= Solution
Write $B(n,n)=\int_0^1[x(1-x)]^{n-1}dx$ and set $v=2x-1$. Since $x(1-x)=(1-v^2)/4$ and $dx=dv/2$,
$$
B(n,n)=2^{1-2n}\int_{-1}^{1}(1-v^2)^{n-1}dv
=2^{2-2n}\int_0^1(1-v^2)^{n-1}dv.
$$
In the last integral put $u=v^2$, giving $dv=du/(2\sqrt u)$. This yields $B(n,n)=2^{1-2n}B(1/2,n)$, and part (a) interchanges the arguments. Hence
$$
\boxed{B(n,n)=\frac{B(n,1/2)}{2^{2n-1}}.}
$$
The <equal-argument beta integral> identity holds for every positive real $n$, not only positive integers.
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