(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.
$\endgroup$ 33 Answers
$\begingroup$From this answer we know the Fourier series development
$$\sin(\sin x)=\sum_{k=0}^\infty J_{2k+1}(1)\sin((2k+1)x).$$
Then by term-wise integration
$$\int\sin(\sin x)\,dx=\sum_{k=0}^\infty\frac{J_{2k+1}(1)}{2k+1}\cos((2k+1)x)+C.$$ The coefficients are quickly decaying
$$0.440051,\\0.00652112,\\0.0000499515,\\2.14618×10^{-7},\\5.8325×10^{-10}, \\1.0891×10^{-12},\\\cdots$$
$\endgroup$ 1 $\begingroup$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 $\displaystyle\int_0^\pi\sin(\sin(x))\,dx = \pi H_o(1) \approx 1.78649$
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.
$\endgroup$ 7 $\begingroup$$\int\sin\sin x~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$
$=-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}x}{(2n+1)!}d(\cos x)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2x)^n}{(2n+1)!}d(\cos x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}x}{(2n+1)!}d(\cos x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}x}{(2n+1)!k!(n-k)!}d(\cos x)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
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