Prove by induction that $I_n = \frac{4^{n+1}n!(n+1)!}{(2n+3)!}$ $I_n$ is defined as:
$$I_n = \int_{0}^{1} \big[x^n \sqrt{1-x}\big] dx$$
Let $p(n)$ be the statement:
$$I_n = \frac{4^{n+1}n!(n+1)!}{(2n+3)!}$$
Prove by mathematical induction $p(n)$ is true for n = 0,1,2,...
It is straightforward to show that the statement $p(0)$ is true and $I_0 = \tfrac{2}{3}$
However I do not no where to begin to prove that $p(k) \Rightarrow p(k+1)$ and trying to integrate $I_n$ is getting me nowhere.
I am not necessarily asking for a general proof, rather some hints about how to tackle it, and whether I need to solve the integration statement.
 A: Let $I_n$ be the integral defined by
$$\begin{align}
I_n&=\int_0^1 x^n\sqrt{1-x}\,dx\\\\
&=\int_0^1 (1-x)^n \sqrt{x}\,dx
\end{align}$$
Integrating by parts with $u=(1-x)^n$ and $v=\frac23 x^{3/2}$ reveals
$$\begin{align}
I_n&=\frac23 n\int_0^1 (1-x)^{n-1}\,x^{3/2}\,dx\\\\
&=\frac23 n I_{n-1}-\frac23 nI_n\\\\
&=\frac{\frac23 n}{1+\frac23 n}I_{n-1}\\\\
&=\frac{2n}{2n+3}I_{n-1} \tag 1
\end{align}$$
Using $I_0=\frac23$ along with $(1)$ and proceeding recursively, we obtain 
$$\begin{align}
I_n&=\frac{2^{n+1}\,n!}{(2n+3)!!}\\\\
&=\frac{2^{n+1}\,n!}{\frac{(2n+3)!}{2^{n+1}\,(n+1)!}}\\\\
&=\frac{4^{n+1}\,n!\,(n+1)!}{(2n+3)!}
\end{align}$$
as was to be shown!
A: Hint: Write $x^n\sqrt{1-x}$ as $x^{n-1}(1-(1-x))\sqrt{1-x}$. That gives:
$$ I(n) = I(n-1) - \int_{0}^{1} x^{n-1}(1-x)^{3/2}\,dx. $$
Use integration by parts on the last integral, and it will reveal itself as a peculiar multiple of $I(n)$.
The previous relation can so be rearranged in order to give a closed formula for $\frac{I(n)}{I(n-1)}$ in terms of $n$ only. Compute $I(0)$ and profit by induction.
