We may prove without using integrals that $$ \frac1{\sqrt{\pi n}}\geqslant 4^{-n}{2n\choose n}\geqslant \frac{1}{\sqrt{\pi(n+1/2)}}.\quad\quad (1) $$ (1) is equivalent to $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\choose n}\right)^2=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)$$ (the identities are straightforward by induction).
Denote $m:=2n+1$. It is not hard to show that $\sin mx=p_m(\sin x)$ for a polynomial of degree $m$ in $\sin x$. The roots of $p_m$ are $\sin \frac{k\pi}{2n+1}$ for $k=-n,\ldots,n$. Thus $$\sin (2n+1)x=(2n+1)\sin x\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{k\pi}{2n+1}}\right)$$ (the multiple $2n+1$ comes from dividing by $x$ and putting $x=0$). Put $x=\frac{\pi}{4n+2}$. Using the inequality $\sin tx\leqslant t\sin x$ (which may be proved, for example, by induction in $t=1,2,\ldots$ from the identity $\sin(t+1)x=\sin tx\cos x+\sin x\cos tx$) for $t=2k$ ($k=1,2,\ldots,n$) we get $$ 1\leqslant (2n+1)\sin\frac{\pi}{4n+2}\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\leqslant \frac{\pi}{2}c_n,\quad c_n\geqslant \frac{2}\pi. $$ Analogously we get $$ \cos 2nx=\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{\pi(2k-1)}{4n}}\right). $$
Dividing by $1-\frac{\sin^2 x}{\sin^2 \frac{\pi}{4n}}$ and substituting $x=\frac{\pi}{4n}$ (for computing the LHS at this point use l'Hôpital rule) we get $$ n\tan\frac{\pi}{4n}\leqslant \prod_{k=2}^n\left(1-\frac1{(2k-1)^2}\right)=\frac1{2d_n},\quad d_n\leqslant \frac1{2n\tan \frac{\pi}{4n}}\leqslant \frac2\pi. $$
UPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $\sum 1/n^2=\pi^2/6$ and $\sum (-1)^{k-1}/(2k-1)=\pi/4$ using these trigonometric things.)