$\int_0 ^ \frac{\pi}{2}[\sin 2x (1+\cos 3x) ]dx$ . Here $[t]$ denotes the greatest integer less than or equal to $t$. $\int_0 ^ \frac{\pi}{2}[\sin 2x (1+\cos 3x) ]dx$ . Here $[t]$ denotes the greatest integer function.
Can anyone give me a hint?

$I = \int_0^{2\pi} [\sin 2x (1+\cos 3x)] = \int_0^{2\pi} [\sin 2(2\pi - x) (1+\cos 3(2\pi - x))]   =  \int_0^{2\pi} [- \sin 2x (1+\cos 3x)$
$ = -\int_0^{2\pi} [\sin 2x (1+\cos 3x)] - 1$
$\therefore 2I = - \int_0^{2\pi}  dx = -2 \pi  \implies I = - \pi$
