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!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    Yes that would probably be the problem then.

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook