Calculate the integral $\int e^{\sin x}\,dx$

Elementary? Not much. One of the applicable methods explained in

While solving the integral of $e^{\sin(x)}$

$\displaystyle \int e^{\sin(x)} dx=I_0(1)x + \frac{\pi}{2}L_0(1) + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right ) $

$I_n(1)$ is rapidly diminishing and the shape of the integral is perfectly explained through this formula.

We have that

$$\int e^{\sin x}\mathrm dx=\int\left(\sum_{k=0}^\infty\frac{\sin^k(x)}{k!}\right)\mathrm dx=\sum_{k=0}^\infty\left(\int\frac{\sin^k(x)}{k!}\mathrm dx\right)=\sum_{k=0}^\infty\frac1{k!}\left(\int\sin^k(x)\mathrm dx\right)$$

where the swap between the summation and the integral is possible because the series converges uniformly for $x\in\Bbb R$. But I dont think that we can go far further than the last expression.

However, from the last expression, we can approach easily to any desired level of error the value of any definite integral.

We knows too that

$$I_n=\frac1n((n-1)I_{n-2}-\cos x(\sin x)^{n-1})$$

for $I_n:=\int\sin^n(x)\mathrm dx$. But as @MrYouMath said in the end we will need numerical methods, and we can use it directly in the original integral, by example using any kind of Simpson's rule.

Copying my answer to the almost duplicate question

See also https://math.stackexchange.com/a/117545/442

Although this indefinite integral has no known closed form, certain definite integrals are known... $$ \int_{-\pi/2}^{\pi/2} e^{\sin x}\;dx = \pi\;I_0(1) , $$ where $I_0$ is a modified Bessel function