Analyzing the Setup
To solve the integral I=∫02π[sin2x(1+cos3x)]dx, we first define the function inside the greatest integer brackets as:
The goal is to evaluate the integral of the floor function [g(x)] over the interval [0,2π].
Applying Symmetry
The limits of integration, 0 to 2π, suggest the use of the King's Property, which states:
Applying this to our function g(x) with a=2π, we substitute x with 2π−x:
sin(2(2π−x))=sin(4π−2x)=−sin2x
cos(3(2π−x))=cos(6π−3x)=cos3x
The Master Equation
Combining these results, we find the symmetry of the function:
g(2π−x)=(−sin2x)(1+cos3x)=−g(x)
We now express the integral I in two ways:
Adding these two equations yields:
2I=∫02π([g(x)]+[−g(x)])dx
Final Calculation
We utilize the property of the greatest integer function, which states that [t]+[−t]=−1 for all $t
otin \mathbb{Z}$. Since the set of points where g(x) is an integer has measure zero, we can treat the integrand as −1 almost everywhere:
Dividing by 2, we arrive at the final result:
I=−π