Can this be computed analytically? 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.
 A: If 
$$I=\int_{\frac12}^1(x(1-x))^n\,dx$$
then the change of variable $x\mapsto 1-x$ shows that
$$I=\int_0^{\frac12}(x(1-x))^n\,dx$$
and hence
$$2I=\int_0^1(x(1-x))^n\,dx.$$
Now use the Beta function to show that
$$I=\frac12\frac{\Gamma^2(n+1)}{\Gamma(2n+1)}=\frac12\frac1{(2n+1)\binom{2n}n}.$$
A: 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.
A: $$I=\frac{(n+1)n!}{2(\frac{n}{2})!}\int_{\frac{1}{2}}^{1}{[x(1-x)]^\frac{n}{2}}dx=\frac{(n+1)!}{2(\frac{n}{2})!}\int_{\frac{1}{2}}^{1}{[x(1-x)]^\frac{n}{2}}dx$$
so if $$(1-x)^\frac{n}{2}=1-\frac{n^2x^2}{2^2.2!}-\frac{3n^3x^3}{2^3.3!}$$ etc so follow up by multiplying by $x^\frac{n}{2}$ to get the summation then switch the order of the summation and integral. The result should then nicely cancel with $\frac{(n+1)!}{2(\frac{n}{2})!}$ 
A: Turn the integral into a recursion, using integration by parts.
$$I_n=\int_0^1 [x(1-x)]^n dx\\
u'=1,u(x)=1,v(x)=[x(1-x)]^n,v'(x)=n(1-2x)[x(1-x)]^{n-1}\\
=x[x(1-x)]^n|_0^1-n\int_0^1(x-2x^2)[x(1-x)]^{n-1}\,dx\\
=-n\int_0^1\left(-\frac12+\frac12(1-2x)+2(x-x^2)\right)[x(1-x)]^{n-1}\,dx$$
The three terms relate $I_n$  to $I_{n-1}$, and you can eventually reduce it to $I_0=1$
