Feynman's trick VIII

Integral found on Instagram

We show the proof of this result found on Instagram using the Feynman's trick again \[\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \ln(\sin x)dx = -\frac{\pi}{32}- \frac{3\pi\ln(4)}{16}\]
Proof \[\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \ln(\sin x)dx = \underbrace{\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x\right) \ln(\sin x)dx}_{A} +\underbrace{\int_{0}^{\frac{\pi}{2}} \left(cos^4 x\right) \ln(\sin x)}_{B}\] Hence \begin{align*} A = \int_{0}^{\frac{\pi}{2}} \left(\sin^4 x\right) \ln(\sin x)dx = & \int_{0}^{1} \frac{w^4 \ln(w)}{\sqrt{1-w^2}}dw \quad (w \mapsto \sin x)\\ =& \int_{0}^{1} \frac{w^4 \frac{d}{dt}\Big|_{t=0+}w^t}{\sqrt{1-w^2}}dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \frac{w^{4+t}}{\sqrt{1-w^2}}dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} w^{4+t} (1-w^2)^{-\frac{1}{2}}dw \\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2}\int_{0}^{1} r^{\frac{t+5}{2}-1}(1-r)^{\frac{1}{2}-1}dr \quad (r \mapsto w^2) \\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2}B\left(\frac{t+5}{2},\frac{1}{2}\right)\\ =& \lim_{t \to 0+}\frac{1}{2}\left[\psi^{0}\left(\frac{t+5}{2}\right)-\psi^{0}\left(\frac{t}{2}+3\right)\right]B\left(\frac{t+5}{2},\frac{1}{2}\right)\\ =& \frac{1}{2}\left[\psi^{0}\left(\frac{5}{2}\right)-\psi^{0}\left(3\right)\right]B\left(\frac{5}{2},\frac{1}{2}\right)\\ =&\frac{1}{2}\left[\frac{8}{3}-\gamma -\ln(4) -\frac{3}{2}+\gamma \right]\frac{3\pi}{8}\\ =& \frac{\pi}{64}(7-6\ln(4)) \end{align*} \begin{align*} B = \int_{0}^{\frac{\pi}{2}} \left(\cos^4 x\right) \ln(\sin x) =& \int_{0}^{1} (1-w^2)^{\frac{3}{2}} \ln(w) dw \quad (w \mapsto \sin x)\\ =& \int_{0}^{1} (1-w^2)^{\frac{3}{2}} \frac{d}{dt}\Big|_{t=0+}w^t dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} (1-w^2)^{\frac{3}{2}} w^t dw \\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2} \int_{0}^{1} (1-r)^{\frac{3}{2}}r^{\frac{t}{2}-\frac{1}{2}} dr \quad (r \mapsto w^2)\\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2} B\left(\frac{t+1}{2},\frac{5}{2}\right)\\ =& \lim_{t \to 0+} \frac{1}{2} \left[ \psi^{0} \left(\frac{t+1}{2}\right) - \psi^{0}\left(\frac{t}{2}+3\right)\right]B\left(\frac{t+1}{2},\frac{5}{2}\right)\\ =& \frac{1}{2}\left[-\gamma-\ln(4) -\frac{3}{2}+\gamma \right]\frac{3\pi}{8}\\ =& -\frac{3\pi }{64}(3+2\ln(4)) \end{align*} Therefore \[A+B = -\frac{\pi}{32}- \frac{3\pi\ln(4)}{16}\] \[\boxed{\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \ln(\sin x)dx = -\frac{\pi}{32}- \frac{3\pi\ln(4)}{16}}\]