Evaluate $\int\sin(\sin x)~dx$

I was skimming the virtual pages here and noticed a limit that made me wonder the following
question: is there any nice way to evaluate the indefinite integral below?

$$\int\sin(\sin x)~dx$$ Perhaps one way might use Taylor expansion. Thanks for any hint, suggestion.

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The Jacobi-Anger expansion gives $\sin(\sin(x))$ in terms of an infinite sine series weighted by Bessel functions: en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion – Bitrex Jan 26 '13 at 16:38
@Bitrex: very interesting. Something new to me. Thank you! – user 1618033 Jan 26 '13 at 16:41
It looks so simple but it looks difficult also. – Nancy Rutkowskie Jan 26 '13 at 17:07
@NancyR: yeah. That's true. – user 1618033 Jan 26 '13 at 17:54
– sdcvvc Mar 4 '13 at 15:05

Maybe you could do something like substitute $u=\sin{x}$ and get

$$\int du \: \frac{\sin{u}}{\sqrt{1-u^2}}$$

You could Taylor expand the denominator and be in a position to integrate even moments of $\sin{u}$ and see if the resulting series is useful.

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OK. Thanks for your suggestion. (+1) – user 1618033 Jan 26 '13 at 16:46
Do you mean $du$? – leo Jan 26 '13 at 17:10
Oops, yes of course – Ron Gordon Jan 26 '13 at 17:32
Out of curiosity if you deifne x to be a number like you would when looking for an answer say $\frac {3^{1/2}}{2}$ can u just draw it on a triangle then look for the angle on the triangle then use that angle and opp/hyp of that angle to find the side of that triangle then integrate that constant over your bounds? – Faust7 May 10 '13 at 13:16
@Faust7: not entirely sure what you are asking. When doing trig substitutions, it is wise to use a right triangle as your guide in relating sines, cosines, etc. But I do not understand "integrating the constant over the bounds." $x$ is a variable and is expressed in terms of another variable, $u$. That's it. – Ron Gordon May 10 '13 at 13:21

For the maclaurin series of $\sin x$ , $\sin x=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$

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

Now for $\int\sin^{2n+1}x~dx$ , where $n$ is any non-negative integer,

$\int\sin^{2n+1}x~dx$

$=-\int\sin^{2n}x~d(\cos x)$

$=-\int(1-\cos^2x)^n~d(\cos x)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$

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

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

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