Write a value

let $x^{3}=t$

Differentiating on both sides we get,

$3 \mathrm{x}^{2} \mathrm{dx}=\mathrm{dt}$

$x^{2} d x=\frac{1}{3} d t$

substituting above equation in $\int x^{2} \sin x^{3} d x$ we get,

$=\int \frac{1}{3} \sin t d t$

$=-\frac{1}{3} \cos t+c$

$=-\frac{1}{3} \cos x^{3}+c$