reduction formulae for $\int_0^1 (x^{\frac{n+2}2}\sqrt{1-x})dx\,\!$ how do I prove that
$\int_0^1 (x^{\frac{n+2}2}\sqrt{1-x})dx\,\!={\frac{n+2}{n+5}}\int_0^1 (x^{\frac{n}2}\sqrt{1-x})dx\,\!$
I couldn't find which functions u and v' to use for the intgeration by parts.
Help please. 

Let $$I_n = \int_0^1 x^{\frac{n+2}2}\sqrt{1-x}dx\,$$
Noticing $\frac{n+2}{2} = \frac{n}{2}+1$, you want to set
$$u = x^{\frac{n+2}{2}} \text{ and } dv = \sqrt{1-x}dx$$
That gives $$du = \frac{n+2}{2}x^{\frac{n}{2}} dx\text{ and } v = -\frac{2}{3}(1-x)\sqrt{1-x}$$
Integrating by parts yields 
$$\begin{align}I_n &= -x^{\frac{n+2}{2}} \frac{2}{3}(1-x)^{3/2}\Bigg|_0^1 + \int_0^1 \frac{n+2}{2}x^{\frac{n}{2}}\frac{2}{3}(1-x)\sqrt{1-x} dx\\~\\&= 0 + \frac{n+2}{3}\int_0^1 x^{\frac{n}{2}}(1-x)\sqrt{1-x}dx\\~\\&=\frac{n+2}{3}\left(\int_0^1x^{\frac{n}{2}} \sqrt{1-x}dx -\int_0^1x^{\frac{n}{2}+1} \sqrt{1-x}dx\right)\\~\\&=\frac{n+2}{3}\left(I_{n-1} - I_n\right) \end{align}$$
