# Easy integral?

#### Drexel28

MHF Hall of Honor
Compute $$\displaystyle \int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right)dx$$

#### simplependulum

MHF Hall of Honor
Compute $$\displaystyle \int_0^{2\pi}e^{\cos(x)}\cos\left(x+\sin(x)\right)dx$$

Is the integrand the derivative of

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

#### tonio

Is the integrand the derivative of

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

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

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

Tonio

#### simplependulum

MHF Hall of Honor
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 .(Happy)

• Bruno J.

#### Drexel28

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

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

Tonio
Just tinkering

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 .(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.

#### Random Variable

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

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