Wallis integrals ($I_n={\int }_0^{\pi /2}{\sin }^{n}xdx$)

The Wallis integrals are $I_n={\int }_0^{\pi /2}{\sin }^{n}xdx$ for integers $n\ge 0$. Integration by parts gives $n{I}_n=(n-1)I_{n-2}$ for $n\ge 2$, with $I_0=\pi /2$, $I_1=1$. HenceBecause $0<\sin x<1$ in the integration interval, $0<I_n<I_{n-1}$ for $n\ge 1$. Combining this with the recurrence gives $2n/(2n+1)<I_{2n+1}/I_{2n}<1$, proving that the ratio tends to one and yielding the Wallis product.