Burnside curve is given by the equation $y^2=x^5-x.$ I intended to find the Area formula in $(2)$ of the the given link. And I found after some mistakes: $$A=2\int_{-1}^0 \sqrt{x^5-x}dx=2\int_0^1 x^{1/2}(1-x^4)^{1/2}dx\stackrel{x^4=u}{=}\frac12\int_0^1 u^{-5/8}(1-u)^{1/2}du=\frac12B(3/8,3/2)$$ $$=\frac12\frac{\Gamma(3/8)\Gamma(3/2)}{\Gamma(3/8+3/2)}=\frac{\sqrt\pi}{4}\frac{\Gamma(3/8)}{\Gamma(15/8)}.$$
Is there an alternative way of evaluating this integral? Also, how can I find $(3)$ in that link?
Thanks in advance.
The gamma function obeys the reflection identity $$\Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin \pi z}, \tag{42}$$ as well as the recursion $$\Gamma(z+1) = z \Gamma(z), \tag{39}$$ from which we obtain $$\Gamma(\tfrac{15}{8}) = \tfrac{7}{8} \Gamma(\tfrac{7}{8}) = \frac{7}{8}\frac{1}{\Gamma(\frac{1}{8})} \cdot \frac{\pi}{\sin \frac{7}{8}\pi}.$$ Hence $$\frac{\sqrt{\pi}}{4} \frac{\Gamma(\frac{3}{8})}{\Gamma(\frac{15}{8})} = \frac{\sqrt{\pi}}{4} \cdot \frac{8}{7\pi} \sin \frac{7\pi}{8} \Gamma(\tfrac{1}{8}) \Gamma(\tfrac{3}{8}).$$
Next we evaluate the trigonometric factor by using the identity $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2},$$ hence $$\sin^2 \frac{7\pi}{8} = \frac{1 - \cos \frac{7\pi}{4}}{2} = \frac{1 - \frac{1}{\sqrt{2}}}{2} = \frac{2-\sqrt{2}}{4}.$$ This gives $$\frac{\sqrt{\pi}}{4} \frac{\Gamma(\frac{3}{8})}{\Gamma(\frac{15}{8})} = \frac{2}{7 \sqrt{\pi}} \sqrt{\frac{2-\sqrt{2}}{4}} \Gamma(\tfrac{1}{8})\Gamma(\tfrac{3}{8}) = \frac{1}{7}\sqrt{\frac{2 - \sqrt{2}}{\pi}} \Gamma(\tfrac{1}{8})\Gamma(\tfrac{3}{8})$$ as claimed.