Find a recursive formula for $\int_0^2x^n(4-x^2)^{1/2}dx$ 
Find a recursive formula for $I_n=\int_0^2x^n(4-x^2)^{1/2}dx$ for $n\ge2$.
Source : Cambridge A-level Further Mathematics 9231_s16_qp_13 Question 6

My work
$\frac{d}{dx}(x^n(4-x^2)^{3/2})=-3x^{n+1}(4-x^2)^{1/2}+nx^{n-1}(4-x^2)^{3/2}$
Integrating each term gives,
$0=-3I_{n+1}+n\int _0^2 x^{n-1}(4-x^2)^{3/2} dx$
I got stuck on this integral : $ n\int _0^2 x^{n-1}(4-x^2)^{3/2} dx$
Integration by parts will give $3I_{n+1}$  which is not helpful. I also tried the substitution $x=2\sin \theta$ but got stuck again with $n\int (2\sin \theta)^{n-1}(18\cos^4 \theta)d\theta$.  The mark scheme was not helpful as well.
The final answer should be $(n+3)I_{n+1}=4nI_{n-1}$. I feel like I'm missing something obvious because the question is not worth many marks.
 A: Let $$A_n = \int_0 ^2 \mathrm{d}x \ x^n (4-x^2)^{1/2}$$
then $$I_n = 4 A_n - A_{n+2}$$
After differentiating the integrand, as you did. The relation one gets is
$$4n A_{n-1} = (n+3) A_{n+1}$$
Now, we use this relation to relate $I_{n-1}$ and $I_{n+1}$
$$I_{n+1} =4 A_{n+1} -A_{n+3}$$
$$I_{n-1} =4 A_{n-1} -A_{n+1}$$
Eliminating $A_{n+3}$ and $A_{n-1}$, we get
$$I_{n-1} = \frac{3}{n}A_{n+1}$$ and
$$I_{n+1} = \frac{12}{n+5}I_{n+1}$$
Diving these you get
$$(n+5)I_{n+1} = 4n I_{n-1}$$
I am not sure if your relation $(n+3) I_{n+1} =4n I_{n-1}$ is correct.
A: Method 1
Rationalization followed by integration by parts yields
$$
\begin{aligned}
I_{n} &=\int_{0}^{2} x^{n}\left(4-x^{2}\right)^{\frac{1}{2}} d x \\
&=\int_{0}^{2} \frac{x^{n}\left(4-x^{2}\right)}{\left(4-x^{2}\right)^{\frac{1}{2}}} d x \\
&=-\int_{0}^{2} x^{n-1}\left(4-x^{2}\right) d \left(4-x^{2}\right)^{\frac{1}{2}} \\
&=-\left[x^{n-1}\left(4-x^{2}\right)^{\frac{3}{2}}\right]_{0}^{2}+\quad \int_{0}^{2}\left[4(n-1) x^{n-2}-(n+1) x^{n}\right](4-x)^{\frac{1}{2}}\\&=4(n-1) I_{n-2}-(n+1) I_{n}
\end{aligned}
$$
Rearranging yields
$$
\boxed{I_{n}=\frac{4(n-1)}{n+2} I_{n-2}}
$$
In particular, $$
\begin{array}{l}
I_{0}=\pi, \quad I_{1}=\dfrac{8}{3}\quad  \Rightarrow  \quad  I_{2}=\dfrac{4}{4} I_{0}=\pi \textrm{ and }  I_{3}=\dfrac{4(2)}{5} \cdot \dfrac{8}{3}=\dfrac{64}{15}.
\end{array}\\\\
$$
Method 2
Integration by parts yields $$
\begin{aligned}
{I}_{n} &=\int_{0}^{2} x^{n}\left(4-x^{2}\right)^{\frac{1}{2}} d x \\
&=-\frac{1}{3} \int_{0}^{2} x^{n-1} d\left(4-x^{2}\right)^{\frac{3}{2}} \\
&=-\left[\frac{x^{2}\left(4-x^{2}\right)^{\frac{3}{2}}}{3}\right]_{0}^{2}+\frac{n-1}{3} \int_{0}^{2} x^{n-2}\left(4-x^{2}\right)^{\frac{3}{2}} d x\\&= \frac{n-1}{3} \int_{0}^{2} x^{n-2}\left(4-x^{2}\right)\left(4-x^{2}\right)^{\frac{1}{2}} d x\\&=\frac{n-1}{3}\left(4 I_{n-2}-I_{n}\right)
\end{aligned}
$$
We can now conclude that $$
\boxed{I_{n}=\frac{4(n-1)}{n+2} I_{n-2}}
$$
