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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 ∏.
     
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