How this comes?

1. May 5, 2012

waqarrashid33

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

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2. May 5, 2012

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.

3. May 5, 2012

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).

4. May 5, 2012

Mentallic

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

Start from there.

5. May 5, 2012

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.

6. May 5, 2012

Thanks....

7. May 5, 2012

DonAntonio

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

DonAntonio

8. May 5, 2012

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 $a=\infty$

9. May 5, 2012

DonAntonio

I don't know how you got that. I get
$$\int_{-T}^T \cos^2(t)dt=\left[\frac{t+\cos t\sin t}{2}\right]_{-T}^T=\frac{1}{2}\left[T+\cos T\sin T-\left(-T-\cos(-T)\sin(-T)\right)\right]=\frac{2T}{2}=T$$
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

10. May 5, 2012

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

$$\left(-T+\cos(-T)\sin(-T)\right)$$

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

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