I couldn't figure out the reduction formula $I_n = \frac{n-1}{n+2} I_{n-2}$ for the integral $I_n=\int_0^1 x^n \sqrt{1-x^2} dx$ for each non-negative integer $n\geq 2$.
I couldn't figure this one out without ending up with powers of 3/2 which didn't fit with the reduction formula.
Let's cheat our way through $\begin{aligned} I_n=\displaystyle\int_0^1x^n\sqrt{1-x^2}\,\mathrm dx\overset{(\ast)^1}{=}\int_0^{\pi/2}\sin^n u\cos^2 u\,\mathrm du&\overset{(\ast)^2}{=}\frac{1}{2}\mathfrak B\left(\frac{n+1}{2},\frac{3}{2}\right)\\ &\overset{(\ast)^3}{=}\frac{1}{2}\dfrac{\Gamma(\frac{n+1}{2})\Gamma(\frac{3}{2})}{\Gamma(\frac{n+4}{2})}\\ &=\frac{1}{2}\dfrac{\Gamma(\frac{n-1}{2}+1)\Gamma(\frac{3}{2})}{\Gamma(\frac{n+2}{2}+1)}\\ &\overset{(\ast)^4}{=}\frac{1}{2}\dfrac{\frac{n-1}{2}\Gamma(\frac{n-1}{2})\Gamma(\frac{3}{2})}{\frac{n+2}{2}\Gamma(\frac{n+2}{2})}\\ &=\frac{n-1}{n+2}\frac{1}{2}\dfrac{\Gamma(\frac{n-1}{2})\Gamma(\frac{3}{2})}{\Gamma(\frac{n+2}{2})}\\ &=\frac{n-1}{n+2}\mathfrak B\left(\frac{n-1}{2},\frac{3}{2}\right)\\ &=\frac{n-1}{n+2}\int_0^{\pi/2}\sin^{n-2} u\cos^2 u\,\mathrm du\\ &=\frac{n-1}{n+2}\int_0^{1}x^{n-2} \sqrt{1-x^2}\,\mathrm dx=\frac{n-1}{n+2}I_{n-2} \end{aligned}$
Where $\mathfrak B(\cdot,\cdot)$ is the beta function, $\Gamma(\cdot)$ is Euler's gamma function and $$\begin{aligned} &(\ast)^1:x=\sin u;\ \mathrm dx=\cos u\,\mathrm du\\ &(\ast)^2:\mathfrak B(n,m)\equiv 2\int_0^{\pi/2}(\sin x)^{2n-1}(\cos x)^{2m-1}\,\mathrm dx\\ &(\ast)^3:\mathfrak B(n,m)\equiv\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}\\ &(\ast)^4:\Gamma(z+1)=z\Gamma(z) \end{aligned}$$
Extablish \begin{align} &\left[x^{n-1}(1-x^2)^{3/2}\right]’ =\left[(n-1)x^{n-2}- (n+2)x^n\right] \sqrt{1-x^2} \end{align} and integrate both sides of the equation to obtain $I_n = \frac{n-1}{n+2} I_{n-2}$.
For any $n\ge 2$, $$ \begin{aligned} I_n&=\int_0^1 x^n \sqrt{1-x^2} d x \\ &=\int_0^1 \frac{x^n\left(1-x^2\right)}{\sqrt{1-x^2}} d x \\ &=-\int_0^1\left(x^{n-1}-x^{n+1}\right) d\left(\sqrt{1-x^2}\right) \\ &= {(n-1) \int_0^1 x^{n-2} \sqrt{1-x^2} d x-(n+1) \int_0^1 x^n \sqrt{1-x^2} dx} \quad \textrm{ (via IBP)} \\ &= {(n-1) I_{n-2}-(n+1) I_n} \\ \therefore I_n&=\frac{n-1}{n+2} I_{n-2} \end{aligned} $$
$$I_n:= \int_0^1x^n \sqrt{1-x^2}dx =\int_0^\frac{\pi}{2}\sin^n(x) \cos^2(x)dx$$
$$= \frac{\sin^{n+1}( x) \cos(x)}{n+1} \bigg|_0^\frac{\pi}{2} + \frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n+2}(x) dx =\frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n}(x) (1-\cos^2(x)) dx$$ $$= \frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n}(x) dx - \frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n}(x) \cos^2(x) dx$$ $\\[5pt]$
$$I_n =\frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n}(x) dx -\frac{I_n}{n+1} $$ $\\[5pt]$ $$I_n \left( \frac{n+2}{n+1} \right)= \frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n}(x) dx = \frac{1}{n+1}\int_0^\frac{\pi}{2}\sin^{n-1}(x) \sin(x) dx $$
$$= - \frac{cos(x) \sin^{n-1}(x)}{n+1} \bigg|_0^\frac{\pi}{2} + \frac{n-1}{n+1} \int _{0}^ \frac{\pi}{2} \sin^{n-2}(x)\cos^2(x) =\frac{n-1}{n+1} I_{n-2} $$
$\\[5pt]$ $$I_n \left( \frac{n+2}{n+1} \right)=\frac{n-1}{n+1} I_{n-2} $$
$$I_n = \frac{n-1}{n+2} I_{n-2}$$
