Integral of $e^{2\sin x}$ I'm trying to find the integral of $e^{2\sin x}$. I tried using integration by parts and u-substitution but neither seems to work in the end. Would someone be able to take me through how to do this?
I was given this question by a classmate who found it but couldn't do it and we're in an extension maths class in high school. Going by the comments, i'm guessing this is beyond the scope of what we can do...
 A: As already said in comments, the antiderivative cannot be expressed in terms of elementary functions. So, may be, you could use a Taylor series, using $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)$$ and replace $y$ by $2\sin(x)$. This will give $$e^{2\sin(x)}=1+2 \sin (x)+2 \sin ^2(x)+\frac{4 }{3}\sin ^3(x)+\cdots$$ and you are then left with a linear combination of integrals $$I_n=\int \sin^n(x)\, dx$$ for which exist at least a reduction formula $$I_n=-\frac 1n\sin^{n-1}(x)\cos(x)+\frac{n-1}n I_{n-2}$$
Edit
egreg made a good point in comments. When I wrote (in my initial answer) that the antiderivative does not exist, it was a bad shortcut to say that the antiderivative cannot be expressed in terms of elementary functions. Since I did not want to hide my mistake (not to say more), I did not modify the text. I change it now and let egreg the merit of pointing my bad wording.
A: Although the integrand does not possess a closed form anti-derivative in terms of elementary functions $($see Liouville's theorem and the Risch algorithm for more information$)$, its definite 
counterpart is $\displaystyle\int_0^\tfrac\pi2e^{a\sin x}~dx~=~\int_0^\tfrac\pi2e^{a\cos x}~dx~=~\frac\pi2\Big[I_0(a)+L_0(a)\Big]$, where I and L are 
the Bessel and Struve functions, respectively.
