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Integration homework

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Find an expression for

    [tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex]

    Hence prove that
    [tex]\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2} [/tex]


    2. Relevant equations

    [tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex]
    and
    [tex]\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2} [/tex]

    3. The attempt at a solution

    I found an expression for [tex]\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta) [/tex] which was [tex]\sum_{k=1}^{n}1-2sin^{2}k\theta[/tex] but couldn't continue because it doesn't look like the appropriate expression.
     
    Last edited: Mar 14, 2013
  2. jcsd
  3. Mar 14, 2013 #2

    SammyS

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    What is 1 - sin2(u) ?
     
  4. Mar 14, 2013 #3
    oh it should have been 1-2sin^2u
    edited*
     
  5. Mar 14, 2013 #4

    SammyS

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    cos(2u) = ?
     
  6. Mar 15, 2013 #5
    Yes, that is how I go 1-2sin^2(u) in the first place (because the integral had sin) but I don't see how it can help me evaluate the integral :(
     
  7. Mar 15, 2013 #6
    Do you know the identity that [itex]cos θ= \frac{e^{iθ} + e^{-iθ}}{2}[/itex]?
     
  8. Mar 15, 2013 #7
    It seems helpful to use this identity to reduce the question to geometric progression, then what it takes is some computation.
     
  9. Mar 15, 2013 #8

    vela

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    Along the same lines as raopeng's hint, but a little bit less tedious:

    ##\cos 2x + \cos 4x + \cdots + \cos (2nx)## is equal to the real part of ##e^{2ix} + e^{4ix} + \cdots + e^{2nix}##
     
  10. Mar 16, 2013 #9
    How does that help me do the integral though?
     
  11. Mar 16, 2013 #10

    vela

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    If you calculate the sum correctly, it'll become clear.
     
  12. Mar 16, 2013 #11
    I have never learnt how to express cis in terms of e so can you please give me a hint on how to do it?
    Thank you!
     
  13. Mar 16, 2013 #12

    vela

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    Not really sure what kind of hint you're looking for.
     
  14. Mar 16, 2013 #13
    How do I find the real part of e^2ix...
     
  15. Mar 16, 2013 #14

    vela

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    Use Euler's formula, keeping in mind that ##x## is real.
     
  16. Mar 17, 2013 #15
    I get
    e^2ix=cos2x+ison2x
    e^4ix=cos4x+ison4x
    .
    .
    .
    e^nix=cosnx+isonnx

    I still don't see how this can help with the integral :(
     
  17. Mar 17, 2013 #16

    vela

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    Hint: ##e^{2ix} + e^{4ix} + \cdots + e^{2nix} = (e^{2ix})^1 + (e^{2ix})^2 + \cdots + (e^{2ix})^n##. You have a geometric progression on your hands, as raopeng noted earlier.
     
  18. Mar 18, 2013 #17
    Ok, so I get [tex]\frac{e^{ix}+e^{-ix}}{2} =cosx [/tex]
    [tex]\frac{e^{ix}-e^{-ix}}{2} =sinx[/tex]
     
  19. Mar 18, 2013 #18
    How can I make use of these identities now?
     
  20. Mar 18, 2013 #19

    vela

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    We're not here to spoon-feed you the solution. You need to show some initiative at solving the problem yourself.
     
  21. Mar 18, 2013 #20
    I know this is dumb but this is what I did :(
    [tex]\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx[/tex]
     
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