Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 42

Find the definite integral $\displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}{2}} \frac{x^2 \sin x}{1 + x^6} dx$

Let the function $\displaystyle f(x) = \frac{x^2 \sin x}{1 + x^6}$. We have that $\displaystyle f(-x) = \frac{(-x)^2 \sin (-x)}{1 + (-x)^6} = \frac{x^2 (- \sin x)}{1 + x^6} = - \frac{x^2 \sin x}{1 + x^6} = -f(x)$

Hence $f$ is an odd function. Thus, it is symmetric to the origin, we can say that the difference of the area below the curve from $\displaystyle \left[ \frac{- \pi}{2}, 0 \right]$ and area below the curve from $\displaystyle \left[ 0, \frac{\pi}{2} \right]$ is zero. Therefore,

$\displaystyle \int^{\frac{\pi}{2}}_{\frac{- \pi}{2}} \frac{x^2 \sin x}{1 + x^6} dx = 0$