Can this be computed analytically?

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

$$I=\frac12\frac{\Gamma^2(n+1)}{\Gamma(2n+1)}=\frac12\frac1{(2n+1)\binom{2n}n}.$$

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.