Anyone recognizing or knowing how to solve this integral? [Cross-posted with minor changes from Math SE; see note 2 below].
I would like to find methods for closed-form (or, failing that, series) solution of 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 $X=\pi/2$, $\pi$ or $2\pi$ and where $a$, $b$ are constants. Substitution ($t=\tan(x/2)$, $\cos(x)$ or $\sin(x)$) or conversion of the products of trigonometric functions to their sum or difference does not seem to lead to an amenable form, as arguments become (too) complicated. I also cannot find any of them listed in tables. 
Any leads? 
NB: 


*

*Numerical or stochastic quadrature is not feasible in my case, as $a$ and $b$ are parameterized.

*As opposed to the Maths SE post, my query here focuses on potential general techniques, as a core for solving these and related (more complicated) integrals of functionals. Different approaches and lines of attack are of greater interest to me than just the result itself. For example, is there a deeper reason why we have a solution for the 5th but not 6th case? Are there methods that work equally on e.g. ln(ln()) as they would on cos(cos())? Etc. 
 A: How about these special cases...
\begin{align}
\int_0^{2\pi} \cos(a \cos(x)) \cos(b \sin(x)) \;dx &= 
2\pi \;J_0\big(\sqrt{a^2+b^2}\;\big)
\\
\int_0^{2\pi} \cos(a \cos(x)) \sin(b \sin(x)) \;dx &= 0
\\
\int_0^{2\pi} \sin(a \cos(x)) \cos(b \sin(x)) \;dx &= 0
\\
\int_0^{2\pi} \sin(a \cos(x)) \sin(b \sin(x)) \;dx &= 0
\end{align}
where $J_0$ is a Bessel function.
And
\begin{align}
\int_0^{\pi} \cos(a \cos(x)) \cos(b \sin(x)) \;dx &= 
\pi \;J_0\big(\sqrt{a^2+b^2}\;\big)
\\
\int_0^{\pi} \cos(a \cos(x)) \sin(b \sin(x)) \;dx &= ?
\\
\int_0^{\pi} \sin(a \cos(x)) \cos(b \sin(x)) \;dx &= 0
\\
\int_0^{\pi} \sin(a \cos(x)) \sin(b \sin(x)) \;dx &= 0
\end{align}
A: For $X = 2\pi,$ at least these are integrals of a meromorphic function over the unit circle. In fact, unless I am sadly mistaken, the only singularity of this function is at the origin, so the residue theorem is your friend (although for two of the integrals, the function is odd, so the integral is automatically zero). In the cases where the function is even, the integral from $0$ to $\pi$ is half of the integral from $0$ to $2\pi.$
