Let $R>0$ be fixed, $n\ge 2$. What is $$\int_{-R}^R(R-x^2)^\frac{n-1}{2}dx?$$
I tried substitution $x=\sqrt{R}\sin(t)$. Then $dx=\sqrt{R}\cos(t)dt$ and $$\int_{-R}^R(R-x^2)^\frac{n-1}{2}dx=\int_{\sqrt{R}\sin(-R)}^{\sqrt{R}\sin(R)}(R(1-\sin^2(t))^\frac{n-1}{2}\sqrt{R}\cos(t)dt$$ $$=R^\frac{n-1}{2}\sqrt{R}\int_{\sqrt{R}\sin(-R)}^{\sqrt{R}\sin(R)}(1-\sin^2(t))^\frac{n-1}{2}\cos(t)dt$$ $$=R^\frac{n-1}{2}\sqrt{R}\int_{\sqrt{R}\sin(-R)}^{\sqrt{R}\sin(R)}(\cos^2(t))^\frac{n-1}{2}\cos(t)dt,$$ However, it looks very complicated and I am not sure to handle the integral limits. I would proceed with partial integration. Is the calculation correct until now?
The limits should be $\pm\frac{\pi}{2}$ (you might want to review integration by substitution), but if you change the bottom one to $0$ then double you get the same result. You can then write the result in terms of the Beta function. It's actually faster to get there with $x=\sqrt{Ry}$, but the proof that $\operatorname{B}(a,\,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ uses the trigonometry anyway.
Your integral is $$R^{n/2}\int_0^1 y^{-1/2}(1-y)^{\tfrac{n-1}{2}}dy=R^{n/2}\operatorname{B}(\tfrac{1}{2},\,\tfrac{n+1}{2}).$$Calling this $R^{n/2}u_n$ gives $u_0=\pi,\,u_n u_{n+1}=\frac{2\pi}{n+1}$ because $\Gamma(\tfrac{1}{2})=\sqrt{\pi}$. Equivalently $u_0=\pi,\,u_1=2,\,\frac{u_{n+2}}{u_n}=\frac{n+1}{n+2}$.