I would like to understand if the following two infinite series can be further expressed in terms of known functions:
$$\sum_{n\geq 0} (-1)^nI_{2n+1}(A)\frac{\cos((2n+1)B)}{2n+1}$$
and
$$\sum_{n\geq 1} (-1)^nI_{2n}(A)\frac{\sin(2nB)}{2n}$$
where $A$ is a positive real constant and $B$ is real. $I_n(\cdot)$ is the nth order of the modified Bessel function of first kind.
Have a look at this question. We have that:
$$ e^{A\sin x}= I_0(A) + 2\sum_{n\geq 0}(-1)^n I_{2n+1}(A)\sin((2n+1) x)+2\sum_{n\geq 1}(-1)^n I_{2n}(A)\cos(2nx) $$ hence by considering the even part and replacing $x$ with $Bx$ we get: $$ \cosh(A\sin(Bx))= I_0(A) + 2\sum_{n\geq 1}(-1)^n I_{2n}(A)\cos(2nBx) $$ and: $$ I_0(A)+2\sum_{n\geq 1}(-1)^{n}I_{2n}(A)\frac{\sin(2nB)}{2nB}=\int_{0}^{1}\cosh(A\sin(Bx))\,dx $$ so, by rearranging, $$\sum_{n\geq 1}(-1)^n I_{2n}(A)\frac{\sin(2nB)}{2n}=\color{red}{-\frac{B}{2} I_0(A)+\frac{B}{2}\int_{0}^{1}\cosh(A\sin(Bx))\,dx} $$ and the other series can be managed in a similar way. From:
$$ \sinh(A\sin(Bx))= 2\sum_{n\geq 0}(-1)^n I_{2n+1}(A)\sin((2n+1)Bx) $$ it follows that: $$ \color{red}{\frac{B}{2}\int_{0}^{1}\sinh(A\sin(Bx))\,dx} = \sum_{n\geq 0}(-1)^n I_{2n+1}(A)\frac{1-\cos((2n+1)B)}{2n+1}.$$