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.
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I do not understand this topic and particularly this example. In the class the result was found out but I still do not get it. How did the teacher came up with this outcome?