[Cambridge/1997] Use the substitution $y = x^3$ to find the value of
$\int_{0}^{2} \frac{x^2}{1 + x^6} \, dx.$

1. Differentiate the given substitution to relate $dx$ and $dy$.

Let $y = x^3$.
$\frac{dy}{dx} = 3x^2 \implies x^2 \, dx = \frac{1}{3} \, dy$

2. Determine the new limits of integration in terms of $y$.

When $x = 0$, $y = 0^3 = 0$.
When $x = 2$, $y = 2^3 = 8$.

3. Express the definite integral entirely in terms of $y$.

$\int_{0}^{2} \frac{x^2}{1 + x^6} \, dx = \int_{0}^{2} \frac{x^2}{1 + (x^3)^2} \, dx$
$= \int_{0}^{8} \frac{1}{1 + y^2} \left( \frac{1}{3} \right) \, dy$
$= \frac{1}{3} \int_{0}^{8} \frac{1}{1 + y^2} \, dy$

4. Evaluate the definite integral using standard integration formulas.

$= \frac{1}{3} \left[ \tan^{-1} y \right]_{0}^{8}$
$= \frac{1}{3} (\tan^{-1} 8 - \tan^{-1} 0)$
$= \frac{1}{3} \tan^{-1} 8$