Indefinite integral via u-substitution

Let $u = x^3+2$. Then $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}\,du$. Substituting into the integral gives $\int x^2\cos(x^3+2)\,dx = \int \cos(u)\cdot \frac{1}{3}\,du = \frac{1}{3}\int \cos(u)\,du.$

Integrating gives $\frac{1}{3}\sin(u) + C$. Substituting back $u = x^3+2$ yields the final answer $\frac{1}{3}\sin(x^3+2) + C.$

The factor that became du after dividing by three is x squared dx. This is because u equals x cubed plus two, so du equals three x squared dx, meaning x squared dx equals one third du. That substitution let the cosine term become a simple function of u alone.