## Integral Inequality

1. The problem statement, all variables and given/known data
Prove

2. Relevant equations

(π^3)/12≤∫_0^(π/2)▒〖(4x^2)/(2-sinx) dx≥(π^3)/6〗

Also look at atachment

3. The attempt at a solution

I can't get round this one, since when you substitute x by 0 is always 0 and I don't know how to get ∏^3/12
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Attached Thumbnails

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 Recognitions: Science Advisor Hint. Try replacing the sin(x) with a simpler "bounding" function. BTW. Your second inequality is the wrong way around.
 I can't get this one. What bounding function. I'm lost....

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## Integral Inequality

 Quote by diorific I can't get this one. What bounding function. I'm lost....
What are the minimum and maximum values of $\displaystyle \frac{1}{2-\sin(x)}$ for $\displaystyle 0\le x\le \frac{\pi}{2}\ ?$
 Ok, that is $\displaystyle \frac{1}{2}$ ≤ $\frac{1}{2-\sin(x)}$ ≤ 1 But then 0 ≤ 4x2 ≤ π2 So then 0 ≤ $\displaystyle \frac{4x^2}{2-sinx}$ ≤ π2 I know this might be wrong, but I don't really know how to continue.
 Don't take the maximum (or the minimum) of the polynomial. By the mean value theorem, there exists C such that $$\int_{0}^{\pi/2}\frac{4x^2}{2-\sin(x)}\,dx=\frac{1}{2-\sin(c)}\int_{0}^{\pi/2}4x^2\,dx$$ and $0\leq c\leq \pi/2$. Take the integral and maximize/minimize the factor by adjusting C appropriately.

 Quote by Millennial Don't take the maximum (or the minimum) of the polynomial. By the mean value theorem, there exists C such that $$\int_{0}^{\pi/2}\frac{4x^2}{2-\sin(x)}\,dx=\frac{1}{2-\sin(c)}\int_{0}^{\pi/2}4x^2\,dx$$ and $0\leq c\leq \pi/2$. Take the integral and maximize/minimize the factor by adjusting C appropriately.
thank you so much.

I finally managed to resolve this!!!

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