Evaluate $\int_{0}^\frac{\pi}{2} \sqrt{1+\sin^2(x)}dx$ 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,
$$\int_{0}^\frac{\pi}{2} \sqrt{1+\sin^2(x)}dx=\frac{1}{4\sqrt{2\pi}}\left(4\Gamma^2\left(\frac{3}{4}\right)+\Gamma^2\left(\frac{1}{4}\right)\right) $$

Hint 1:

 $$\frac{\mathrm d}{\mathrm dx}\sin(x)=\cos(x)=\sqrt{1-\sin^2(x)}$$

Hint 2:

 Conjugate the "numerator" so that the only radical is in the denominator.

Hint 3:

 Perform a simple substitution so that you get a linear function inside the radical.

All steps shown:

 $$\int_0^{\pi/2}\sqrt{1+\sin^2(x)}~\mathrm dx=\int_0^{\pi/2}\sqrt{\frac{1+\sin^2(x)}{1-\sin^2(x)}}\cos(x)~\mathrm dx=\int_0^1\sqrt{\frac{1+x^2}{1-x^2}}~\mathrm dx\\=\int_0^1\frac{1+x^2}{\sqrt{1-x^4}}~\mathrm dx=\frac14\int_0^1\frac{x^{-3/4}+x^{-1/4}}{\sqrt{1-x}}~\mathrm dx=\frac14B\left(\frac14,\frac12\right)+\frac14B\left(\frac34,\frac12\right)$$

