Compute $\displaystyle \int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right) dx$
$\displaystyle (e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x $ $\displaystyle =e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=$... $\displaystyle e^{\cos x}\cos(x+\sin x)$ !! Oh, dear hollie mollie: yes, it is!
A question: how did you come up with it? Some program or some insight?
Tonio
I write $\displaystyle e^{\cos(x)} \cos( x + \sin(x) )$
$\displaystyle = Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ] $
$\displaystyle = Re[ e^{e^{ix}} e^{ix} ] $
then substitute $\displaystyle u = e^{ix} $
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 .
Just tinkering
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.
let $\displaystyle I(a) = \int^{2 \pi}_{0} e^{a \cos x} \cos (x + a \sin x) \ dx = Re \Big(\int^{2 \pi}_{0} e^{ae^{ix}}e^{ix} \ dx \Big)$
then $\displaystyle I'(a) = Re \Big(\int^{2 \pi}_{0} e^{2ix}e^{ae^{ix}} \ dx\Big) $
let $\displaystyle u = ae^{ix} $
EDIT: then $\displaystyle I'(a) = Re \Big( -\frac{i}{a^{2}} \int^{a}_{a} ue^{u} \ du \Big) = Re(0) = 0 $
so $\displaystyle I(a) = C $
but $\displaystyle I(0) = 0$
which means $\displaystyle C=0$
so $\displaystyle I(a) = 0 $
and $\displaystyle \int^{2 \pi}_{0} e^{\cos x} \cos (x + \sin x) \ dx = I(1) = 0 $