1. **State the problem:** We need to evaluate the integral $$\int x^2 \sec^2(x^3) \, dx.$$\n\n2. **Identify the substitution:** Notice the inner function inside the secant squared is $x^3$. Let $$u = x^3.$$ Then, $$\frac{du}{dx} = 3x^2 \implies du = 3x^2 \, dx.$$\n\n3. **Rewrite the integral in terms of $u$:** From the substitution, $$x^2 \, dx = \frac{du}{3}.$$ So the integral becomes $$\int x^2 \sec^2(x^3) \, dx = \int \sec^2(u) \frac{du}{3} = \frac{1}{3} \int \sec^2(u) \, du.$$\n\n4. **Recall the integral formula:** The integral of $$\sec^2(u)$$ with respect to $$u$$ is $$\tan(u) + C.$$\n\n5. **Integrate and substitute back:**\n$$\frac{1}{3} \int \sec^2(u) \, du = \frac{1}{3} \tan(u) + C = \frac{1}{3} \tan(x^3) + C.$$\n\n**Final answer:** $$\int x^2 \sec^2(x^3) \, dx = \frac{1}{3} \tan(x^3) + C.$$