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?

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  $0.045$%).
