TITLE: How to solve integral recursive relation - $I_n=\int_0^1(x-x^2)^ndx$
QUESTION [8 upvotes]: Let $I_n=\int_0^1(x-x^2)^ndx$. Prove that $I_n=\frac{1}{4}\cdot\frac{2n}{2n+1}I_{n-1}$.
This sounds like a rather easy exercise, but no matter how hard I try, I can't quite put my finger on it (I tried integration by parts with $\int_0^1x'(x-x^2)^ndx$, but couldn't manage to get too far). Could you help me out?

REPLY [12 votes]: Integrating by parts in two different ways gives
$$
\begin{array}{rllr}
I_n
&=\int_0^1x^n(1-x)^n\,\mathrm{d}x&=\hphantom{\frac{n}{n+1}}\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{x(1-x)}\,\mathrm{d}x&\qquad(1)\\
&=\frac{n}{n+1}\int_0^1x^{n+1}(1-x)^{n-1}\,\mathrm{d}x&=\frac{n}{n+1}\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{x^2}\,\mathrm{d}x&(2)\\
&=\frac{n}{n+1}\int_0^1x^{n-1}(1-x)^{n+1}\,\mathrm{d}x&=\frac{n}{n+1}\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{(1-x)^2}\,\mathrm{d}x&(3)\\
\end{array}
$$
Since $2\color{red}{x(1-x)}+\color{red}{x^2}+\color{red}{(1-x)^2}=\color{red}{1}$, if we add $2(1)+\frac{n+1}{n}(2)+\frac{n+1}{n}(3)$, we get
$$
2I_n+\frac{n+1}{n}I_n+\frac{n+1}{n}I_n=\int_0^1x^{n-1}(1-x)^{n-1}\color{red}{1}\,\mathrm{d}x=I_{n-1}\tag{4}
$$
Thus, solving $(4)$ for $I_n$ yields
$$
I_n=\frac{n}{4n+2}I_{n-1}\tag{5}
$$