1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrate problem

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\int^{2\pi}_0{cos(x)cos(nx)dx}=?,n\in Z[/itex]

    2. Relevant equations

    3. 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.
  2. jcsd
  3. May 27, 2012 #2
    Hi swarog46, welcome to PF!! :biggrin:

    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 :smile:

    PS : I believe by 'zero solve' you mean zero as the solution...
  4. May 27, 2012 #3
    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]
  5. May 27, 2012 #4
    Did you read this?

    Your integral itself is incorrect.
  6. May 27, 2012 #5
    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 ∏.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook