We can use the reciprocal modulus transformation to characterize $E\left(x,\frac{1}{2}\right)$ as a sum of two incomplete beta functions.
Recall the reciprocal modulus transformation (proof in the appendix):
$$E\left(\phi,\frac{1}{p} \right) = \frac{1}{\sqrt{p}}E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)-\frac{1-\sqrt{p}^2}{\sqrt{p}}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)$$
If we put $p=2$
and using
$$\frac14 \text B_{\sin^2(2x)}\left(\frac12,\frac34\right)=\text E(x,2) $$
and
$$\frac14\text B_{\sin^2(2x)}\left(\frac12,\frac14\right)=\text F(x,2)$$
\begin{align*}E\left(\phi,\frac{1}{2} \right) = &\frac{1}{\sqrt{2}}E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right),2\right)+\frac{1}{\sqrt{2}}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right),2\right)\\
=& \frac{1}{4\sqrt{2}}B_{\sin^2\left(2\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right)\right)}\left(\frac12,\frac34\right)+ \frac{1}{4\sqrt{2}} B_{\sin^2\left(2\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right)\right)}\left(\frac12,\frac14\right)\\
=& \frac{1}{4\sqrt{2}}B_{1-\cos^4(\phi)}\left(\frac12,\frac34\right)+ \frac{1}{4\sqrt{2}} B_{1-\cos^4(\phi)}\left(\frac12,\frac14\right)
\end{align*}
So the incomplete elliptic integral of the second kind with $\displaystyle k=\frac{1}{2}$ is also related to the incomplete beta function through the following sum:
$$\boxed{E\left(\phi, \frac{1}{2} \right) = \frac{1}{4\sqrt{2}}B_{1-\cos^4(\phi)}\left(\frac12,\frac34\right)+ \frac{1}{4\sqrt{2}} B_{1-\cos^4(\phi)}\left(\frac12,\frac14\right)} $$
Edit (Appendix): Proof of the reciprocal modulus transformation.
We will use the notation from Wolfram:
$$F(\alpha,m ) = \int_{0}^{\alpha} \frac{1}{\sqrt{1-m\sin^2\alpha}}$$
Typically, the reciprocal modulus transformation is used to transform modulus exceeding the unity to the standard modulus form $0<p\leq 1$. However, nothing prevent us to use it the other way around:
\begin{align*} E\left(\phi,\frac{1}{p} \right) = &\int_{0}^{\phi} \frac{1}{\sqrt{1-\frac{1}{p}\sin^2\theta }} d\theta \\
=& \int_{0}^{\arcsin\left(\frac{\sin\phi}{\sqrt{p}}\right)} \frac{\sqrt{p}\cos^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi \quad \left( \sin\varphi \mapsto \frac{\sin\theta}{\sqrt{p}}\right)\\
=& \sqrt{p}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) - \sqrt{p}\int_{0}^{\arcsin\left(\frac{\sin\phi}{\sqrt{p}}\right)} \frac{\sin^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi
\end{align*}
The last integral is known as $D(\alpha,p)$ and has the solution.
$$ D(\alpha,p) = \int_{0}^{\alpha} \frac{\sin^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi = \frac{F(\alpha,p)-E(\alpha,p)}{p}$$
(To prove this just apply the standard definition to the right hand side.)
So
\begin{align*} E\left(\phi,\frac{1}{p} \right) =& \sqrt{p}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) - \sqrt{p}\int_{0}^{\arcsin\left(\frac{\sin\phi}{\sqrt{p}}\right)} \frac{\sin^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi\\
=& \sqrt{p}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) - \frac{F\left(\arcsin\left(\frac{\phi}{\sqrt{p}}\right),p\right)-E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)}{\sqrt{p}}\\
=& \frac{1}{\sqrt{p}} E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)-\frac{1-p}{\sqrt{p}}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)
\end{align*}
Best Answer
We have $$\int_0^{\pi/2} \log^{2n} (\tan \frac{x}{2}) K(\sin x) dx = \int_0^1 \frac{2 \log^{2n} t}{1+t^2} K(\frac{2t}{1+t^2}) dt$$ Their values can be extracted by differentiating $$\tag{*}\int_0^1 \frac{t^{4a} + t^{-4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt = \frac{\pi}{8}\cot(\pi(\frac{1}{4}-a))\frac{\Gamma \left(a+\frac{1}{4}\right)^2}{\Gamma \left(a+\frac{3}{4}\right)^2} \quad -1/4<\Re(a)<1/4$$
I give two different proofs of $(*)$, I come up with the longer proof first.
First proof of $(*)$: using quadratic transformation $(1+t^2)K(t^2) = K(2t/(1+t^2))$, we have $$ \begin{aligned}\int_0^1 \frac{t^{4a}}{1+t^2} K(\frac{2t}{1+t^2}) dt &= \frac{1}{2} \int_0^1 t^{2a-1/2} K(t) dt \\ &= \frac{1}{2} \frac{\pi}{1+4a}{_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1) \end{aligned}$$ here we noted that $K(t)$ is a $_2F_1$. Using third formulas here gives $${_3F_2}(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} + a; 1, \frac{5}{4} + a; 1) = \frac{4 a+1}{4 a-1} {_3F_2(\frac{1}{2}, \frac{1}{2}, \frac{1}{4} - a; 1, \frac{5}{4} - a; 1)} +\frac{\Gamma \left(\frac{1}{4}-a\right) \Gamma \left(a+\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}-a\right) \Gamma \left(a+\frac{3}{4}\right)}$$ we see the $a$ in $_3F_2$ becomes $-a$, giving $(*)$. Q.E.D.
Second proof of $(*)$: It's much longer. See edit history.