Closed forms of $\int_A^B \sin(\sin(ax))dx$ and $\int_A^B \sin(\sin(ax)) \cdot \sin(\sin(bx))dx$ I would like to know if:


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*$\int_A^B \sin(\sin(ax))dx$ has a closed-form? The solution of Maple requires the presence of Struve functions in its expression. But at least Maple is able to solve it, so even if complicated I guess it has an analytical form.

*$\int_A^B \sin(\sin(ax)) \cdot \sin(\sin(bx))dx$ has a closed-form? this one Maple was not able to solve. I tried to find a similar expression in this tables of Bessel functions but there are none looking like these one. And with Liouville's theorem and the Risch algorithm I am a bit lost in the differential algebra to prove that it is not integrable.
 A: For $\int_A^B\sin\sin ax~dx$ ,
$\int_A^B\sin\sin ax~dx$
$=\int_A^B\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}ax}{(2n+1)!}~dx$
$=-\int_A^B\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}ax}{a(2n+1)!}~d(\cos ax)$
$=\int_A^B\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2ax)^n}{a(2n+1)!}~d(\cos ax)$
$=\int_A^B\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}ax}{a(2n+1)!}~d(\cos ax)$
$=\int_A^B\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}ax}{a(2n+1)!k!(n-k)!}~d(\cos ax)$
$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}ax}{a(2n+1)!k!(n-k)!(2k+1)}\right]_A^B$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!(\cos^{2k+1}aB-\cos^{2k+1}aA)}{a(2n+1)!k!(n-k)!(2k+1)}$
For $\int_A^B\sin\sin ax\sin\sin bx~dx$ ,
$\int_A^B\sin\sin ax\sin\sin bx~dx$
$=\dfrac{1}{2}\int_A^B\cos(\sin ax-\sin bx)~dx-\dfrac{1}{2}\int_A^B\cos(\sin ax+\sin bx)~dx$
$=\dfrac{1}{2}\int_A^B\sum\limits_{n=0}^\infty\dfrac{(-1)^n(\sin ax-\sin bx)^{2n}}{(2n)!}~dx-\dfrac{1}{2}\int_A^B\sum\limits_{n=0}^\infty\dfrac{(-1)^n(\sin ax+\sin bx)^{2n}}{(2n)!}~dx$
