Prove Integral Inequality: π^3/12≤∫_0^(π/2)

[diorific]

Homework Statement

Prove

Homework Equations

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

Also look at atachment

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

 

[uart]

Hint. Try replacing the sin(x) with a simpler "bounding" function.

BTW. Your second inequality is the wrong way around.

 

[diorific]

I can't get this one. What bounding function. I'm lost...

 

[SammyS]

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 $\displaystyle 0\le x\le \frac{\pi}{2}\ ?$

 

[diorific]

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.

 

[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.

 

[diorific]

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!