1. **State the problem:** Evaluate the integral $$\int x^2 \sec^2(x^3) \, dx$$. 2. **Identify the substitution:** Notice the inner function inside the secant squared is $x^3$. Let $$u = x^3$$. 3. **Compute the differential:** Then $$du = 3x^2 \, dx \implies dx = \frac{du}{3x^2}$$. 4. **Rewrite the integral:** Substitute $u$ and $dx$ into the integral: $$\int x^2 \sec^2(u) \cdot \frac{du}{3x^2} = \int \frac{\cancel{x^2}}{3 \cancel{x^2}} \sec^2(u) \, du = \frac{1}{3} \int \sec^2(u) \, du$$. 5. **Integrate:** Recall that $$\int \sec^2(u) \, du = \tan(u) + C$$. 6. **Substitute back:** Replace $u$ with $x^3$: $$\frac{1}{3} \tan(x^3) + C$$. **Final answer:** $$\int x^2 \sec^2(x^3) \, dx = \frac{1}{3} \tan(x^3) + C$$.