Find: ∫ x^2 · sin^-1(x^3 / 2) d x

$$\begin{aligned} & \text { Let } x^{\frac{3}{2}}=t \\ & \Rightarrow \frac{3}{2} x^{\frac{1}{2}} d x=d t \end{aligned} $$ The given integral becomes $\displaystyle \frac{2}{3} \int t \sin ^{-1} t d t$ $$\begin{aligned} & =\frac{2}{3}\left[\sin ^{-1} t \times \frac{t^{2}}{2}-\int \frac{1}{\sqrt{1-t^{2}}} \times \frac{t^{2}}{2} d t\right] \\ & =\frac{1}{3}\left[\sin ^{-1} t \times t^{2}+\int \frac{1-t^{2}-1}{\sqrt{1-t^{2}}} d t\right] \\ & =\frac{1}{3}\left[\sin ^{-1} t \times t^{2}+\int \sqrt{1-t^{2}} d t-\int \frac{1}{\sqrt{1-t^{2}}} d t\right] \\ & =\frac{1}{3}\left[t^{2} \sin ^{-1} t+\frac{t}{2} \sqrt{1-t^{2}}+\frac{1}{2} \sin ^{-1} t-\sin ^{-1} t\right]+\mathrm{C} \end{aligned} $$