Relating $\sin(x)^{2a+1}$ on interval $(0,\pi/2)$ to a factorial form The integral:

$$\int_0^{\pi/2} \sin(x)^{2a+1}dx $$

has a closed form solution in terms of factorials:

$$[(2^a)(a!)]^2 [(2a+1)!]^{-1}$$

How does this come about?
 A: I assume $a$ is an integer. 
Here comes a naive solution: use induction, together with integration by part. Denote by $f(a)=[2^a a!]^2((2a+1)!)^{-1}$
$a=0: \int_0^{\pi/2}\sin x dx=1=f(0)=(2^00!)^2(1!)^{-1}$
Assume $a=k$ is true, for $a=k+1$,
$$
\begin{split}
\int_0^{\pi/2}(\sin x)^{2(k+1)+1}dx&=\int_0^{\pi/2}(\sin x)^{2k+1}(1-\cos 2x)/2 dx\\
&=\frac{1}{2}\int_0^{\pi/2}(\sin x)^{2k+1}dx-\frac{1}{4}((\sin x)^{2k+1}\sin 2x|_0^{\pi/2}\\
&-\int_0^{\pi/2}\sin 2xd(\sin x)^{2k+1})\\
&=\frac{1}{2}f(k)+\frac{2k+1}{2}\int_0^{\pi/2}(\sin x)^{2k+1}\cos x^2dx\\
&=\frac{1}{2}f(k)+\frac{2k+1}{2}\int_0^{\pi/2}(\sin x)^{2k+1}(1-\sin x^2)dx\\
&=(k+1)f(k)-\frac{2k+1}{2}\int_0^{\pi/2}(\sin x)^{2(k+1)+1}dx
\end{split}
$$
That is,
$$
f(k+1)=(k+1)f(k)-\frac{2k+1}{2}f(k+1)
$$
or
$$
f(k+1)=2\cdot\frac{k+1}{2k+3}f(k)
$$
It is not difficult to compute that:
$$
f(k+1)=[2^{k+1} (k+1)!]^2((2(k+1)+1)!)^{-1}
$$
A: In fact the answer follows in one shot by using the beta function

$$ \beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,\mathrm{d}\theta=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0. $$

So in your case we have
$$ 2x-1 = 2a+1 \implies x = a+1 \\ 2y-1 = 0 \implies y=\frac{1}{2} $$
which gives the solution

$$ I = \frac{1}{2}\frac{\Gamma(a+1)\Gamma(1/2)}{\Gamma(a+3/2)} $$

