= Wallis integrals
{c}
{title2=$I_n=\int_0^{\pi/2}\sin^n x\,dx$}
The <Wallis integrals> are $I_n=\int_0^{\pi/2}\sin^n x\,dx$ for integers $n\geq0$. <Integration by parts> gives $nI_n=(n-1)I_{n-2}$ for $n\geq2$, with $I_0=\pi/2$, $I_1=1$. Hence
$$
I_{2n}=\frac{(2n)!}{2^{2n}(n!)^2}\frac\pi2,\qquad
I_{2n+1}=\frac{2^{2n}(n!)^2}{(2n+1)!}.
$$
Because $0<\sin x<1$ in the integration interval, $0<I_n<I_{n-1}$ for $n\geq1$. 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>.
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