Since
$$
\log(\sin(t/2))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(kt)}k
$$
and for $\lambda\ge0$,
$$
\int_{-\infty}^\infty\frac{\cos(\lambda t)}{1+t^2}\,\mathrm{d}t=\pi e^{-\lambda}
$$
we get
$$
\begin{align}
\int_0^\infty\frac{\log(\sin^2(t/2))}{a^2+t^2}\,\mathrm{d}t
&=-\int_{-\infty}^\infty\left(\log(2)+\sum_{k=1}^\infty\frac{\cos(kt)}k\right)\frac{\mathrm{d}t}{a^2+t^2}\\
&=-\frac{\pi\log(2)}a-\sum_{k=1}^\infty\frac{\pi e^{-ak}}{ak}\\
&=-\frac{\pi\log(2)}a+\frac\pi{a}\log(1-e^{-a})\\[6pt]
&=\frac\pi{a}\log\left(\frac{1-e^{-a}}2\right)
