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. 
 A: 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 
$$\begin{align}I_n &= -x^{\frac{n+2}{2}} \frac{2}{3}(1-x)^{3/2}\Bigg|_0^1 + \int_0^1 \frac{n+2}{2}x^{\frac{n}{2}}\frac{2}{3}(1-x)\sqrt{1-x} dx\\~\\&= 0 + \frac{n+2}{3}\int_0^1 x^{\frac{n}{2}}(1-x)\sqrt{1-x}dx\\~\\&=\frac{n+2}{3}\left(\int_0^1x^{\frac{n}{2}} \sqrt{1-x}dx -\int_0^1x^{\frac{n}{2}+1} \sqrt{1-x}dx\right)\\~\\&=\frac{n+2}{3}\left(I_{n-1} - I_n\right) \end{align}$$
A: Recall that when $\Re(a),\Re(b)>0$ $$\int_0^1x^{a-1}(1-x)^{b-1}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$thus 
\begin{align}
\int_0^1x^{\frac{n+2}{2}}\sqrt{1-x}dx&=\int_0^1x^{\frac{n}{2}+2-1}(1-x)^{\frac32-1}dx\\
&=\frac{\Gamma(\frac{n}{2}+2)\Gamma(\frac32)}{\Gamma(\frac{n}{2}+2+\frac32)}\\
&=\frac{(\frac{n}{2}+1)\Gamma(\frac{n}{2}+1)\Gamma(\frac32)}{(\frac{n}{2}+1+\frac32)\Gamma(\frac{n}{2}+1+\frac32)}\\
&=\frac{n+2}{n+5}\frac{\Gamma(\frac{n}{2}+1)\Gamma(\frac32)}{\Gamma(\frac{n}{2}+1+\frac32)}\\
&=\frac{n+2}{n+5} \int_0^1 x^{\frac n2+1-1}(1-x)^{\frac32-1}dx\\
&=\frac{n+2}{n+5} \int_0^1 x^{\frac n2}\sqrt{1-x}dx\\
\end{align}
