Explanation

(b)
$\int_{0}^{2\pi}{(\sin{x + |\sin{x)dx}}}$
$= \int_{0}^{\pi}{(\sin{x + \sin{x)dx + \int_{\pi}^{2\pi}{(\sin{x - \sin{x)\ dx}}}}}}$
$= {\int_{0}^{\pi}{2\sin{x\ dx + \int_{\pi}^{2\pi}0}}\ dx = 2\lbrack - \cos{x\rbrack}}_{0}^{\pi} + 0$
$= - 2(\cos{\pi - \cos{0) = - 2( - 1 - 1) = 4}}$