$\begin{aligned} & I=\int_0^a x^2 \sin x^3 d x ; \text { Put } x^3=t \Rightarrow x^2 d x=\frac{d t}{3} \\ & \therefore I=\frac{1}{3} \int_0^{a^3} \sin t d t=-\frac{1}{3}[\cos t]_0^{a^3}=-\frac{1}{3}\left[\cos a^3-1\right] \\ & =\frac{1}{3}\left[\cos a^3 -1\right] \\ & \end{aligned}$