Where does the fraction 1/2 come from in this problem?

[waqarrashid33]

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

 

[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]

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.

 

[chiro]

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.

 

[waqarrashid33]

Thanks...

 

[DonAntonio]

Thanks...

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

DonAntonio

 

[Mentallic]

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 $a=\infty$

 

[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 $a=\infty$

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

 

[Mentallic]

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)

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

This should be

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

 

[DonAntonio]

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