Integrate x^2sin(x) between -pi and pi

It is possible to solve this question using integration by parts. However, we note that sin(x) is an odd function, meaning that sin(-x) = -sin(x). Thus x²sin(x) is also an odd function. This means that the area under x²sin(x) from 0 to pi is equal to the area under x²sin(x) from -pi to 0. Hence the integral of x²sin(x) between -pi and pi is 0.