Context $\begin{align} K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}\tag{1} \end{align}$ and $\begin{align} E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2t}dt\tag{2} \end{align}$ the complete elliptic integrals of the first and second kind respectively. I have a proof of: $\begin{align}\int_{0}^{1}\frac{E(\frac{x}{\sqrt{x^2+8}})}{\sqrt{8-7x^2-x^4}}dx=\frac{1}{3}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1+\cos^2x}}\int_{0}^{\pi/2}{\sqrt{1+\cos^2x}}dx\tag{3} \end{align}$. But it uses series expansions that I want to avoid. The way I have to get $(3)$ is proving first that: $\begin{align} \int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{dkdx}{\left(\sin^2{k}\sin^2{x}+8 \right)^{3/2}}=\frac{E(k_{1})K(k_{1})}{24\pi^2}\\=\frac{1}{96\pi}+\frac{\Gamma(\frac{1}{4})^4}{768\pi^3}.\tag{4} \end{align}$ where $k_{1}=\frac{1}{\sqrt{2}}$. Then integrating on the left hand of $(4)$ we arrive to the left hand of $(3)$.
Question
Can we prove $(3)$ in another way in the sense that we can arrive to it making only integral manipulations?
