By using the Maclaurin series $K(k)==\frac{\pi}2\sum_{n=0}^\infty c_n ^2 k^{2n}$ where $c_{n}={2n\choose n}2^{-2n}$ we have $$\int_0^{\frac{\pi}2}K(\sin t)dt\\ =\frac{\pi}2\sum_{n=0}^\infty c_n ^2\int_0^{\frac{\pi}2}\sin^{2n}t\,dt\\ =\frac{\pi^2}4\sum_{n=0}^\infty c_n ^3. $$ But, how can I show that $$\int_0^{\frac{\pi}2}K(\sin t)dt=\frac{\Gamma(\tfrac14)^4}{16\pi}?$$ I don't know any techniques other than inserting series.
Thanks.
Here is an elementary proof based on Decoupling Double Integrals.
$$I=\int_0^{\pi/2}\int_0^{\pi/2}\frac{1}{\sqrt{1-(\sin x)^2(\sin y)^2}}dxdy$$
Substitution $(\sin x)(\sin y)=\sin z$
$$I=\int_0^{\pi/2}\int_0^{y}\frac{1}{\sqrt{(\sin y)^2-(\sin z)^2}}dzdy$$
Use $(\sin x)^2-(\sin y)^2=\sin(x-y)\sin(x+y)$
Substitution $y-z=a$ and $y+z=b$
$$I=\frac{1}{2}\int_0^{\pi/2}\frac{1}{\sqrt{\sin a}}\int_a^{\pi-a}\frac{1}{\sqrt{\sin b}}dbda$$
Substitution $$s=\int_a^{\pi-a}\frac{1}{\sqrt{\sin b}}db$$
$$ds=-\frac{2}{\sqrt{\sin a}}da$$
$$I=-\frac{1}{4}\int_0^{\pi/2}s\left(\frac{d}{da}s\right)da=\frac{s^2|_{a=0}}{8}$$
$$I=\frac{1}{2}\left(\int_0^{\pi/2}\frac{1}{\sqrt{\sin x}}dx\right)^2$$
$$I=\frac{\Gamma^4(1/4)}{16\pi}$$
Bonus
$$\int_0^1k^{-1/2}Kdk=\frac{\Gamma^4(1/4)}{16\pi}$$
