<s>[INST] Proving an integral equation 
Question:

I have had a hard time solving and understanding the solution to this problem.
Given Solution:

Questions like why (and how) this step:

Or why $5I_{3}$ = $8I_{1}$
 
Have been puzzling me. Can anyone please give a clear answer to why and how each step is done. Thanks!
 [/INST] The point is that, since
$$
\frac{d}{dx}\left[x^n(4-x^2)^{3/2}\right] = 4n x^{n-1}\sqrt{4-x^2} - (n+3)x^{n+1}\sqrt{4-x^2},
$$
and the fudamental theorem of calculus tells us that
$$
\int_0^2\frac{d}{dx}\left[x^n(4-x^2)^{3/2}\right]dx = 2^n(4-2^2)^{3/2}-0^n(4-2^2)^{3/2} = 0,
$$
we must have that
$$
\int_0^2\left[4n x^{n-1}\sqrt{4-x^2} - (n+3)x^{n+1}\sqrt{4-x^2}\right]dx = 0.
$$
Since integration is linear, the integral can be separated into
$$
4n\int_0^2 x^{n-1}\sqrt{4-x^2}dx - (n+3)\int_0^2 x^{n+1}\sqrt{4-x^2}dx = 0,
$$
that is,
$$
(n+3)\int_0^2 x^{n+1}\sqrt{4-x^2}dx = 4n\int_0^2 x^{n-1}\sqrt{4-x^2}dx,
$$
and so we have $(n+3)I_{n+1} = 4n I_{n-1}$.
$5I_3 = 8I_1$ just comes from setting $n=2$.
</s>