$\int_0 ^ \frac{\pi}{2}[\sin 2x (1+\cos 3x) ]dx$ . 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?

$$y=\sin(2x)(1+\cos(3x)),x\in[0,\pi/2]$$note that $0\leq 2x\leq \pi \to 0\leq \sin(2x)\leq 1\\1+\cos(3x)\geq0\\$
now find max,min of $y=\sin(2x)(1+\cos(3x)) $ on $x\in[0,\pi/2]$ it seems $0\leq y<1 $so $[y]=0$ so $\int_0^{\pi/2}[y]dx\to 0
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https://www.desmos.com/calculator/inubfmxu8p

