Question

The question is in the attached picture​

Answer 1

Answer: $\frac{1}{3} \sin x^{3}+C$

Step-by-step explanation:

Let $u=x^3$. Then, $3x^2 dx = du \longrightarrow x^2 dx =\frac{1}{3}du$

So, we can rewrite the original integral as

$\frac{1}{3} \int \cos u \text{ } du=\frac{1}{3} \sin u+C=\frac{1}{3} \sin x^{3}+C$