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how to find the average of sin square theta?

Physics
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Don't get it :S
me neither...how can u find the average of just one quantity?
You need \(<\sin^2\theta>\) right?

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not a quantity.
*just one term
average must be 1/2 ? :P
how you can say that?
@Vaidehi09 basically that means average value of \(\sin^2\theta\) in its period - which is \(\pi\). (the graph for rekindling old memories:)|dw:1343221614291:dw|
is this graph applicable for that one? @apoorvk
You can try integration
please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?
\[1\div(b-a)\int\limits_{a}^{b}\sin^2\theta d \theta \]
For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),
take a = 0 and b = 2pi
no bro @apoorvk it should be 2pi
@apoorvk i cant understand why the period is pi.
\[\int\limits_0^{\pi}\sin^2\theta.d\theta\] and then I divide this by the difference of the limit, that is \(\pi -0=\pi\). Basically like the sum of the values by the 'domain covered'.
@Vaidehi09, do you understand that the period sinx of 2pi?
yes
okay. thanks to all.
now, @Vaidehi09 when I square the function - two changes happen to curve - 1. The curve's shape changes. 2. The negative part of the graph is flipped up in the positive axis. So, whereas the period was 2 pi for sinx, the \(\sin^2x\) graph repeats itself after every \(\pi\)-distance. Hence the period is pi.
OmG ! Expert ! xD Keep it up brother.
Not really :P
@apoorvk ah yes..of course! i was getting confused with the -ve part and what kind of change will happen to the shape of the graph....but i get it now. thanks a lot! :))
No worries :]
I am new to calc.. so I am not sure about my replies :P
@apoorvk good Explaination.

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