1. Reduction Formula

Given that $\displaystyle I_{n} = \int_{0}^{1}{x^n(1-x)^{\frac{1}{2}}dx}$ where $\displaystyle n \geq 0$, prove that for $\displaystyle n \geq 1$

$\displaystyle I_{n} = \frac{2n}{2n+3} I_{n-1}$
The first thing that comes to mind is $\displaystyle x = \sin^2u$ so $\displaystyle dx = 2 \sin u \cos u du$

giving $\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}{sin^{2n} \cos u \ 2 \sin u \cos u \ du }$

$\displaystyle \Rightarrow 2 \int_{0}^{\frac{\pi}{2}}{sin^{2n+1} \cos^2 u \ du }$

my first thought is to use integration by part, with $\displaystyle du = sin^{2n+1} \cos u$ and $\displaystyle v = \cos u$

this doesn't get me very far sadly....

2. Just do integration by parts without the substitution:
$\displaystyle u=x^n$, $\displaystyle dv=\sqrt{1-x}dx$
Thus $\displaystyle du=nx^{n-1}dx$, $\displaystyle v=-\frac{2}{3}(1-x)^{\tfrac{3}{2}}$
and:
$\displaystyle I_{n}=\int_{0}^{1}x^n\sqrt{1-x}\,dx$
$\displaystyle I_{n}=-\frac{2}{3}x^n(1-x)^{\tfrac{3}{2}}|_{x=0}^1-\int_{0}^{1}(nx^{n-1})\cdot\left(-\frac{2}{3}(1-x)^{\tfrac{3}{2}}\right)\,dx$
$\displaystyle I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)^{\tfrac{3}{2}}\,dx$
$\displaystyle I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)\sqrt{1-x}\,dx$
$\displaystyle I_{n}=\frac{2n}{3}\int_{0}^{1}(x^{n-1}-x^n)\sqrt{1-x}\,dx$
$\displaystyle I_{n}=\frac{2n}{3}\left(\int_{0}^{1}x^{n-1}\sqrt{1-x}\,dx-\int_{0}^{1}x^n\sqrt{1-x}\,dx\right)$
$\displaystyle I_{n}=\frac{2n}{3}(I_{n-1}-I_{n})$
and algebra can do the rest.

--Kevin C.