Very indirect subsitution

Start with the numerator integral. I make the substitution $u=\arccos(e^t)$ . So, $\int_{-\infty}^0 \frac{te^t}{\sqrt{1-e^{2t}}}dt = -\int_{\pi/2}^0 \ln(\cos(u))du=\int_0^{\pi/2} \ln(\cos(u))du= \int_0^{\pi/2} \ln(\cos(0+\pi/2-x))dx=\int_0^{\pi/2} \ln(\sin(x))dx$ . $$ Now here’s what I did for the denominator integral: $\int_0^\pi \ln(1+\cos(x))dx=\int_0^\pi \ln(\frac{1+\cos(x)}{1}\frac{1-\cos(x)}{1-\cos(x)})dx=\int_0^\pi \ln(\sin^2(x))dx-\int_0^\pi \ln(1-\cos(x))dx$ $=2\int_0^\pi \ln(\sin(x))dx -\int_0^\pi \ln(1-\cos(0+\pi-x))dx=4\int_0^{\pi/2} \ln(\sin(x))dx-\int_0^\pi \ln(1+\cos(x))dx$ . i.e., $\int_0^\pi \ln(1+\cos(x))dx=4\int_0^{\pi/2} \ln(\sin(x))dx-\int_0^\pi \ln(1+\cos(x))dx$ or $\int_0^\pi \ln(1+\cos(x))dx$ $=2\int_0^{\pi/2} \ln(\sin(x))dx$ . $$ Thus, the denominator is twice as large as the numerator, which leads to the answer, 1/2.