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.

• Almost surely, the primitive is nonelementary. May 15 '18 at 11:36
• Why were you so sure? May 15 '18 at 12:14
• Because of Liouville's theorem. en.wikipedia.org/wiki/…
– user65203
May 15 '18 at 12:32

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

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

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

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;

$\sum_{n=1}^\infty\sum_{k=1}^n\frac{(-1)^{k-n}\cdot 2^{1-2n}\sqrt\pi\sin(x)^{-1+2k}}{(-1+2k)\cdot\Gamma(k)\cdot\Gamma(\frac12+n)\cdot\Gamma(1-k+n)}$

• 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?
– user284001
May 15 '18 at 11:51
• 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.
– user284001
May 15 '18 at 11:56
• @Kevin thank you. I've rectified my mistake. May 15 '18 at 12:00
• 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.) May 15 '18 at 12:10

$\int\sin\sin x~dx$

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

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

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

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

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

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