TITLE: Can this be computed analytically?
QUESTION [6 upvotes]: For even values of $n$, can the following be proven analytically?
$$(n+1)\binom{n}{\frac{n}2}\int_{\frac12}^1[x(1-x)]^{\frac{n}2}dx=\frac12$$
I cannot seem to compute this analytically. Is it possible? I can compute this numerically for various values of $n$:
In [202]: from sympy import binomial as bnm; from numpy import power as pw
In [203]: wp=lambda n=10: (n+1)*bnm(n,int(n/2))*I(lambda x:pw(x*(1-x),int(n/2)),1/2,1)
In [204]: wp(n=18),wp(20),wp(26),wp(32),wp(42),wp(48),wp(52)
Out[204]: 
    (0.500000000000000,
     0.500000000000000,
     0.500000000000000,
     0.500000000000000,
     0.500000000000000,
     0.500000000000001,
     0.500000000000000)

Please note that this question is not a duplicate of Prove: $\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx$ for $0 \leq k \leq n$
The reason that this question is not a duplicate of that question is the different integration limits. The different integration limits completely alter the question as well as the solution.

REPLY [2 votes]: Since you're only interested in even $n$, you may as well ask if
$$(2m+1){2m\choose m}\int_{1/2}^1(x(1-x))^m\,dx={1\over2}$$
Letting $x=(1+u)/2$ and keeping only the integral on the left hand side changes the identity to prove into
$$I(m)=\int_0^1(1-u^2)^m\,du={4^m\over(2m+1){2m\choose m}}$$
This can be proved by induction.  We certainly have 
$$I(0)=\int_0^1du=1={4^0\over(2\cdot0+1){0\choose0}}$$
and integration by parts gives
$$\begin{align}
I(m+1)&=\int_0^1(1-u^2)^{m+1}\,du\\
&=u(1-u^2)^{m+1}\Big|_0^1+2(m+1)\int_0^1u^2(1-u^2)^m\,du\\
&=2(m+1)\int_0^1(1-(1-u^2))(1-u^2)^m\,du\\
&=2(m+1)(I(m)-I(m+1))
\end{align}$$
hence 
$$(2m+3)I(m+1)=2(m+1)I(m)={2(m+1)4^m\over(2m+1){2m\choose m}}={4^{m+1}\over{2(m+1)\choose m+1}}$$
The final equality here is verified by observing that
$${1\over4}{2m+2\choose m+1}={(2m+2)(2m+1)(2m)!\over4(m+1)^2(m!)^2}={2m+1\over2(m+1)}{2m\choose m}$$
This completes the induction.