A trig integral with a tangent substitution

Here's a completely different method. For $a,b\in\Bbb C$ such that $\operatorname{Re}a,\operatorname{Re}b>-1$, we define $$I(a,b)=\int_0^{\pi/2}\sin(x)^{a}\cos(x)^bdx$$ Sub.: $$t=\sin(x)^2\Rightarrow 2t^{-1/2}(1-t)^{-1/2}dt=dx$$ Hence $$I(a,b)=2\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}dt=2\int_0^1t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}dt$$ Then recall the definition of the Beta function $$\mathrm B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ Where $$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt$$ is the Gamma function, which extends the definition of factorial to all complex $s$ ($\Gamma(s)=(s-1)!$).

We then see immediately that $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ Choosing $a=2n$ and $b=0$, and using $\Gamma(1/2)=\sqrt\pi$: $$J_n=I(2n,0)=\frac{\sqrt\pi}2\frac{\Gamma(n+1/2)}{\Gamma(n+1)}=\frac{\sqrt\pi}{2n}\frac{\Gamma(n+1/2)}{\Gamma(n)}\qquad \text{assuming}\ \ n>0$$ Then we recall the Legendre duplication formula: $$\Gamma(s+1/2)=2^{1-2s}\sqrt\pi \frac{\Gamma(2s)}{\Gamma(s)}$$ $$\frac{\Gamma(s+1/2)}{\Gamma(s)}=2^{1-2s}\sqrt\pi \frac{\Gamma(2s)}{\Gamma^2(s)}$$ Then using $\Gamma(s)=(s-1)!$ and the definition of ${2s\choose s}$, we have $$\frac{\Gamma(2s)}{\Gamma^2(s)}=\frac{s}2{2s\choose s}$$ so $$\frac{\Gamma(s+1/2)}{\Gamma(s)}=\frac{s\sqrt\pi}{4^s}{2s\choose s}$$ And $$J_n=\frac{\pi}{2^{2n+1}}{2n\choose n}$$ And from symmetry about $x=\pi/2$, we have that $$\int_0^\pi \sin(x)^{2n}dx=\frac\pi{4^n}{2n\choose n}$$ QED