Finding the best upper bound of $\int_a^b\frac{\sin^2x}x\,dx$ by hand

Question

Q: Without using any series expansions, find, by hand, the best upper bound of $$\int_a^b\frac{\sin^2x}x\,dx.$$

Attempt:

An exact expression would not be expected since we would be dealing $\text{Si}(x)$ after integrating by parts.

Putting $f(x)=\sin^2x$ and $g(x)=1/x$ into C-S inequality for integrals, we have that $$\int_a^b\frac{\sin^2x}x\,dx\le\sqrt{\left(\int_a^b\frac1{x^2}\,dx\right)\left(\int_a^b\sin^4x\,dx\right)}$$ and using the elementary identity that $$\int\sin^4x\,dx=\frac{3x}8-\frac{\sin2x}4+\frac{\sin4x}{32}+C$$ we get $$\begin{align}\int_a^b\frac{\sin^2x}x\,dx&\le\frac1{4\sqrt2}\sqrt{\left(\frac1a-\frac1b\right)\left(12(b-a)-8(\sin2b-\sin2a)+\sin4b-\sin4a\right)}\\&\le\frac1{4\sqrt{2ab}}\sqrt{12(a-b)^2-8(a-b)(\sin2a-\sin2b)-(\sin4a-\sin4b)}\end{align}$$

This measures how well the upper bound fits the integral at different limits.

Can it be improved?

Answer 1

My first upper bound would be $\ln(b)-\ln(a)$. Of course this ignores the variation in $\sin^2$. It would be interesting to compare this with the C-S bounds for various values of a and b.