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Homework Help: 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]
     
  22. Mar 18, 2013 #21
    I know, you are not allowed to solve the question form me but if the student is stuck shouldn't you give some hints? :(
     
  23. Mar 18, 2013 #22
    It is already very close! Remember we have the identity that [itex]sinθ=\frac{e^iθ-e^{-iθ}}{2i}[/itex], how does this relate to the integration you wrote?

    BTW if you find such problems difficult, given that you haven't learned about Euler's Equation, it is perhaps more productive to have a better grasp of the material. It is indeed better than asking as it will improve your understanding on the topic a lot more. And all the joy of revelation after solving a problem will be the greatest when you figure it out completely on your own :)
     
  24. Mar 18, 2013 #23
    The denominator is equal to 2isinx and the numerator =2isin(2n+1)x?
    How can I integrate the integral I have in post 20 though, it looks very hard :(
     
  25. Mar 18, 2013 #24
    yes and it would equal integrating a sum of cos because you just prove that these two expressions are equivalent. And integrating separately the cos in the series is an easy job.
     
  26. Mar 18, 2013 #25
    But why is
    [tex]\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx[/tex]
    equal to the cos2x series? Isn't this the imaginary part of e^2ix+e^4ix+...+e^2nix and the cos2x series the real part?
     
    Last edited: Mar 18, 2013
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