Let $I(b)$ is the following integral
$$I(b)=\int_0^b \sin(\sin(x)) \, dx.$$
There are some $b$ value for that we know a closed-form of $I(b)$ in term of Struve function $\mathbf{H}_n(x)$. For example
$$\begin{align} I(\pi/2) = & \, \frac{\pi}{2}\mathbf{H}_0(1) \\ I(\pi) = & \, \pi\mathbf{H}_0(1) \\ I(3\pi/2) = & \, \frac{\pi}{2}\mathbf{H}_0(1) \\ I(2\pi) = & \, 0. \end{align}$$
I know that the closed-form of $I(k\pi/2)$ comes from the integral form of Struve function, but perhaps there are other closed forms of this type of integrals.
Question. Is there a closed-form of $I(\pi/4)=\int_0^{\pi/4} \sin(\sin(x)) \, dx$?
$\int_0^b\sin(\sin x)~dx$
$=\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$
$=-\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}x}{(2n+1)!}d(\cos x)$
$=\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2x)^n}{(2n+1)!}d(\cos x)$
$=\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}x}{(2n+1)!}d(\cos x)$
$=\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}x}{(2n+1)!k!(n-k)!}d(\cos x)$
$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}\right]_0^b$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!(\cos^{2k+1}b-1)}{(2n+1)!k!(n-k)!(2k+1)}$