Easy integral?

**Drexel28** · May 11th 2010, 09:10 PM

Compute $\int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right) dx$

**simplependulum** · May 11th 2010, 11:46 PM

Originally Posted by Drexel28

Compute $\int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right) dx$

Is the integrand the derivative of

$e^{\cos(x)} \sin( \sin(x) )$ ?

**tonio** · May 12th 2010, 03:48 AM

Originally Posted by simplependulum

Is the integrand the derivative of

$e^{\cos(x)} \sin( \sin(x) )$ ?

$(e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x$ $=e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=$ ... $e^{\cos x}\cos(x+\sin x)$ !! Oh, dear hollie mollie: yes, it is!

A question: how did you come up with it? Some program or some insight?

Tonio

**simplependulum** · May 12th 2010, 05:30 AM

I write $e^{\cos(x)} \cos( x + \sin(x) )$

$= Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ]$

$= Re[ e^{e^{ix}} e^{ix} ]$

then substitute $u = e^{ix}$

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 .

**Drexel28** · May 12th 2010, 02:21 PM

Originally Posted by tonio

$(e^{\cos(x)} \sin( \sin(x) ))'=-\sin x \,e^{\cos x}\sin(\sin x)+e^{\cos x}\cos(\sin x) \cos x$ $=e^{\cos x}\left(\cos(\sin x)\cos x-\sin x\,\sin(\sin x)\right)=$ ... $e^{\cos x}\cos(x+\sin x)$ !! Oh, dear hollie mollie: yes, it is!

A question: how did you come up with it? Some program or some insight?

Tonio

Just tinkering

Originally Posted by simplependulum

I write $e^{\cos(x)} \cos( x + \sin(x) )$

$= Re[ e^{\cos(x)}e^{ix} e^{i\sin(x)} ]$

$= Re[ e^{e^{ix}} e^{ix} ]$

then substitute $u = e^{ix}$

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 .

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

We know that $\oint_{|z|=1}e^zdz=0$ and thus $\int_0^{2\pi}e^{e^{it}}\left(e^{it}\right)'dt=\int _0^{2\pi}e^{\cos(t)}\left(\cos(\sin(t))+i\sin(\sin (t))\right)\left(-\sin(t)+i\cos(t)\right)=0$ , 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.

**Random Variable** · May 12th 2010, 04:32 PM

let $I(a) = \int^{2 \pi}_{0} e^{a \cos x} \cos (x + a \sin x) \ dx = Re \Big(\int^{2 \pi}_{0} e^{ae^{ix}}e^{ix} \ dx \Big)$

then $I'(a) = Re \Big(\int^{2 \pi}_{0} e^{2ix}e^{ae^{ix}} \ dx\Big)$

let $u = ae^{ix}$

EDIT: then $I'(a) = Re \Big( -\frac{i}{a^{2}} \int^{a}_{a} ue^{u} \ du \Big) = Re(0) = 0$

so $I(a) = C$

but $I(0) = 0$

which means $C=0$

so $I(a) = 0$

and $\int^{2 \pi}_{0} e^{\cos x} \cos (x + \sin x) \ dx = I(1) = 0$

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