Answer: [tex]\frac{1}{3} \sin x^{3}+C[/tex]
Step-by-step explanation:
Let [tex]u=x^3[/tex]. Then, [tex]3x^2 dx = du \longrightarrow x^2 dx =\frac{1}{3}du[/tex]
So, we can rewrite the original integral as
[tex]\frac{1}{3} \int \cos u \text{ } du=\frac{1}{3} \sin u+C=\frac{1}{3} \sin x^{3}+C[/tex]
