Integral question with square roots 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:
$$\int_0^1\frac{\sqrt{x}}{\sqrt{1-x^6}}dx $$
 A: 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.
A: 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?
