I disagree with correct solution in the OP and will present my own analysis for your scrutiny. First, let's reduce the clutter and set $a=E_n$. We seek to find $a$ such that
$$2\int_0^{a^{1/4}}\sqrt{a-x^4}dx=\left(n+\frac{1}{2}\right)$$
Now let $x^4=at$ or $x=(at)^{1/4}$, then
$$
dx=\frac{a^{1/4}}{4}t^{-3/4}dt\\
\sqrt{a-x^4}=\sqrt{a}\sqrt{1-t}\\
x=a^{1/4}\to t=1
$$
Substituting and rearranging we get
$$\frac{a^{3/4}}{2}\int_0^1 t^{-3/4}(1-t)^{1/2}dt=\left(n+\frac{1}{2}\right)$$
Introducing the complete beta function,
$$B(\nu,\mu)=\int_0^1 t^{\nu-1}(1-t)^{\mu-1}dt=\frac{\Gamma(\nu)\Gamma(\mu)}{\Gamma(\nu+\mu)}$$
Clearly, $\nu=1/4$ and $\mu=3/2$ and we can then show that
$$a=\left[ \frac{2\left(n+\frac{1}{2}\right)\Gamma(7/4)}{\Gamma(1/4)\Gamma(3/2)}\right]^{4/3}$$
At the OP's suggestion, we can substitute
$$
\Gamma(7/4)=(3/4)\Gamma(3/4)\\
\Gamma(3/2)=\sqrt{\pi}/2\\
\Gamma(1/4)\Gamma(3/4)=\pi\sqrt{2}
$$
and demonstrate that
$$E_n=\left[ \frac{2\Gamma(3/4)^2\left(n+\frac{1}{2}\right)}{\pi\sqrt{2\pi}}\right]^{4/3}$$
So, it would appear that the original correct solution was missing a factor of $\pi$ in the denominator.
The solution I present here has been validated numerically for $n\in\mathbb{R^+}$, i.e., not just $n\in\mathbb{Z}$.
$$f(\alpha, \beta, t) = \int_0^t x^{\alpha -1} (t-x)^{\beta -1}\;dx
= \big(t^{\alpha - 1} * t^{\beta -1}\big)(t)$$
so using the fact that convolution is multiplicative in the Laplace domain,
\begin{align}
\mathcal{L}\{f(\alpha, \beta, t)\}(s)
&= \mathcal{L}\{t^{\alpha-1}\}(s) \cdot \mathcal{L}\{t^{\beta - 1}\}(s)\\
&= \frac{\Gamma(\alpha)}{s^{\alpha}}\cdot \frac{\Gamma(\beta)}{s^{\beta}}\\
&= \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \cdot\underbrace{\frac{\Gamma(\alpha+\beta)}{s^{\alpha+\beta}}}_{\mathcal{L\{t^{\alpha+\beta - 1}\}(s)}}
\end{align}
Taking inverse Laplace we then find
$$f(\alpha,\beta,t) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} t^{\alpha+\beta -1}$$
hence at $t=1$:
$$B(\alpha,\beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$
Added To address the extra question in the comments,
$$
\Gamma(\tfrac12)\cdot\Gamma(\tfrac12) = \Gamma(\tfrac12+\tfrac12)\cdot B(\tfrac12,\tfrac12)
= (1)\int_0^1 x^{-1/ 2} (1-x)^{-1/ 2}\;dx
= \int_0^1 \frac{dx}{\sqrt{x(1-x)}}$$
We can evaluate this integral by noting the symmetry about $x=\tfrac12$, so substitute $x =\frac{1+u}{2}$ to shift the symmetry about $u=0$
$$\Gamma(\tfrac12)^2 = \int_{-1}^1 \underbrace{\frac{du}{\sqrt{1-u^2}}}_{\text{even function}}
= 2\int_0^1 \frac{du}{\sqrt{1-u^2}}
= 2\cdot \left.\sin^{-1}(u)\right|_0^1
= 2 \cdot \left(\frac{\pi}{2} - 0\right) = \pi$$
Hence $\Gamma(\tfrac12) = \sqrt{\pi}$.
Best Answer
Hint: If $u = \frac{t}{1-t}$, then $u(1-t)=t$, that is, $$\begin{align*} u-ut &=t \\ \Rightarrow u &= ut+t \\ \Rightarrow u &=t(1+u).\end{align*}$$ So $t= \frac{u}{1+u}$. Try substituting this now.