I feel like I'm very close, so I would only like a hint. I'm only using real methods with the main thing I'm trying to connect the integral to is the Beta function. With a bunch of substitutions, I have boiled the integral down to
$$\int_{0}^\frac{\pi}{2} \sqrt{1+\sin^2(x)}dx=\sqrt{2}\int_{0}^\infty \frac{\sqrt{1+x^4}}{(1+x^2)^2}dx=\sqrt{2}\int_{0}^\infty \frac{2x^2(x^4-x^2+2)}{(1+x^2)^3\sqrt{1+x^4}}dx$$
I feel like there is some substitution that could convert the integral into something in terms of the Beta function but I cannot figure it out for the life of me.
For reference,
$$\int_{0}^\frac{\pi}{2} \sqrt{1+\sin^2(x)}dx=\frac{1}{4\sqrt{2\pi}}\left(4\Gamma^2\left(\frac{3}{4}\right)+\Gamma^2\left(\frac{1}{4}\right)\right) $$
Best Answer
Hint 1:
Hint 2:
Hint 3:
All steps shown: