Answer
$\frac{3(1+\sqrt 2)}{5}.$
Work Step by Step
We have
\begin{align*}
\int_{\pi/2}^{\pi} \sin\left(\frac{5\theta -\pi}{6}\right) d \theta &=\frac{6}{5}\int_{\pi/2}^{\pi}\frac{5}{6} \sin\left(\frac{5\theta -\pi}{6}\right) d \theta \\
&=-\frac{6}{5 }\cos\left(\frac{5\theta -\pi}{6}\right)|_{\pi/2}^{\pi}\\
&=-\frac{6}{5 }\cos\left(\frac{4\pi}{6}\right)+\frac{6}{5 }\cos\left(\frac{3\pi}{12}\right)\\
&=\frac{3(1+\sqrt 2)}{5}.
\end{align*}
