# Homework Help: Integrate problem

1. May 27, 2012

### swarog46

1. The problem statement, all variables and given/known data
$\int^{2\pi}_0{cos(x)cos(nx)dx}=?,n\in Z$

2. Relevant equations

3. The attempt at a solution
I think that this eq has zero solve. However my teacher says that its incorrect.

$cos(x) cos(nx) = 1/2 (cos (x-nx) + cos (x+nx))$
$\int^{2\pi}_0{cos(x)cos(nx)dx}=\frac{1}{2n}(sin(2 \pi (1-n))+sin(2 \pi (1+n)))$ which gives 0 for n=1, but when i give n=1 before start integrating, then i've got non zero solve.

2. May 27, 2012

### Infinitum

Hi swarog46, welcome to PF!!

The step in the quote is wrong. How did you do this? Try substitution of cos(1-n)x = t? and similarly for the other term? This will give you a (n-1) and (n+1) in the denominator. Recheck that step

PS : I believe by 'zero solve' you mean zero as the solution...

3. May 27, 2012

### swarog46

of course i mean that=)

Whatever in the result for n=1 i get \int=0.
and for $\int^{2\pi}_0{cos(x)cos(1*x)dx}=\int^{2\pi}_0{cos^{2}(x)}=\pi$

4. May 27, 2012

### Infinitum

Did you read this?

Your integral itself is incorrect.

5. May 27, 2012

### algebrat

One way to believe this, is that for n>1, cos(nx) is "even" about the midpoint ∏, while cos(x) is "odd" about the midpoint ∏, so it integrates to zero.

Also $\int \sin^2x+\cos^2x=\int 1$, which is a quick way to rememeber your integral of $\cos^2 x$ is ∏.