# How this comes?

by waqarrashid33
Tags: None
 P: 77 Let me know how 1/2 comes from it.see attachemet Attached Thumbnails
 HW Helper P: 3,326 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.
 P: 77 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).
HW Helper
P: 3,326

## How this comes?

 Quote by 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).
Yes that would probably be the problem then.
$$\int1+\cos(2t)dt=t+\frac{1}{2}\sin(2t)$$

Start from there.
P: 4,542
 Quote by 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).
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.
 P: 77 Thanks....
P: 606
 Quote by waqarrashid33 Thanks....

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

DonAntonio
HW Helper
P: 3,326
 Quote by DonAntonio Interesting: you don't even need $\,\,T\to\infty\,\,$. It is 1/2 for any $\,\,T\neq 0\,$. DonAntonio
Not quite, the final steps of the solution are to simplify $$\frac{1}{2}\left(1+\lim_{T\to a}\frac{\sin(2T)}{2T}\right)$$

and that expression is only equal to 1/2 if $$\lim_{T\to a}\frac{\sin(2T)}{2T}=0$$ which only happens for $a=\infty$
P: 606
 Quote by Mentallic Not quite, the final steps of the solution are to simplify $$\frac{1}{2}\left(1+\lim_{T\to a}\frac{\sin(2T)}{2T}\right)$$ and that expression is only equal to 1/2 if $$\lim_{T\to a}\frac{\sin(2T)}{2T}=0$$ which only happens for $a=\infty$

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

DonAntonio
HW Helper
P: 3,326
 Quote by DonAntonio I don't know how you got that. I get $$\int_{-T}^T \cos^2(t)dt=\left[\frac{t+\cos t\sin t}{2}\right]_{-T}^T$$
How did you get that?

$$\cos^2t=\frac{1}{2}\left(1+\cos(2t)\right)$$

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

 Quote by DonAntonio $$\left(-T-\cos(-T)\sin(-T)\right)$$
This should be

$$\left(-T+\cos(-T)\sin(-T)\right)$$
P: 606
 Quote by Mentallic How did you get that? $$\cos^2t=\frac{1}{2}\left(1+\cos(2t)\right)$$ Oh ok I see what you have, after integrating you converted sin(2t) to 2sin(t)cos(t) This should be $$\left(-T+\cos(-T)\sin(-T)\right)$$

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