# 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

$$\frac{8}{3}\int\limits_0^1 (1-y)^{\frac{1}{3}}y^{-\frac{1}{3}}\ dy$$

• Wouldn't that be $t=(\frac{x}{2})^3=\frac{x^3}{8}$? Dec 1, 2015 at 4:49
• Oh, right, right, right. Dec 1, 2015 at 4:55
• en.wikipedia.org/wiki/Beta_function Dec 1, 2015 at 9:58

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