Show that $$\int_{-\pi/3}^{\pi/3} \log \vert 8 \sin(t/2) (1 + \sin t)^2 \vert dt = 0.$$
WolframAlpha claims that this is true. I've tried manipulating the integrand a bunch and various trig identities, but it hasn't simplified things. It looks vaguely like Jensen's formula, but I'm not sure how to use this.
It also sort of looks like one of those integrals that can be evaluated using complex analysis and contour shifting, but I don't see how.
Note that (using $t \to -t$ on the negative interval):
$\int_{-\pi/3}^{\pi/3} \log \vert 8 \sin(t/2) (1 + \sin t)^2 \vert dt =2\pi \log 2+2\int_{0}^{\pi/3}\log {\sin(t/2)}dt+ 4\int_{0}^{\pi/3}\log {\cos t}dt$
Using now $t/2 \to t$ we get
$\int_{0}^{\pi/3}\log {\sin(t/2)}dt=2\int_{0}^{\pi/6}\log {\sin t}dt=2\int_{\pi/3}^{\pi/2}\log {\cos t}dt$,
so putting all together the original integral becomes:
$2\pi \log 2+4\int_{0}^{\pi/2}\log {\cos t}dt=4\int_{0}^{\pi/2}\log {(2\cos t)}dt=0$ by the well known result (and easy to prove by simple manipulations once one argues that the integral is indeed convergent) that
$\int_{0}^{\pi/2}\log {(2\sin t)}dt=\int_{0}^{\pi/2}\log {(2\cos t)}dt=0$