Evaluate the Integral: $$\int_{0}^{2\pi}\ {e^{\sin t}}\ dt $$
I've found results for this which involve an infinite series which becomes increasingly complicated and other series that involve more integrals. Ideally the solution should be a well defined constant.
With Modified Bessel functions $$I_\alpha(x)=\dfrac{1}{\pi}\int_0^\pi e^{x\cos t}\cos\alpha t\ dt$$ then \begin{align} \int_{0}^{2\pi} e^{\sin t}\ dt &= \int_{0}^{2\pi} e^{\cos t}\ dt \\ &= \int_{0}^{\pi} e^{\cos t}\ dt+\int_{0}^{\pi} e^{-\cos t}\ dt \\ &= \pi I_0(1)+\pi I_0(-1) \\ &= 2\pi I_0(1) \\ &= 2\pi\sum_{n=0}^\infty\dfrac{1}{2^{2n}n!^2} \end{align}
Taylor series of $$e^x = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$$ For $$x = \sin t$$
$$e^{\sin t} = \sum\limits_{n=0}^{\infty}\frac{\sin^n t}{n!}$$ Let's integrate $$\int\limits_{0}^{2\pi} e^{\sin t} \ dt = \sum\limits_{n=0}^{\infty} \frac{1}{ n!}\int\limits_{0}^{2\pi}\sin^n t \ dt$$
The integral could be computed using the method mentioned here, i.e. using integration by parts as they did here. You will end up with a relation of linking $I_n = \int\limits_{0}^{2\pi}\sin^n t \ dt$ with $I_{n-2}$, and by induction or recursion, you could show that $$\int\limits_{0}^{2\pi}\sin^n t \ dt = \pi \frac{n-1}{n}\frac{n-3}{n-2} \ldots \frac{3}{4} \qquad n = 2k$$ and zero for odd $n$.
Use Simpson's rule to approximate this integral.
Below is the formula for using Simpson's $\displaystyle\frac{1}{3}$ rule
$$\int _{a}^{b}f(x)\,dx \approx \frac {\Delta x}{3}\left(f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+2f(x_{4})+\cdots +4f(x_{n-1})+f(x_{n})\right)$$
Where $\Delta x=\frac{b-a}{n}$ and $x_i=a+i\Delta x$.
$n$ is the number of equal intervals you want to take. The bigger $n$ is, the better is the accuracy of your solution.
So if we take $n=6$, $b=2\pi$, $a=0$, we have $\displaystyle\Delta x=\frac{2\pi-0}{6}=\frac{\pi}{3}$, hence we have end points as $\displaystyle a=0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}, 2 \pi=b$.
Using Simpson's $\displaystyle\frac{1}{3}$ rule, we have:
$\begin{align*}\int_{t=0}^{2\pi}\ {e^{\sin t}}dt&\approx\frac{\Delta x}{3}\left(f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+2f(x_{4})+4f(x_{5})+f(x_{6})\right)\\&\approx\frac{\pi}{3*3}\left(f(0)+4f\left(\frac{\pi}{3}\right)+2f\left(\frac{2\pi}{3}\right)+4f\left(\frac{3\pi}{3}\right)+2f\left(\frac{4\pi}{3}\right)+4f\left(\frac{5\pi}{3}\right)+f(2\pi)\right)\\&\approx\frac{\pi}{9}\left(1+9.50977070094466+4.75488535047233+...+1.68248010421646+1\right)\\&\approx7.95464392016463\end{align*}$
And there you have it: $$\int_{t=0}^{2\pi}\ {e^{\sin t}}dt\approx7.95\text{ (correct to 2 decimal places).}$$
Like I have mentioned above if we keep increasing the number of intervals ($n$), the above approximation will get closer and closer to the actual answer.
$$I=\int_0^{2\pi}e^{\sin(x)}dx$$ we know that: $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ so: $$e^{\sin(x)}=\sum_{n=0}^{\infty}\frac{\sin^n(x)}{n!}$$ this is not very nice. We know that: $$\int_{0}^{2\pi}e^{\sin(x)}dx=\int_0^{2\pi}e^{\cos(x)}dx$$ therefore: $$I^2=\left(\int_0^{2\pi}e^{\sin(x)}dx\right)\left(\int_0^{2\pi}e^{\cos(x)}dx\right)=\int_0^{2\pi}\int_0^{2\pi}e^{\sin(x)+\cos(y)}dxdy=\int_0^{2\pi}\int_0^{2\pi}re^{\sin(r\cos(\theta))+\cos(r\sin(\theta))}drd\theta$$ I was thinking a substition could be made so that the inside of the $\sin(r\cos(\theta))$ and $\cos(r\sin(\theta))$ would be more similar but nothing appears to be possible.@Nosrati appears to have the easiest solution
If you cannot use Bessel function directly, use, as in answers and comments $$e^{\sin (t)} = \sum\limits_{n=0}^{\infty}\frac{\sin^n( t)}{n!}$$ and the fact that $$\int_0^{2\pi} \sin^n(t)\,dt=\frac{\sqrt{\pi } \left(1+(-1)^n\right) \Gamma \left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}$$ that is to say $0$ if $n$ is odd. So $$\int_0^{2\pi}e^{\sin (t)}\,dt = 2 \sqrt{\pi }\sum\limits_{n=0}^{\infty} \frac{ \Gamma \left(n+\frac{1}{2}\right)}{(n!)^2}$$
