John Wallis and calculating ratios I was asked the following question:

Using Wallis's method, calculate the values $n=0,\frac12,1,\frac32,2,\frac52$ in row $p=\frac32$ of his ratio table.

The Wallis rereferred to here is John Wallis of England (1616-1703). The ratio table looks like this: $$\begin{matrix}p/n&0&1&2&3&4&5&6&7&...\\0&1&1&1&1&1&1&1&1&...\\1&1&2&3&4&5&6&7&8&...\\2&1&3&6&10&15&21&28&36&...\\3&1&4&10&20&35&56&84&120&...\\...&...&...&...&...&...&...&...&...&...&\\\end{matrix}$$
I don't understand how to read this table or how to actually go about solving the question. The table's goal is to find the ratio of the area of the unit square to the area aenclosed in the first quadrant by the curve $y=(1-x^{\frac1p})^n$.
 A: The ratio table represents the ratio of the unit square over the area under the stated curve, $y = (1-x^{1/p})^n$.  We shall assume $p, n \geqslant 0$ and adopt the convention that the area is $1$ when $p=0$.The area of the unit square is always $1$, while the area under the curve is
\begin{align}
A_{p,n} = \int_0^1 (1-x^{1/p})^n ~ dx
\end{align}
and the relevant Wallis number in the table is simply the reciprocal, $W_{p,n} = 1 / A_{p,n}$.  It would seem by convention the area $A_{p,n}$ is taken to be $1$ when $p=0$.
To evaluate the integral, you can begin by making the substitution $x = \sin^{2p} \theta$ for $0 \leqslant \theta \leqslant \pi/2$, so that $dx = 2p\sin^{2p-1}\theta \cos \theta ~d\theta$
\begin{align}
A_{p,n} &= \int_0^{\pi/2} (1-\sin^2\theta)^n \cdot 2p \sin^{2p-1}\theta \cos \theta d\theta \\
&= 2p\int_0^{\pi/2} \cos^{2n+1} \theta \sin^{2p-1}\theta ~d\theta.
\end{align}
In this case you are only interested in a particular instance, $p=3/2, n=0,\frac{1}{2},1, \cdots$, giving
$$A_{p,n} = 2p \int_0^{\pi/2}\sin^2\theta \cos^{2n+1}\theta ~d\theta$$ where $m=2n+1=1,2,3,\cdots$.  Integrate once by parts,
\begin{align}
A_{p,n} = 2p \int_0^{\pi/2} \cos^{2n+1} \theta - \cos^{2n+3}\theta  ~ d\theta .
\end{align}
These integrals are standard, see the Wikipedia entry for Wallis's integral, giving,
\begin{align}
\begin{array}
~A_{3/2,0} = 1 & A_{3/2,1/2} = \frac{3\pi}{16} \\
A_{3/2,1} = \frac{6}{15} & A_{3/2,3/2} = \frac{3\pi}{32} \\
\cdots & \cdots\\
n=k, k=0,1,\cdots & n=k+\frac{1}{2},k=0,1,\cdots\\
A_{3/2,n}=3\frac{2^{2k}k!^2}{(2k+1)!}\frac{1}{2k+3} &
A_{3/2,n}=3\frac{(2k+2)!}{2^{2k+3}(k+1)!^2}\frac{1}{k+2} \frac{\pi}{2}
\end{array}
\end{align}
and the corresponding Wallis ratios are the reciprocals of these numbers.

More generally, you can evaluate $A_{p,n}$ for $p$ and $n$ when either is a multiple of $\frac{1}{2}$ by considering the integral,
$$I_{r,s} = \int_0^{\pi/2} \sin^r\theta \cos^s \theta~d\theta.$$
in which $r, s$ are non-negative integers;  first notice $I_{r,s}=I_{s,r}$.  Evaluate the special cases when $r=0,1, s \geqslant 1$ and $s=0,1, r \geqslant 1$.  Then use integration by parts to obtain the recurrence relation,
$$I_{r,s}=\frac{s-1}{r+1} I_{r+2,s-2}$$
where by symmetry we can take $r \geqslant s \geqslant 2$.  Using the special cases this should give all values for $I_{r,s}$.  Then $A_{p,n} = 2p I_{2p-1,2n+1}$.
