1. **State the problem:** We want to find the integral of the function $$ (3x^2 + 2x) \sin(x^3 + x^2) \, dx $$. 2. **Identify the method:** This integral suggests using substitution because the derivative of the inside function of sine, $$ x^3 + x^2 $$, appears as a factor outside. 3. **Set the substitution:** Let $$ u = x^3 + x^2 $$. 4. **Find $$ du $$:** Differentiate $$ u $$ with respect to $$ x $$: $$$ \frac{du}{dx} = 3x^2 + 2x $$$ So, $$$ du = (3x^2 + 2x) \, dx $$$ 5. **Rewrite the integral:** Substitute $$ u $$ and $$ du $$ into the integral: $$$ \int (3x^2 + 2x) \sin(x^3 + x^2) \, dx = \int \sin(u) \, du $$$ 6. **Integrate:** The integral of $$ \sin(u) $$ is: $$$ -\cos(u) + C $$$ 7. **Back-substitute:** Replace $$ u $$ with the original expression: $$$ -\cos(x^3 + x^2) + C $$$ **Final answer:** $$$ \int (3x^2 + 2x) \sin(x^3 + x^2) \, dx = -\cos(x^3 + x^2) + C $$$