How to solve the integral $I_n=\int_{0}^{1}x^n\sqrt{1-x}$? Let $I_n=\int_{0}^{1}x^n\sqrt{1-x}$. I need to find a recurrence relation, but I do not see any way to do it. Can you help me?
 A: Observe that
$$\begin{align}
I_n &=\int_0^1 x^n\sqrt{1-x}dx\\
&=\int_0^1 x^n(1-x)\frac{dx}{\sqrt{1-x}}\\
&=\int_0^1 x^n\frac{dx}{\sqrt{1-x}}-\int_0^1 x^{n+1}\frac{dx}{\sqrt{1-x}}\\
&= J_n-J_{n+1}
\end{align}$$
... if we define $J_n$ as follows:
$$J_n := \int_0^1 x^n\frac{dx}{\sqrt{1-x}}$$
It follows by IBP that
$$\begin{align}
J_n
&=\int_0^1 x^n\frac{dx}{\sqrt{1-x}}\\
&=\bigg[-2x^n\sqrt{1-x}\bigg]_0^1 +2n\int_0^1 x^{n-1}\frac{dx}{\sqrt{1-x}}\\
&=2n\int_0^1 x^{n-1}\sqrt{1-x}dx\\
&=2nI_{n-1}\\
\end{align}$$
So we have
$$I_n=J_n-J_{n+1}$$
and
$$J_n=2nI_{n-1}$$
Thus, we have
$$I_n=2nI_{n-1}-2(n+1)I_n$$
$$(2n+3)I_n=2nI_{n-1}$$
and so
$$\color{green}{I_n=\frac{2n}{2n+3}I_{n-1}}$$
A: By integration by parts, for $n \geq 1$ 
\begin{align*}
I_n
&= \left[ -\frac{2}{3}x^n (1-x)^{3/2} \right]_{0}^{1} + \frac{2n}{3} \int_{0}^{1} x^{n-1}(1-x)^{3/2} \, dx \\
&= \frac{2n}{3} \int_{0}^{1} x^{n-1}(1-x)\sqrt{1-x} \, dx \\
&= \frac{2n}{3}(I_{n-1} - I_n).
\end{align*}
Solving this in terms of $I_n$, we obtain
$$ I_n = \frac{2n}{2n+3} I_{n-1}.\tag{*} $$

Remark. Using the gamma function, it follows from the beta function identity that
$$ I_n = B\left( n+1, \tfrac{3}{2} \right) = \frac{n!(\frac{1}{2})!}{(n+\frac{3}{2})!}. $$
Using this we also confirm that $\text{(*)}$ holds.
A: Let $\sqrt{1-x}=\cos t, x=\sin^2t\implies dx=2\sin t\cos t\ dt$
$$I_n=\int_0^1x^n\sqrt{1-x}dx=\int_0^{\pi/2}\sin^{2n}t\cos t(2\sin t\cos t\ dt)$$
$$=2\int\cos^2t\sin^{2n+1}t\ dt=B(x,y)$$
where $2x-1=2\iff x=?,2y-1=2n+1\iff$
See Beta Function
