# Integral Inequality

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

## The Attempt at a Solution

#### Attachments

• Inequality.jpg
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uart
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....

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

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!!!