Integral to Gamma function For this function i need to convert it to either a Gamma or a Beta function
$$\int_0^1 \frac{3}{(1-x^3)^\frac{1}{3}} \,dx$$
I know I need to make the substitution $x^3=z$ but I am unsure where to go from here
 A: For simplicity, define
$$\mathcal I := \int_0^1 \frac{3}{(1-x^3)^\frac{1}{3}} \,dx$$
Make the $u$-substitution $u=x^3$ as you prescribed. Then:

*

*$x = u^{1/3}$

*$dx = \frac 1 3 u^{-2/3} du$
Insert both of these definitions into the integral and pull out constants via linearity. (They happen to cancel.) We find that
$$\mathcal I = \int_0^1 u^{-2/3} (1-u)^{-1/3}du$$
Recall that the definition of the beta function is given by
$$\text B(x,y) := \int_0^1 t^{x-1}(1-t)^{y-1}dt$$
$\mathcal I$ now matches this same form for appropriate $x,y$:

*

*$-2/3 = x-1 \implies x = 1/3$

*$-1/3 = y-1 \implies y = 2/3$
Thus,
$$\mathcal I = \text B \left( \frac 1 3, \frac 2 3 \right) \approx 3.6276$$
WolframAlpha numerically confirms this answer: their answer for the integral and for the beta function are at the links. If you want to convert to a gamma function version, note the identity
$$\text B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
In this case, then (since $\Gamma(1/3 + 2/3) = \Gamma(1) = 1$),
$$\mathcal I = \Gamma \left( \frac 1 3 \right) \Gamma \left( \frac 2 3 \right)$$
We can even simplify this result further by use of Euler's reflection formula and get a better closed form. The reflection formula is given by
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$
Take $z = 1/3$ and how it might be utilized for $\mathcal I$ is clear. Then
$$\mathcal I = \frac{\pi}{\sin(\pi/3)} = \frac{\pi}{\sqrt{3} / 2} = \frac{2 \pi}{\sqrt 3}$$
(Credits to user Ty who pointed out the simplification owing to Euler's reflection formula in the comments!)
