I summarize the proof of the relation between the gamma and the beta functions as described in section 20.7, p. 680-681 of Angus Taylor's Advanced Calculus, Blaisdell Publishing Company, 1955, and the $B(p,p)$ functional equation in exercise 8, p. 683, from which one can derive the gamma function duplication formula.
An alternative derivation of
$\Gamma (p)\Gamma (q)=\Gamma (p+q)B(p,q)\qquad (1)$
uses 3 substitutions. The $1^{st}$ one, $t=u/(1+u)$, converts the beta $B(p,q)$ integral from
$B(p,q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}\ \mathrm{d}t\qquad (2)$
into
$B(p,q)=\int_{0}^{\infty }\frac{u^{p-1}}{(1+u)^{p+q}}.\ \mathrm{d}u\qquad (3)$
The $2^{nd}$ one, $t=vu$, transforms the gamma function integral
$\Gamma (p)=\int_{0}^{\infty }t^{p-1}e^{-t}\ \mathrm{d}t\qquad (4)$
into
$\Gamma (p)=\int_{0}^{\infty }v^{p}u^{p-1}e^{-vu}\ \mathrm{d}u.$
Hence
$\Gamma (p)\Gamma (q)=\int_{0}^{\infty }u^{p-1}\left( \int_{0}^{\infty }v^{p+q-1}e^{-v-vu}\ \mathrm{d}v\right) \ \mathrm{d}u.$
By $(4)$ and making the $3^{rd}$ change of variables, $v=w/(1+u)$, we obtain
$\int_{0}^{\infty }v^{p+q-1}e^{-v-vu}\ \mathrm{d}v=\frac{\Gamma (p+q)}{% (1+u)^{p+q}},\qquad (5)$
and by $(3)$,
$\Gamma (p)\Gamma (q)=\Gamma (p+q)\int_{0}^{\infty }\frac{u^{p-1}}{% (1+u)^{p+q}}\ \mathrm{d}u=\Gamma (p+q)B(p,q).\qquad\square$
The substitution $u=4t(t-1)$ in $B(p,p)=2\int_{0}^{1/2}\left[ t(t-1)\right] ^{p-1}\ \mathrm{d}t$ yields the functional equation $B(p,p)=2^{1-2p}B(p,1/2).$