Explanation

(b)
We know that sin $\frac{\pi}{6} = \frac{1}{2}$
$\sin{\left( \pi + \frac{\pi}{6} \right) = \sin{\frac{7\pi}{6} = - \frac{1}{2}}}$
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$\sin\ \frac{11\pi}{6} = \sin\left( 2\pi - \frac{\pi}{6} \right)$
$= - \sin{\frac{\pi}{6} = - \frac{1}{2}}$
$\sin\ \frac{9\pi}{6} = \sin{\frac{3\pi}{6} = - 1}$
Hence, we divide the interval $\pi\ to\ 2\pi$ as
$\left( \pi,\frac{7\pi}{6} \right),\left( \frac{7\pi}{6},\frac{11\pi}{6} \right),\left( \frac{11\pi}{6},2\pi \right)$
$\sin x = \left( 0, - \frac{1}{2} \right),\left( - \frac{1}{2}, - 1 \right),\left( - \frac{1}{2},0 \right)$
$2sinx = (0, - 1),( - 1, - 2),( - 1,0)$
$\therefore\int_{\pi}^{2\pi}{\lbrack 2\sin{x\rbrack dx = \int_{\pi}^{7\pi/6}{- 1dx + \int_{7\pi/6}^{11\pi/6}{- 2\ dx}}}}$
$+ \int_{11\pi/6}^{2\pi}{- 1dx}$
$= - \frac{5\pi}{3}$