Anyone recognizing or knowing how to solve this integral?

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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.)

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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.