Without doing any calculus of an integral containing a sine, verify that $\displaystyle \int_0^{\frac{\pi}{2}}xsin(x)\leq\frac{\pi^2}{8}$. My idea was by bounding the integral to get something greater but still lesser than $\displaystyle \frac{\pi^2}{8}$. So I thought about bounding $\displaystyle \int_0^{\frac{\pi}{2}}xsin(x)$ this way : $\displaystyle \int_0^{\frac{\pi}{2}}xsin(x)\leq \int_0^{\frac{\pi}{2}}x=\frac{\pi}{2}$, which is unfortunately greater than $\displaystyle \frac{\pi^2}{8}$. I have no other idea of how to procede. Any idea is welcome.