Integral question with square roots

Question

I can't find the solution for this integral. I don't know why I didn't succeed in solving it.
I will be really glad if someone can help me find the answer to the following:

$$\int_0^1\frac{\sqrt{x}}{\sqrt{1-x^6}}dx $$

Answer 1

The change of variable $t=x^6$ yields $x=t^{1/6}$ and $\mathrm dt=6x^5\mathrm dx=6t^{5/6}\mathrm dx$ hence $$ I=\int_0^1\frac{\sqrt{x}}{\sqrt{1-x^6}}\mathrm dx=\int_0^1\frac{t^{1/12}}{(1-t)^{1/2}}\frac{\mathrm dt}{6\cdot t^{5/6}}, $$ that is, $$ I=\frac16\int_0^1t^{-3/4}(1-t)^{-1/2}\mathrm dt=\frac16\mathrm{Beta}\left(\frac14,\frac12\right), $$ where, the Beta function $\mathrm{Beta}$ is defined, for every positive $(a,b)$, as $$ \mathrm{Beta}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$ Varous expressions of $I$ in terms of values of the Gamma function $\Gamma$ follow.

Answer 2

Observe that $$\int_0^1 \cfrac{\sqrt{x}}{\sqrt{1-x^6}}dx = \int_0^1 \cfrac{\sqrt{x}}{\sqrt{1-(x^3)^2}}dx = \cfrac{1}{3}\int_0^1 \cfrac{\sqrt{x}}{x^2} \cfrac{3x^2}{\sqrt{1-(x^3)^2}}dx$$ Which is $$\cfrac{1}{3} \int_0^1 x^{-3/2}\cfrac{3x^2}{\sqrt{1-(x^3)^2}}dx.$$ The change $t = x^3$ yields now $$\cfrac{1}{3}\int_0^1 \cfrac{1}{\sqrt{t}} \cfrac{dt}{\sqrt{1-t^2}}$$ Yet another change: $t = \sin s \Rightarrow dt = \cos s ds$, which gives $$\cfrac{1}{3} \int_0^{\pi / 2} \cfrac{ds}{\sqrt{\sin s}}.$$ Which turns out to be... an elliptic integral of the first kind. It does not look like you can express this integral in terms of simple functions in a reasonable way. What is your background on integration?