Integrate $ \int_{0}^1 \frac{dx}{\sqrt{1-x^{6}}}$

Question

Integrate $$ \int_{0}^1 \frac{dx}{\sqrt{1-x^{6}}}$$

This should be equal to $$ \frac{\Gamma \Big(\frac{1}{3}\Big)^{3}}{2^{\frac{7}{3}} \times\pi}$$

I am getting $$\frac{\Gamma \Big(\frac{1}{6}\Big)\Gamma \Big(\frac{1}{2}\Big)}{6\times \Gamma \Big(\frac{2}{3}\Big)}$$ How to proceed$?$

Try the Gamma function's duplication & reflection formulae. – J.G. Sep 29 '19 at 14:35 — J.G., Sep 29 '19 at 14:35
I tried! But didn't get the answer as mentioned. – Mathaddict Sep 29 '19 at 14:38 — Mathaddict, Sep 29 '19 at 14:38
Related: https://math.stackexchange.com/a/1631760/44121 – Jack D'Aurizio Sep 29 '19 at 16:19 — Jack D'Aurizio, Sep 29 '19 at 16:19

J.G. · Accepted Answer · 2019-09-29T15:17:53.993

Call your integral $I$, and define $a_n:=\Gamma\left(\frac{n}{6}\right)$. Like you, I get$$I=\frac{a_1a_3}{6a_4}=\frac{\sqrt{3}a_1a_2a_3}{12\pi}=\frac{a_1a_2}{4\sqrt{3\pi}}.$$We must equate this to $\frac{a_2^3}{2^{7/3}\pi}$, i.e. show $$\frac{a_2^2}{a_1}=2^{1/3}\sqrt{\frac{\pi}{3}}.$$The duplication formula with $z=\frac16$, followed by the reflection formula with $s=\frac13$, obtains$$a_2=\frac{1}{2^{2/3}\sqrt{\pi}}a_1a_4=a_1\frac{2^{1/3}\sqrt{\pi}}{\sqrt{3}a_2},$$which rearranges to the required result.

Integrate $ \int_{0}^1 \frac{dx}{\sqrt{1-x^{6}}}$

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