At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .(Happy)

At first i did it by integration by parts but the steps are quite long so i was not sure at that time ... Now , i am sure it is true with your help .(Happy)

There is another method that involves complex analysis, and it is the way I was hoping someone would do this.

We know that \(\displaystyle \oint_{|z|=1}e^zdz=0\) and thus \(\displaystyle \int_0^{2\pi}e^{e^{it}}\left(e^{it}\right)'dt=\int_0^{2\pi}e^{\cos(t)}\left(\cos(\sin(t))+i\sin(\sin(t))\right)\left(-\sin(t)+i\cos(t)\right)=0\), in particular the real an imaginary parts of that integral must equal zero. Do a little work with trig identities and you will get our integral is the imaginary part of it.