## Calculus 8th Edition

$$\int_{-\pi/3}^{\pi/3}x^4\sin xdx=0$$
Denote the function under the integral $f(x)=x^4\sin x$. Now we have $$f(−x)=(-x)^4\sin(-x)=x^4(-\sin x)=-x^4\sin x=−f(x)$$ i.e. the function $f$ is odd. Knowing that the integral of an odd function with the symmetric integration bounds (one bound is $−a$ and the other is $a$) is equal to $0$ we have that: $$\int_{-\pi/3}^{\pi/3}x^4\sin xdx=0$$