What is the value of $\int_{0}^{2\pi}(x-\pi)^2 (\sin x) dx$? What is the value of $\int_{0}^{2\pi}(x-\pi)^2 (\sin x) dx$?

AFAIK : $f(x)$ is odd function $(x-\pi)^2$ should be even because of square, and it's odd because of $(\sin x)$.

Can you explain in formal way please?


Hint: Take, $y=x-\pi$. So, now the $f(y)=y^2\sin(y+\pi)=-y^2\sin y$ . So, this is your odd function. Current upper and lower limit of your integral will be $\pi$ and $-\pi$ respectively.
