<s>[INST] Reduction Formulae Integral $x(1-x^3)$ The question asks us:
"If 

$$u_n=\int_0^1x(1-x^3)^ndx$$ 

show that 

$$u_n= \frac {3n}{3n+2}u_{n-1}$$

I've tried integration by parts using a coefficient of $1, x$ and even tried reducing the $1-x^3$ term into its factors but with no progress. Any help would be greatly appreciated.
 [/INST] \begin{align}
u_n &= \int_0^1 \frac{1}{2}x^2 n(1-x^3)^{n-1} 3x^2 dx \\
    &= \frac{3}{2}n\int_0^1 x x^3 (1-x^3)^{n-1} dx \\
    &= \frac{3}{2} n \int_0^1 x(x^3-1+1)(1-x^3)^{n-1}dx \\
    &= -\frac{3}{2}n u_n + \frac{3}{2}n u_{n-1}
\end{align}
Therefore,
$$
u_n = \frac{3n}{2+3n} u_{n-1}
$$
</s>