Steps for Solving an Inequality Proof

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the question and how i tried to solve it are in this link

http://img155.imageshack.us/img155/4710/55108660qi1.gif
 
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You need a little bit more. It is true that
x- \frac{x^3}{3!}+ \frac{x^5}{5!}> x- \frac{x^3}{3!}
but you have to fit
sin(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}
into that. Notice that because this is an alternating series with decreasing terms, the entire "tail" (all terms past "n") is less than the difference between the n-1 and n terms.
 
what is the steps for the solution?
 
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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