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How this comes?

  1. May 5, 2012 #1
    Let me know how 1/2 comes from it.see attachemet

    Attached Files:

  2. jcsd
  3. May 5, 2012 #2


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    Have you tried anything for yourself?

    Solve the integral and see if you can simplify the limit to transform the expression into a form that you know the answer to.
  4. May 5, 2012 #3
    i tried a lot but answer goes wrong..
    i didn't touched with calculus since long time...May be it is because of this...
    For the integral of 1+cos(2t) gives 2T+sin(2T).
  5. May 5, 2012 #4


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    Yes that would probably be the problem then.

    Start from there.
  6. May 5, 2012 #5


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    Try solving your improper integral (for 1 + cos(2t) you will get anti-derivative t + 1/2sin(2t)) and then expand into F(T) - F(-T), collect your constant outside of the integral (1/2T) bring it together and you will get a limit expression in terms of T where T goes to infinity.

    There are limit theorems you can use to solve this also.
  7. May 5, 2012 #6
  8. May 5, 2012 #7

    Interesting: you don't even need [itex]\,\,T\to\infty\,\,[/itex]. It is 1/2 for any [itex]\,\,T\neq 0\,[/itex].

  9. May 5, 2012 #8


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    Not quite, the final steps of the solution are to simplify [tex]\frac{1}{2}\left(1+\lim_{T\to a}\frac{\sin(2T)}{2T}\right)[/tex]

    and that expression is only equal to 1/2 if [tex]\lim_{T\to a}\frac{\sin(2T)}{2T}=0[/tex] which only happens for [itex]a=\infty[/itex]
  10. May 5, 2012 #9

    I don't know how you got that. I get
    [tex]\int_{-T}^T \cos^2(t)dt=\left[\frac{t+\cos t\sin t}{2}\right]_{-T}^T=\frac{1}{2}\left[T+\cos T\sin T-\left(-T-\cos(-T)\sin(-T)\right)\right]=\frac{2T}{2}=T[/tex]
    as [itex]\,\,\cos(-T)\sin(-T)=-\cos T\sin T\,\,[/itex] , and then
    [tex]\frac{1}{2T}\int^T_{-T}\cos^2 t\,dt=\frac{1}{2}[/tex]
    like that, without limit...

  11. May 5, 2012 #10


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    How did you get that?


    Oh ok I see what you have, after integrating you converted sin(2t) to 2sin(t)cos(t)

    This should be

    Last edited: May 5, 2012
  12. May 5, 2012 #11

    Yes indeed. So much worrying about the change of sign in the sine of -T that I forgot I had that minus sign out of the parentheses. Thanx.

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