How to solve this definite integral? $\int_0^1\frac{1}{\sqrt{1+x^4}}\,{\rm d}x$

Question

$$I=\int_0^1\frac{1}{\sqrt{1+x^4}}\,{\rm d}x$$

I am trying this question by substituting $$x=\sqrt{\tan(y)}$$ And finally arrived at a form of $$\sqrt{2\operatorname{csc}(2y)}$$ But how to proceed further?

Answer 1

We have that, $$x = \sqrt{\tan{\theta}} \Rightarrow \text{d}x = \frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}}}\text{d}\theta.$$ Therefore, $$\int_0^\frac{\pi}{4} \frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}} \sec{\theta}} \text{d}\theta = \int_0^\frac{\pi}{4} \frac{\text{d}\theta}{2\sqrt{\sin{\theta}\cos{\theta}}} = \int_0^\frac{\pi}{4} \frac{\text{d}\theta}{\sqrt{2\sin{2\theta}}} = \frac{1}{2\sqrt{2}} \int_0^\frac{\pi}{2} \frac{\text{d}\phi}{\sqrt{\sin{\phi}}}=\frac{1}{2\sqrt{2}} \int_0^\frac{\pi}{2} \frac{\text{d}\phi}{\sqrt{\sin{\phi}}} = \frac{1}{4\sqrt{2}} \text{B}\left(\frac{1}{4}, \frac{1}{2}\right)$$ Then the definition of the Beta Function can be applied to reduce it in terms of the Gamma Function.

From: https://artofproblemsolving.com/community/c7h1713030p11055627

Answer 2

Let $x=\frac1u$,

$$I=\int_0^1\frac{1}{\sqrt{1+x^4}}dx=\int_1^\infty\frac{1}{\sqrt{1+u^4}}du\Longrightarrow 2I=\int_0^\infty\frac{1}{\sqrt{1+u^4}}du$$

Let $t=u^4$

$$I=\frac18\int_0^\infty\frac{t^{-\frac34}}{(1+t)^\frac12}dt=\frac18B\left(\frac14,\frac14\right)=\frac{\Gamma\left(\frac14\right)^2}{8\sqrt\pi}$$