Challenging Integral: Solving ∫ [x(8-x^3)^1/3] dx from 0 to 2

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Hi all,
Can you guys please help me with the following integration problem

2
∫ [x (8-x3)^1/3 ] dx
0
Thanks in advance.
 
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What have you attempted for the problem so far?
 
I have substituted u^3 = 8 - x^3 then 3 u^2 du = - 3 x^2 dx
But now the problem has only x so in order to substitute for dx I have to divide and multiply by x and that means I can not eliminate cube root.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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