# Homework Help: Using Cauchy integral formula to compute real integral?

1. Oct 21, 2012

### Mixer

1. The problem statement, all variables and given/known data

Compute the following integral around the path S using Cauchys integral formula for derivatives:

$\int$ez / z2

Integral path S is a basic circle around origin.

Then, use the result to compute the following integral

$\int$ ecos (x) cos(sin (x) - x) dx from 0 to ∏

2. Relevant equations

Cauchy integral formula: http://en.wikipedia.org/wiki/Cauchy's_integral_formula

3. The attempt at a solution

I was able to compute the first part and got 4∏i. But I'm stuck in second part. How I'm supposed to use the result I got in first part to compute the second? Any hint for me?

2. Oct 21, 2012

### jackmell

Well, you know $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ and you got $\cos(a-b)$ in there. Ok, keep that in mind.

Now, what do you get when you let $z=e^{it}$ in the expression:

$$\int \frac{e^z}{z^2}dz$$

and then split the integral into a real and imaginary part?

3. Oct 21, 2012

### Gib Z

Hint 1: $\int_{\gamma} f(z) dz = \int^b_a f( \gamma(t) ) \gamma'(t) dt.$

Hint 2: $\cos(\sin x - x) = \operatorname{Re} e^{i\sin x - ix}.$

4. Oct 21, 2012

### Mixer

Thank you very much!

I actually had a typo in my first post. So the integral I'm supposed to compute is

$\int$e2 cos (t)cos(sin(t) - t) dt from 0 to ∏

I was, however, able to compute the integral you gave me to this kind of form (this is an imaginary part of the integral:

$\int$e2 cos (t) cos (sin(t) - t) dt from 0 to 2∏

is equal to 4∏ (the result I got in first part)

Now, how can I get the correct value when integrating from 0 to ∏ ?