Countour integration (cosx)^2n

[Kate2010]

Homework Statement

Use contour integrals and justify your steps, find \int cos²ⁿx dx where the integral is from 0 to 2pi.

Homework Equations

The Attempt at a Solution

My first thought was that this integral would be zero, using the residue theorem with residue 0, as I'm integrating round a closed curve. I think this is wrong but I don't know why.

I then decided to write f(x) = cos ²ⁿx = [e^ix + e^-ix / 2i]²ⁿ
I could use the substitution z=e^ix to get f(z) = [1/2i (z + 1/z)]²ⁿ, this would have a singularity at z=0 but as z is an exponential it is never 0.

I thought about using a keyhole contour with f(z) as above, but I'm really not sure.

 

[vela]

Homework Statement

Use contour integrals and justify your steps, find \int cos²ⁿx dx where the integral is from 0 to 2pi.

Homework Equations

The Attempt at a Solution

My first thought was that this integral would be zero, using the residue theorem with residue 0, as I'm integrating round a closed curve. I think this is wrong but I don't know why.

I then decided to write f(x) = cos ²ⁿx = [e^ix + e^-ix / 2i]²ⁿ
I could use the substitution z=e^ix to get f(z) = [1/2i (z + 1/z)]²ⁿ, this would have a singularity at z=0 but as z is an exponential it is never 0.

I thought about using a keyhole contour with f(z) as above, but I'm really not sure.

It doesn't matter that the e^ix is never 0 on the contour. All that matters is that the singularity lies inside the closed contour.

Don't forget to convert dx into dz as well. It'll introduce another factor of z.