Calculating $ \int_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^{6}}}\mathrm{d}x $

Question

I can't calculate the Integral:

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

any help would be great!

p.s I know it converges, I want to calculate it.

Answer 1

Use substitution $u=x^6$ and $B$ function: $$\int_0^1\frac{\sqrt{x}}{\sqrt{1-x^6}}dx=\frac{1}{6}\int_0^1u^{-\frac{3}{4}}(1-u)^{-\frac{1}{2}}du=\frac{1}{6}\int_0^1u^{\frac{1}{4}-}(1-u)^{\frac{1}{2}-1}du=\frac{1}{6}B(x,y)=\frac{1}{6}\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{4})}$$

Answer 2

Doing the substitution first $x^3 = \cos\theta$ , then $\theta/2 = t$, would lead us to a simpler integral: $$ \int^{\frac{\pi}{2}}_0 \frac{1}{3\sqrt{\cos\theta}} d\theta = \int^{\frac{\pi}{4}}_0 \frac{2}{3\sqrt{1 - 2\sin^2t}} dt $$ and this is elliptic integral of the first kind, hence no closed form. For the numerical value please refer to vadim123's comment.