Does an elementary indefinite integral for 2^sin(x) exist? Wolframalpha says no, and I've had little luck elsewhere, maybe because search engines aren't the best at making sense of the "^" symbol.
And if not, are any numerical methods known for solving this integral?
 A: There is no known closed form I am aware of, but it may be worth noting that
$$J_0(-i\ln(2))=\int_a^{a+2\pi}2^{\sin(x)}~\mathrm dx$$
where $J$ is a Bessel function. As far as numerical goes you can use any standard method.
A: Notice that:
$$2^{\sin(x)}=\left(e^{ln(2)}\right)^{\sin(x)}=e^{\ln(2)\sin(x)}$$
And you will find articles like this that show this is non-elementary.
You can represent it as a series of integrals by using the fact that:
$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!},z\in\mathbb{C}$$
A: $\int 2^{\sin x}~dx$
$=\int e^{\ln2\sin x}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
For $n$ is any natural number,
$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
For $n$ is any non-negative integer,
$\int\sin^{2n+1}x~dx$
$=-\int\sin^{2n}x~d(\cos x)$
$=-\int(1-\cos^2x)^n~d(\cos x)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$
$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
$=x+\sum\limits_{n=1}^\infty\dfrac{x\ln^{2n}2}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}2\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{x\ln^{2n}2}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\ln^{2n}2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\ln^{2n+1}2\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
