1 minute left for this integral

The function $f\colon x\longmapsto x^8\sin(x)$ is odd since $x\longmapsto\sin(x)$ is odd and $x\longmapsto x^8$ is even.
Hence:
$$I:=\int_{-\pi}^{\pi}f(x)\,\mathrm{d}x\overset{u=-x}{= }\int_{\pi}^{-\pi}f(-u)\,(-\mathrm{d}u)=\int_{-\pi}^{\pi}(-f(u))\,\mathrm{d}u=-I$$So $I=0$.