$I=\underset{0}{\overset{2\pi }{\int }}\left[\sin 2x\left(1+\cos 3x\right)\right]dx \dots \left(1\right)$
$I=\underset{0}{\overset{2\pi }{\int }}\left[\sin \left(2\mathrm{\pi }-2x\right)\left(1+\cos \left(2\mathrm{\pi }-3x\right)\right)\right]dx$ 

Applying $\left(\underset{0}{\overset{a}{\int }}f\left(x\right)=\underset{0}{\overset{a}{\int }}f\left(a-x\right)dx\right)$

$$$I=\underset{0}{\overset{2\pi }{\int }}\left[-\sin 2x\left(1+\cos 3x\right)\right]dx \dots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$

$2I=\underset{0}{\overset{2\pi }{\int }}\left(\left[\sin 2x\left(1+\cos 3x\right)\right]+\left[-\sin 2x\left(1+\cos 3x\right)\right]\right)dx$
$\Rightarrow 2I=\underset{0}{\overset{2\pi }{\int }}-1dx \{\because \left[x\right]+\left[-x\right]=\left\{\begin{array}{c}0 :x\in \mathrm{I}\\ -1 :x\notin \mathrm{I}\end{array} \right.$

$\Rightarrow 2I={\left(-x\right)}_0^{2\pi }$

$\Rightarrow 2I=-2\pi$

$\Rightarrow I= -\pi$