Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$, and let $\tilde{h}$ denote its restriction to $\left]0, \frac{1}{2}\right]$. Show that:
$$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^{\frac{1}{2}} h(t)\, dt.$$
Deduce that:
$$\lim_{n \rightarrow +\infty} \sum_{k=1}^{2n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^1 h(t)\, dt.$$