Antisymmetry about the points 1/2

Question

The first (trigonometric) integral gives 0, because the integrand in the interval (0, 1) is antisym- metric about the point 1/2.

What does this exactly mean for the following equation?

Equation

Answer 1

$$I = \int_0^1 \pi\sin^2(\pi t)\cos(\pi t)dt$$

Using the substitution $t = 1-u$, we have

$$I = \int_0^1\pi \sin^2(\pi-\pi u)\cos(\pi-\pi u)du = -I$$

$$\implies 2I=0 \implies I=0$$