Estimate Error: How Did He Get 4?

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The function is f(x) = sin(2x). Find f'(x) and then f''(x). What is the largest possible value of f''(x)?
 
so because sin 2x cannot be more than 1 between the pi/2 and 0??
 
sin(x) oscillates between -1 and 1. Check on the TI.

Also, sin(nx) , such as sin(3x) , just increases the frequency I think. You can graph that too. It just squooshed the oscillations closer to the y-axis. So they are still in between -1 and 1.
 
o ok i got it thanks
 
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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