Integrating sin2θ: How to Prove <sin2θ>=1/2 and <cos2θ>=1/2?

gennarakis
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How can <sin2θ>=1/2 and <cos2θ>=1/2

How is the proof made?Integrate sin2θ from -Infinity to +Infinity?
 
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gennarakis said:
How can <sin2θ>=1/2 and <cos2θ>=1/2

How is the proof made?Integrate sin2θ from -Infinity to +Infinity?

You only need to integrate over one cycle. Every other cycle will be the same, right?
 
Average value on what interval?
 
I just integrated from 0 to 2Pi changed sin2θ=(1-cos2θ)/2 but the result is Pi and not 1/2...:confused:
 
gennarakis said:
I just integrated from 0 to 2Pi changed sin2θ=(1-cos2θ)/2 but the result is Pi and not 1/2...:confused:

You forgot to divide by 2*PI to get the average over the interval...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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