How to find the indefinite integral of sin(sin(x))dx? (I got this function by mistake, when I miswrote other function. Now I'm curious how to find the antiderivative of what I miswrote)
I have no a clue how to calculate it and neither does Wolfram Alpha or any other site that I tried. Trig formulas from school course don't seem to be useful too.
 A: From this answer https://math.stackexchange.com/a/877417/65203 and https://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion we know the Fourier series development
$$\sin(\sin x)=2 \sum_{k=0}^\infty J_{2k+1}(1)\sin((2k+1)x).$$
Then by term-wise integration
$$\int\sin(\sin x)\,dx=2 \sum_{k=0}^\infty\frac{J_{2k+1}(1)}{2k+1}\cos((2k+1)x)+C.$$
The coefficients are quickly decaying
$$0.440051,\\0.00652112,\\0.0000499515,\\2.14618×10^{-7},\\5.8325×10^{-10}, \\1.0891×10^{-12},\\\cdots$$
A: Like many other functions the indefinite   integral does not have a nice closed form.But the indefinite integral can be calcualted numerically or as a Taylor series .
For $\displaystyle\int_0^\pi\sin(\sin(x))\,dx = \pi H_0(1) \approx 1.78649$
where $H$ is the  is the Struve function . This only works for this particular indefinite integral .
One could also try a  fast convergent series such as;
$$\sum_{n=1}^\infty\sum_{k=1}^n\frac{(-1)^{k-n}\cdot 2^{1-2n}\sqrt\pi\sin(x)^{-1+2k}}{(-1+2k)\cdot\Gamma(k)\cdot\Gamma(\frac12+n)\cdot\Gamma(1-k+n)}$$
EDIT: As advised  in the  comments this page is where I had first learned  of the answer and the value is from it.
A: $\int\sin\sin x~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$
$=-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}x}{(2n+1)!}d(\cos x)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2x)^n}{(2n+1)!}d(\cos x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}x}{(2n+1)!}d(\cos x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}x}{(2n+1)!k!(n-k)!}d(\cos x)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
