Integrate $\int_{-1}^{1}x^2\sin^{2019}(x)e^{-x^4}dx$

Question

I'm following a course on analysis and I am supposed to compute the following integral:

$\int_{-1}^{1}x^2\sin^{2019}(x)e^{-x^4}dx$.

I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks

Answer 1

Just substitute $x=-t$ and you will get that: $$I=\int_{-1}^{1}x^2\sin^{2019}(x)e^{-x^4}dx=-\int_{-1}^{1}t^2\sin^{2019}(t)e^{-t^4}dt=-I$$ $$I=-I\Rightarrow I=0$$