Explanation

(c)
$f(x) = \left| \begin{array}{r}
\sin{x + \sin{2x + \sin{3x}}}\ \ \ \ \ \sin{2x\ \ \ \ \ \ \ \ \ \sin{3x}} \\
\ \ \ \ \ \ \ \ 3 + 4\sin{x\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\sin x \\
1 + \sin{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sin{x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1}}
\end{array} \right|$
[Applying $C_{1} \rightarrow C_{1} - (C_{2} + C_{3})\rbrack$
$= \left| \begin{matrix}
\sin x & \sin{2x} & \sin{3x} \\
0 & 3 & 4\sin x \\
0 & \sin x & 1
\end{matrix} \right|$
$= \sin{x(3 - 4\sin^{2}{x)}}$
$= \sin{3x}$
Then, $\int_{0}^{\pi/2}{f(x)dx = \int_{0}^{\pi/2}{\sin{3x\ dx}}}$
$= - \left\lbrack \frac{\cos{3x}}{3} \right\rbrack_{0}^{\pi/2}$
=$\left( - \frac{1}{3} \right)\left\lbrack \cos\left( 3 \times \frac{\pi}{2} \right) - \cos 0 \right\rbrack$
$= \left( - \frac{1}{3} \right)(0 - 1)$
$= \frac{1}{3}$