Finding bounds for a function I would like to show that
$$\frac{1}{\pi^2}<\int_{\pi/2}^\pi\frac{\sin x}{x^3}<\frac{3}{2\pi^2}.$$
We know that $\frac{\sin x}{x^3}\leq\frac{1}{x^3}$ and integrating gives $\int_{\pi/2}^\pi\sin x/x^3\leq3/(2\pi^2)$. I don't know how to make $\leq$ into $<$. On the other hand, $\frac{1}{\pi^3}\leq\frac{\sin x}{x^3}$  and integrating gives $1/\pi^3\leq\int_{\pi/2}^\pi\sin x/x^3$. But $1/\pi^3<1/\pi^2$ so this clearly is not what I want.
Any suggestions in how to fix these issues?
 A: 
I don't know how to make $≤$ into $<$

Notice that $\frac{\sin(x)}{x^3}$ and $\frac{1}{x^3}$ are only equal at $\frac{\pi}{2}$. So if we split $\left[\frac{\pi}{2} , \pi \right]$ as $\left[\frac{\pi}{2} , \frac{3\pi}{4} \right] \cup \left[\frac{3\pi}{4}  , \pi \right]$ for example, we can thus assert that
\begin{align}
\frac{\sin(x)}{x^3}\le \frac{1}{x^3} \text{ on }\left[\frac{\pi}{2} , \frac{3\pi}{4} \right]\\
\frac{\sin(x)}{x^3}\mathbin{\color{red}{<}} \frac{1}{x^3} \text{ on }\left[\frac{3\pi}{4}, \pi \right]
\end{align}
And hence
\begin{align*}
\int_{\frac{\pi}{2}}^\pi\frac{\sin x}{x^3} \, \mathrm{d}x &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\frac{\sin x}{x^3} \, \mathrm{d}x + \int_{\frac{3\pi}{4}}^{\pi}\frac{\sin x}{x^3} \, \mathrm{d}x \\
& \le \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\frac{1}{x^3} \, \mathrm{d}x + \int_{\frac{3\pi}{4}}^{\pi}\frac{\sin x}{x^3} \, \mathrm{d}x \\
& =\frac{10}{9 \pi^2} + \int_{\frac{3\pi}{4}}^{\pi}\frac{\sin x}{x^3} \, \mathrm{d}x\\
&\mathbin{\color{red}{<}}\frac{10}{9 \pi^2} + \int_{\frac{3\pi}{4}}^{\pi}\frac{1}{x^3} \, \mathrm{d}x\\
& = \frac{3}{2\pi^2}
\end{align*}
as desired.

On the other hand $\frac{1}{\pi^3}\leq\frac{\sin x}{x^3}$.

This is not the inequality you want. Notice that on $\left[\frac{\pi}{2}, \pi \right]$ we know $\sin(x)$ is concave, which means it can be bounded from below by the line passing through the endpoints. This line turns out to be
$$
y\ =\ -\frac{2}{\pi}\left(x-\pi\right)
$$
So we can thus say that
$$
\frac{-\frac{2}{\pi}\left(x-\pi\right)}{x^{3}} \le \frac{\sin(x)}{x^3} \text{ on }\left[\frac{\pi}{2} , \pi \right]
$$
And since $ \int_{\frac{\pi}{2}}^\pi\frac{-\frac{2}{\pi}\left(x-\pi\right)}{x^{3}}  \, \mathrm{d}x =\frac{1}{\pi^2} $ this gives the desired result.
Lastly, if you're again worried about the $\le$ instead of $<$, then you can use the same trick of splitting the interval and then evaluating the integrals separately.
A: Just for the fun.
Without any special function, we can have a more than decent approximation of the integral using, as an approximation,
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed, more than $\large 1,400$ years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
This gives
$$\int_{\frac \pi 2}^\pi \frac {sin(x)}{x^3}\,dx \sim  \int_{\frac \pi 2}^\pi \frac{16 (\pi -x)}{x^2 \left(5 \pi ^2-4 (\pi -x) x\right)}\,dx$$
$$5 \pi ^2-4 (\pi -x) x=4 (x-a)(x-b)$$ where
$$a=\left(\frac{1}{2}-i\right) \pi \qquad \text{and} \qquad b=\left(\frac{1}{2}+i\right) \pi$$
$$\frac{16 (\pi -x)}{4(x-a)(x-b)}=-\frac{4 (a b-\pi ( a+  b))}{a^2 b^2 x}+\frac{4 (\pi -a)}{a^2 (a-b) (x-a)}-\frac{4
   (\pi -b)}{b^2 (a-b) (x-b)}+\frac{4 \pi }{a b x^2}$$ Simple integrals leading to
$$\int_{\frac \pi 2}^\pi \frac{16 (\pi -x)}{x^2 \left(5 \pi ^2-4 (\pi -x) x\right)}\,dx=\frac{4 \pi ^2}{25}\left(20-11 \pi -8 \log (2)+2 \log (5)+22 \tan ^{-1}(2)\right)$$ which is $0.121155$ while the exact value is $0.121210$ (so a relative error of  $0.045$%).
