Integrating cos(x)cos(nx): Zero or Non-Zero?

[swarog46]

Homework Statement

[itex]\int^{2\pi}_0{cos(x)cos(nx)dx}=?,n\in Z[/itex]

Homework Equations

The Attempt at a Solution

I think that this eq has zero solve. However my teacher says that its incorrect.

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

 

[Infinitum]

Hi swarog46, welcome to PF!

[itex]\int^{2\pi}_0{cos(x)cos(nx)dx}=\frac{1}{2n}(sin(2 \pi (1-n))+sin(2 \pi (1+n)))[/itex]

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...

 

[swarog46]

of course i mean that=)

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

 

[Infinitum]

Did you read this?

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

Your integral itself is incorrect.

 

[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 [itex]\int \sin^2x+\cos^2x=\int 1[/itex], which is a quick way to rememeber your integral of [itex]\cos^2 x[/itex] is ∏.