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How this comes?by waqarrashid33
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#1
May512, 03:13 AM

P: 77

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



#2
May512, 03:21 AM

HW Helper
P: 3,516

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
May512, 03:33 AM

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). 


#4
May512, 03:46 AM

HW Helper
P: 3,516

How this comes?
[tex]\int1+\cos(2t)dt=t+\frac{1}{2}\sin(2t)[/tex] Start from there. 


#5
May512, 03:46 AM

P: 4,572

There are limit theorems you can use to solve this also. 


#6
May512, 05:33 AM

P: 77

Thanks....



#7
May512, 08:25 AM

P: 606

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


#8
May512, 08:40 AM

HW Helper
P: 3,516

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] 


#9
May512, 10:42 AM

P: 606

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 


#10
May512, 11:28 AM

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P: 3,516

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


#11
May512, 01:17 PM

P: 606

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 


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