Proving \frac{2}{Pi}x<sin x for 0<x<\frac{Pi}{2}

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\frac{2}{Pi}x&lt;sin x for 0<x<\frac{Pi}{2}

I dint figure out how to write Pi in symbol. I don't have any idea what to do with this one.
 
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For \pi use [ itex ]\pi[ /itex ]. In general clicking on LaTex in any post will show the code for it.

To show that \left(\frac{2}{\pi}\right)x\le sin(x) for x&lt; \frac{\pi}{2}, look at the function f(x)= sin(x)- 2x/\pi. It is 0 at both x= 0 and x= \pi/2. It's only critical point is where f&#039;(x)= cos(x)- 2/\pi= 0 or cos(x)= 2/\pi. Since 2\pi is less than one, that occurs at some point between 0 and \pi/2. Further, f"(x)= -sin(x) which is negative for all x between 0 and \pi/2
 
OK thanks a lot.
 
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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