Feynman's trick III

Integral involging log-sine

We prove the following integral using the famous Feynman's trick \[\int_{0}^{\frac{\pi}{2}} \ln^2(\sin x) dx = \frac{\pi \ln^2 2}{2} + \frac{\pi^3}{24}\]
Proof
\begin{align*} \int_{0}^{\frac{\pi}{2}} \ln^2(\sin x) dx =& \int_{0}^{\frac{\pi}{2}} \frac{d^2}{dt^2}\Big|_{t=0+} \sin^t(x) dx \\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{\frac{\pi}{2}} \sin^t(x) dx \quad (w \mapsto \sin x)\\ =& \frac{d^2}{dt^2} \Big|_{t=0+}\int_{0}^{1} w^t(1-w^2)^{-\frac{1}{2}} dw \quad (y \mapsto w^2)\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \frac{1}{2}\int_{0}^{1} y^{\frac{t}{2}-\frac{1}{2}}(1-y)^{-\frac{1}{2}} dw \\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \frac{1}{2} B \left(\frac{t+1}{2},\frac{1}{2}\right)\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{1+t}{2}\right)}{\Gamma\left(1+ \frac{t}{2}\right)}\\ \end{align*} Then \begin{align*} \frac{d}{dt}\Big|_{t=0+} \left[ \frac{d}{dt}\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{1+t}{2}\right)}{\Gamma\left(1+ \frac{t}{2}\right)} \right]=& \frac{d}{dt}\Big|_{t=0+} \frac{\sqrt{\pi} \Gamma(\frac{1+t}{2})\left(\psi^{0}(\frac{1+t}{2}) -\psi^{0}(\frac{t}{2}+1)\right)}{4\Gamma(\frac{t}{2}+1)}\\ =& \lim_{t \to 0+} \frac{\sqrt{\pi}\Gamma(\frac{1+t}{2})}{8\Gamma(\frac{t}{2}+1)} \left[\psi^0 (\frac{t}{2}+1)^2-2\psi^0 (\frac{t+1}{2})\psi^0 (\frac{t}{2}+1)+ \psi^0 (\frac{t+1}{2})^2-\psi^1 (\frac{t}{2}+1)+\psi^1 (\frac{t+1}{2})\right] \end{align*} Taking the limit as $\displaystyle t \to 0+$ \[\boxed{\int_{0}^{\frac{\pi}{2}} \ln^2(\sin x) dx = \frac{\pi \ln^2 2}{2} + \frac{\pi^3}{24}}\]