Reduction formula of $I_n=\int_{0}^{1}(1-\sqrt{x})^ndx$ How to find the reduction for 
$$I_n=\int_{0}^{1}(1-\sqrt{x})^n\mathrm dx$$
I try:
$I_n=uv-\int u^{'}v=x(1-\sqrt{x})^n-\frac{n}{2}\int \sqrt{x}(1-\sqrt{x})^{n-1}\mathrm dx$
This is not working, any other suggestion to deal with $I_n$
 A: With $y:=1-\sqrt{x}$, $I_n=\int_0^12(y^n-y^{n+1})dy=\frac{2}{(n+1)(n+2)}$. This implies the reduction formula $\frac{I_{n+1}}{I_n}=\frac{n+1}{n+3}$.
A: I think you can use the beta function here if that's easier! 
Let $$I=\int_0^1(1-\sqrt{x})^ndx\space\text{and enforce the substitution $x=u^2$}$$
Then we have $dx=2u \space du\implies$ $$I=2\int_0^1u(1-u)^ndu$$
Recall for $\text{Re}(x)>0\space\text{and}\space\text{Re}(y)>0,$ $$B(x,y)=\int_0^1u^{x-1}(1-u)^{y-1}du$$In our case, $x=2$ and $y=n+1$. Hence $$\frac{I}{2}=B(2,n+1)=\frac{n!}{(n+2)!}\implies I=\frac{2n!}{(n+2)!}=\frac{2}{(n+1)(n+2)}$$
The above formula for $I$ is valid for $n>-1.$
 This may not be what you're looking for as a reduction formula, but I hope this is interesting! 
See the Wiki page for the Beta function to see that $B(2,n+1)=\frac{n!}{(n+2)!}$. 
A: You have started out correctly except that you have changed $-\sqrt x$ to $\sqrt x$.  In the second term write $\sqrt x $ as $1- (1-\sqrt x)$ and split the integral into two parts. $I_n=\frac n 2\int_0^{1} ((1-(1-\sqrt x))(1-\sqrt x)^{n-1}dx=\frac n 2I_{n-1}-\frac n 2 I_n$ so $(1+\frac n 2)I_n=\frac n 2I_{n-1}$. So $I_n=\frac n {n+2}I_{n-1}$. 
