$$\int_0^{1/\sqrt{2}}\frac{K\left[\sqrt{1-k^2}\right]}{\sqrt{1-2k^2}\sqrt{1-k^2}}dk=\frac{\Gamma^2(1/8)\Gamma^2(3/8)}{32\pi}$$
With $K$ as the Complete Elliptic Integral of the First Kind,
$$K:=K(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-(k\sin t)^2}}dt$$
Define,
$$I(k)=\int_0^{\pi/2}\frac{K\left[\sqrt{1-(k\sin t)^2}\right]}{\sqrt{1-(k\sin t)^2}}dt$$
I was interested in this Integral because while researching, I found out that this one has a really interesting Fourier-Legendre Expansion.
Denote,
$$P_n(x)=\text{LegendreP}[1-2x^2]$$
(The notation might look a bit weird but to me it made more sense this way)
or we can have this closed form,
$$P_n(x)=\sum_{r=0}^n(-1)^r\frac{(n+r)!}{(n-r)!}\frac{x^{2r}}{r!r!}$$
So we have this expansion,
$$I(k)=\frac{\pi^2}{2}\sum_{n=0}^{\infty}\binom{2n}{n}^2\frac{P_{2n}(k)}{2^{4n}}$$
And we also have the following,
$$P_{2n}(1/\sqrt{2})=\binom{2n}{n}\frac{(-1)^n}{2^{2n}}$$
So this allows us the evaluation,
$$I(1/\sqrt{2})=\frac{\pi^2}{2}\sum_{n=0}^{\infty}\binom{2n}{n}^3\frac{(-1)^n}{2^{6n}}$$
which is related to the Elliptic Singular Value $k_8$.
That would give us,
$$I(1/\sqrt{2})=\frac{\Gamma^2(1/8)\Gamma^2(3/8)}{16\pi\sqrt{2}}$$
or,
$$\int_0^{\pi/2}\frac{K\left[\sqrt{1-(\sin t)^2/2}\right]}{\sqrt{1-(\sin t)^2/2}}dt=\frac{\Gamma^2(1/8)\Gamma^2(3/8)}{16\pi\sqrt{2}}$$
I was wondering if we can prove this using Integral Manipulations alone.
We have this alternate form also,
$$\int_0^{1/\sqrt{2}}\frac{K\left[\sqrt{1-k^2}\right]}{\sqrt{1-2k^2}\sqrt{1-k^2}}dk=\frac{\Gamma^2(1/8)\Gamma^2(3/8)}{32\pi}$$
There happens to be a formula for (Denote $a'=\sqrt{1-a^2})$,
$$\int_0^1\frac{K'}{\sqrt{1-a^2k^2}\sqrt{1-b^2k^2}}dk=K\left[\sqrt{\frac{1-ab-a'b'}{2}}\right]K\left[\sqrt{\frac{1+ab-a'b'}{2}}\right]$$
But not sure how we can use it here.
Best Answer
We may define $$ K_{4}(x) =\frac{K\left (\sqrt{\frac{2x}{1+x} } \right ) }{ \sqrt{1+x} } =\frac{\pi}{2}\,_2F_1\left ( \frac14,\frac34;1;x^2 \right ). $$ And consider the contour integral on the quarter circle in the first quadrant, giving $$ \left ( \int_0^1+\int_{1}^{\infty} \right ) \frac{K_4(x)}{\sqrt{1-x^2} }\text{d}x =i\int_{0}^{\infty} \frac{K_4(ix)}{\sqrt{1+x^2} }\text{d}x. $$ Now that $$ \int_0^1 \frac{K_4(x)}{\sqrt{1-x^2} }\text{d}x =\sqrt{2} \int_{0}^{1} \frac{K^\prime(x)}{\sqrt{1+x^2} } \text{d}x=\text{solvable}, $$ $$ \int_1^\infty \frac{K_4(x)}{\sqrt{1-x^2} }\text{d}x = \int_{1}^{\infty} \frac{1}{\sqrt{2x} } \left ( K\left ( \sqrt{\frac{1+x}{2x}} \right ) +i K\left ( \sqrt{\frac{1-x}{2x}} \right ) \right ) \frac{i}{\sqrt{x^2-1} } \text{d}x=\frac{i}{\sqrt{2} } \int_{0}^{1} \frac{K\left ( \sqrt{\frac{1+x}{2} } \right ) +iK\left ( \sqrt{\frac{1-x}2} \right ) }{\sqrt{x\left ( 1-x^2 \right ) } } \text{d} x,$$ $$ \int_{0}^{\infty} \frac{K_4(ix)}{\sqrt{1+x^2} } \text{d}x =\int_{0}^{\infty} \frac{K\left ( \sqrt{\frac12-\frac{1}{2\sqrt{1+x^2} } } \right ) }{ \left ( 1+x^2 \right )^{3/4} }\text{d}x =\int_{0}^{1} \frac{K\left ( \sqrt{\frac{1-x}{2} } \right ) }{ \sqrt{x\left ( 1-x^2 \right ) } }\text{d}x,$$ therefore your integral is just $$ \int_{0}^{1} \frac{K\left (\sqrt{\frac{1+x}{2} } \right ) }{ \sqrt{x\left ( 1-x^2 \right ) } }\text{d}x =\int_{0}^{\frac{1}{\sqrt{2}} } \frac{K\left ( \sqrt{1-x^2} \right ) }{ \sqrt{1-x^2}\sqrt{1-2x^2} }\text{d}x $$ which could be solved by comparing the real and imaginary part. This method could be used to verify the incredible identity:$$ \int_{0}^{\frac{1}{\sqrt{2} } } K(k)\left ( K(k)+K^\prime(k) \right )^2 \text{d}k=\frac{\Gamma\left ( \frac14 \right )^8 }{128\pi^2}. $$