Finding Intersection of Curves: sin(x) and 2|x|/pi

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Calculate the are of the finite region enclosed by the curves ## y = \dfrac{2|x|}{\pi} ## and ## y = sin(x) ##

I understand that the integral is ## sin(x) - \dfrac{2x}{\pi} ## however I'm having troubles finding where they intersect, I can only find x= 0, how do I solve ## sin(x) = \dfrac{2x}{\pi} ## ? I can see by observation that 0, pi/2, -pi/2 work but is there a method to find these instead of observation?
 
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hi phospho! :smile:
phospho said:
I can see by observation that 0, pi/2, -pi/2 work but is there a method to find these instead of observation?

sorry, not without a computer or special tables :redface:

but what's wrong with observation? :wink:
 
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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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