Sum of squares of binomial coefficients

I'm assuming you meant $K \left(\frac{1}{\sqrt{2}} \right)$ because that's what you have on the right side of the equation.

$$ \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\frac{1}{2} \sin^{2} x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2-\sin^{2}x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+ \cos^{2} x}} \ dx$$

$$= \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1+u^{2}}} \frac{du}{\sqrt{1-u^{2}}} = \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-u^{4}}} du $$

$$= \frac{\sqrt{2}}{4} \int_{0}^{1} t^{-\frac{3}{4}} (1-t)^{\frac{-1}{2}} \ dt = \frac{\sqrt{2}}{4} B \left( \frac{1}{4}, \frac{1}{2} \right) = \frac{\sqrt{2}}{4} \frac{\sqrt{\pi} \ \Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}$$