Evaluating $\int_0^2 x(8-x^3)^{\frac{1}{3}}\ dx$ What substitution would you use to get from
$$\int\limits_0^2 x(8-x^3)^{\frac{1}{3}}\ dx$$
to
$$\int\limits_0^1 (1-t)^{a-1}t^{-a}\ dt, \ a\in(0,1)\ ?$$
I know how to evaluate the second integral and I thought that if I substitute $t={x^3\over8}$ I would reduce this to the form above, but what I get is
$$\frac{8}{3}\int\limits_0^1 (1-y)^{\frac{1}{3}}y^{-\frac{1}{3}}\ dy$$

$$\frac { 8 }{ 3 } \int _{ 0 }^{ 1 }{ \left( { \left( 1-y \right)  }^{ \frac { 1 }{ 3 }  }{ y }^{ \frac { -1 }{ 3 }  } \right) dx } \\ =B\left( \frac { 4 }{ 3 } ,\frac { 2 }{ 3 }  \right) \\ =\frac { \Gamma \left( \frac { 4 }{ 3 }  \right) \Gamma \left( \frac { 2 }{ 3 }  \right)  }{ 2 } $$
