How this comes?

waqarrashid33

Let me know how 1/2 comes from it.see attachemet

Attached Thumbnails

 

Mentallic

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.

waqarrashid33

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).

Mentallic

Yes that would probably be the problem then.
[tex]\int1+\cos(2t)dt=t+\frac{1}{2}\sin(2t)[/tex]

Start from there.

chiro

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.

waqarrashid33

Thanks....

DonAntonio

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

DonAntonio

Mentallic

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]

DonAntonio

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...

DonAntonio

Mentallic

How did you get that?

[tex]\cos^2t=\frac{1}{2}\left(1+\cos(2t)\right)[/tex]

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

This should be

[tex]\left(-T+\cos(-T)\sin(-T)\right)[/tex]

DonAntonio

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.

DonAntonio