Interesting question. We may start from the definition of the Beta function:
$$B(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1}\ \text{d}t$$
and rewrite it when $m\to 2m$:
$$B(2m, n) = \int_0^1 t^{2m-1}(1-t)^{n-1}\ \text{d}t$$
$$B(2m, n) = \int_0^1 t^{2m-1} \frac{(1-t)^n}{1-t}\ \text{d}t$$
Now, since the range of integration is $[0, 1]$, we are allowed to make use of the geometric series
$$\frac{1}{1-t} = \sum_{k = 0}^{+\infty} t^k$$
Hence
$$B(2m, n) = \int_0^1 t^{2m-1} (1-t)^n \sum_{k = 0}^{+\infty} t^k\ \text{d}t = \sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1} (1-t)^n t^k\ \text{d}t$$
And easily write:
$$\sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1+k} (1-t)^n\ \text{d}t$$
Calling now
$$2m-1+k = a ~~~~~~~~~~~ n = b$$
We notice that the integral is well known:
$$ \int_0^1 t^a (1-t)^b\ \text{d}t \equiv B(a+1, b+1)$$
Then we end up with the partial result (re-expanding $a$ and $b$):
$$B(2m, n) = \sum_{k = 0}^{+\infty} B(2m+k, n+1)$$
That series does exist and it does converge to a known result:
$$\sum_{k = 0}^{+\infty} B(2m+k, n+1) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
What you end up with is a sort of recursive relation for the Beta function:
$$B(2m, n) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
BUT
The above expression can be simplified!
$$\frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)} \equiv \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
What we obtained is actually nothing than what we would have obtained by simply substituting at the beginning $m\to 2m$ in the Gamma function / Beta function definition.
$$B(2m, n) = \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
This really suggest that such a particular duplication formula for the Beta function may not exist at all, since all you need is the Gamma function and ITS duplication formula, through which you can evaluate $\Gamma(2m)$.
Seems like that this is the only "duplication formula" for the beta function.
(Also, I found nothing on reviews or literature).
Using $\cos^2 r = 1 - \sin^2 r$ in what you've already determined, you get
$$\begin{equation}\begin{aligned}
\sigma_{n+1} & = n\int_0^\pi \sin^{n-1}r\cos^2 r \, dr \\
& = n\int_0^\pi \sin^{n-1}r(1 - \sin^2 r) \, dr \\
& = n\int_0^\pi (\sin^{n-1}r - \sin^{n+1} r) \, dr \\
& = n\int_0^\pi \sin^{n-1}r \, dr - n\int_0^\pi \sin^{n+1}r \, dr \\
& = n\sigma_{n-1} - n\sigma_{n+1}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This leads to
$$\begin{equation}\begin{aligned}
(n+1)\sigma_{n+1} & = n\sigma_{n-1} \\
\sigma_{n+1} & = \frac{n}{n+1}\sigma_{n-1}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Best Answer
Let $ k\in\mathbb{N} : $
$$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=\int_{0}^{\frac{\pi}{2}}{\sin^{k}{\theta}\,\mathrm{d}\theta}+\int_{\frac{\pi}{2}}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta} $$
Then, making the substitution $ \left\lbrace\begin{aligned}\theta &=\pi-x \\ \mathrm{d}\theta &=-\,\mathrm{d}x \end{aligned}\right. $ in the second term, we get that : $$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=2\int_{0}^{\frac{\pi}{2}}{\sin^{k}{\theta}\,\mathrm{d}\theta} $$
Using the substitution $ \left\lbrace\begin{aligned}u&=\sin^{2}{\theta}\\ \mathrm{d}\theta &=\frac{\mathrm{d}x}{2\sqrt{x}\sqrt{1-x}}\end{aligned}\right. $, we get the following : $$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=\int_{0}^{1}{x^{\frac{k-1}{2}}\left(1-x\right)^{-\frac{1}{2}}\,\mathrm{d}x}=\beta\left(\frac{k+1}{2},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)} $$