four integrals:
$$\int^X_0 \cos(a \cos(x)) \cos(b \sin(x)) dx$$
$$\int^X_0 \cos(a \cos(x)) \sin(b \sin(x)) dx$$
$$\int^X_0 \sin(a \cos(x)) \cos(b \sin(x)) dx$$
$$\int^X_0 \sin(a \cos(x)) \sin(b \sin(x)) dx$$
where $X = \pi$ or $2\pi$ and $a$, $b$ are constants. I cannot find them in tables, and the usual substitutions ($t=\tan(x/2)$, $t=\sin(x)$, $t=\cos(x)$) or use of sum/difference formulas for converting products/sums of trigonometric functions do not lead to 
recognizable forms either. Any leads? (NB: $a$ and $b$ are multidimensional parameters that need separate assignment and evaluation; also further nested integration is needed afterwards, so quadrature or statistical calculation methods are not feasible in my case.)

Applying trig. rules for the product of function, in each case one obtains sum of terms like $\cos(a\cos(x)+b\sin(x))$ (or sines, or with negative signs...).
They can be transformed into $\cos(\sqrt{a^2+b^2}\cos(x-\phi))$. Now, integrating on $(0,2\pi)$ which is a period of the function one can drop the $\phi$ shift and obtain things like
$$\int_0^{2\pi}\cos(\sqrt{a^2+b^2}\cos(x))\,dx =2\pi J_0(\sqrt{a^2+b^2})$$
If $X=\pi$, this should be halfed as the function to be integrated are even.
