Official solution
\[I = \int\limits_0^{2\pi } {\left[ {\sin 2x(1 + \cos 3x)} \right]} dx\] .... (i)
\[ \therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx\]
\[ \because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx\]
By (i) + (ii)
\[ \Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx} \]
\[ \Rightarrow 2I = - (x)_0^{2\pi }\]
\[ \Rightarrow \] I = --\[\pi \]