= Solution
Within the requested interval, the <function> is symmetric about $x=\pi/2$ because $\sin(\pi-x)=\sin x$. Its <derivative> $\cot x$ is positive before $\pi/2$, zero there, and negative afterwards. The negative second <derivative> shows that this is a strictly <concave function>. Consequently its \b[unique maximum is $(\pi/2,0)$]. It tends to $-\infty$ at both ends, giving vertical <asymptotes> \b[$x=0$ and $x=\pi$].
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-log-sine.png]
{title=The logarithm of sine between its two vertical asymptotes}
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