$ \int_{0}^{2} (2x - x^2)^n dx $ recurrence relation Given
$$ I_n =  \int_{0}^{2} (2x - x^2)^n dx $$


*

*Compute $I_2$


I simply expanded it into $$ \int_0^2 4x^2 - 4x^3 + x^4 dx $$ 
and computed it.


*Show that


$$ (2n+1)I_n = 2nI_{n-1} $$
I first tried doing integration by parts by writing it as $ \int_0^2 (x)' (2x-x^2)^n dx$ but that led nowhere and I also tried splitting it into $\int_0^2 (2x+x^2)^{n-1} (2x+x^2) dx$ and that didn't work either. How do I do this?


*Compute
$$ \lim_{n \rightarrow \infty} I_n $$

 A: We have
$$I_n=\int_0^2(x)'(2x-x^2)^ndx=\underbrace{x(2x-x^2)^n\Bigg|_0^2}_{=0}-2n\int_0^2(x-x^2)(2x-x^2)^{n-1}dx\\=-2n\underbrace{\int_0^2(2x-x^2)(2x-x^2)^{n-1}dx}_{=I_n}+n\underbrace{\int_0^2(2x-2)(2x-x^2)^{n-1}dx}_{=0}+2n\underbrace{\int_0^2(2x-x^2)^{n-1}dx}_{=I_{n-1}}$$
A: Alternatively, if you know about the Euler Beta Function, use $x=2u$ and just write
$$ I_n =\int_0^2 (2x-x^2)^n dx = 2 \cdot 4^n \int_0^1 (u-u^2)^n du = 2 \cdot 4^n \int_0^1 u^n (1-u)^n du = 2 \cdot 4^n \frac{(n!)^2}{(2n+1)!}  $$
It is easy to verify your relation between $I_n$ and $I_{n-1}$ from there.
A: you can rewrite $$I_n = \int_0^2 (2x-x^2)^n\, dx = \int_{-1}^1 (1 - x^2)^n\,dx = 2\int_0^1(1-x^2)^n\, dx = 2\int_0^{\pi/2} \sin^{2n+1}t\, dt$$
if we integrate 
$\begin{align}
J_n &= \int_0^{\pi/2} \sin^{2n+1}t\, dt \\  &= \int_0^1 \sin^{2n} t \sin t \,dt\\
&=   -\sin^{2n} t \cos t |_0^{\pi/2} + 2n\int_0^1 sin^{2n-1} t \cos^2 t \,dt\\
&=   2n\int_0^1 sin^{2n-1} t (1-\sin^2 t )\,dt\\
&= 2nJ_{n-1} - 2nJ_{n}
\end{align}$
so we have the recurrence relations $$(2n+1)J_n = 2nJ_{n-1}, (2n+1)I_n = 2nI_{n-1}$$
