Upper bound for $\frac{b^n}{n!}\int_0^\pi x^n(\pi -x)^n \sin x\, dx,\, n\in\mathbb{N}_0,\, b\in\mathbb{N}$

Question

In this article it is argued that $$\frac{b^n}{n!}\int_0^\pi x^n(\pi -x)^n \sin x\, dx\le \frac{\pi b^n}{n!}\left(\frac{\pi}{2}\right)^{2n},\, n\in\mathbb{N}_0,\, b\in\mathbb{N},$$ since $x(\pi -x)\le \frac{\pi ^2}{4}$. But I don't understand why the author didn't use a better bound, namely $\frac{2b^n}{n!}\left(\frac{\pi}{2}\right)^{2n}$, which can be seen by evaluating $$\frac{b^n}{n!}\left(\frac{\pi}{2}\right)^{2n}\int_0^\pi \sin x\, dx.$$ Maybe I'm missing something.