Answer
$$\int x^2\sin{x^3}dx=-\frac{\cos x^3}{3}+C$$
Work Step by Step
$$A=\int x^2\sin{x^3}dx$$
We set $a=x^3$. We then have $$da=3x^2dx$$ $$x^2dx=\frac{1}{3}da$$
Therefore, $$A=\frac{1}{3}\int \sin ada$$ $$A=-\frac{\cos a}{3}+C$$ $$A=-\frac{\cos x^3}{3}+C$$
