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(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.

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From this answer https://math.stackexchange.com/a/877417/65203 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

$$0.440051,\\0.00652112,\\0.0000499515,\\2.14618×10^{-7},\\5.8325×10^{-10}, \\1.0891×10^{-12},\\\cdots$$

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    $\begingroup$ I think you should add a note that this isn't a closed form. $\endgroup$ – Jam May 15 '18 at 12:12
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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.

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  • $\begingroup$ You answer here is almost identical to the wording in this Quora post from Feb 2017 here quora.com/What-is-int-sin-sin-x-mathrm-dx-equal-to I wodner if this is where you got the information form? $\endgroup$ – higgs May 15 '18 at 11:51
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    $\begingroup$ @Kevin yes, i had read and made a note of it. Should i delete my post and link to that post? please advice $\endgroup$ – The Integrator May 15 '18 at 11:54
  • $\begingroup$ I think technically the link should feature since it is a repost from another website but I am not that familiar with the house rules on posting an answer that is somebody else's work. $\endgroup$ – higgs May 15 '18 at 11:56
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    $\begingroup$ @Kevin thank you. I've rectified my mistake. $\endgroup$ – The Integrator May 15 '18 at 12:00
  • $\begingroup$ You can mark the post community wiki. Then upvotes don't give you any reputation. (You could also include a link to the wiki page on which functions have closed form indefinite integrals using elementary functions.) $\endgroup$ – Ethan Bolker May 15 '18 at 12:10
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$\int\sin\sin x~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$

$=-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}x}{(2n+1)!}d(\cos x)$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2x)^n}{(2n+1)!}d(\cos x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}x}{(2n+1)!}d(\cos x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}x}{(2n+1)!k!(n-k)!}d(\cos x)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$

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