Step 1: Substitution

Let $t = x^{3/2}$. Then, $\frac{dt}{dx} = \frac{3}{2}x^{1/2}$, which implies $dx = \frac{2}{3}x^{-1/2} dt$.

Step 2: Rewrite the integral

Substitute $x^{3/2}$ with $t$ and $dx$ with $\frac{2}{3}x^{-1/2} dt$ in the integral:

$\int x^2 \sin^{-1}(x^{3/2}) dx = \int x^2 \sin^{-1}(t) \cdot \frac{2}{3}x^{-1/2} dt = \frac{2}{3} \int x^{3/2} \sin^{-1}(t) dt$

Since $t = x^{3/2}$, we have:

$\frac{2}{3} \int t \sin^{-1}(t) dt$

Step 3: Integration by parts

Use integration by parts: $\int u dv = uv - \int v du$. Let $u = \sin^{-1}(t)$ and $dv = t dt$. Then, $du = \frac{1}{\sqrt{1-t^2}} dt$ and $v = \frac{t^2}{2}$.

So, $\int t \sin^{-1}(t) dt = \sin^{-1}(t) \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{\sqrt{1-t^2}} dt = \frac{t^2}{2} \sin^{-1}(t) - \frac{1}{2} \int \frac{t^2}{\sqrt{1-t^2}} dt$

Step 4: Evaluate the remaining integral

To evaluate $\int \frac{t^2}{\sqrt{1-t^2}} dt$, let $t = \sin(\theta)$. Then $dt = \cos(\theta) d\theta$.

$\int \frac{t^2}{\sqrt{1-t^2}} dt = \int \frac{\sin^2(\theta)}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) d\theta = \int \frac{\sin^2(\theta)}{\cos(\theta)} \cos(\theta) d\theta = \int \sin^2(\theta) d\theta$

Using the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$, we have:

$\int \sin^2(\theta) d\theta = \int \frac{1 - \cos(2\theta)}{2} d\theta = \frac{1}{2} \int (1 - \cos(2\theta)) d\theta = \frac{1}{2} (\theta - \frac{1}{2} \sin(2\theta)) + C = \frac{1}{2} (\theta - \sin(\theta)\cos(\theta)) + C$

Since $t = \sin(\theta)$, $\theta = \sin^{-1}(t)$ and $\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - t^2}$.

So, $\int \frac{t^2}{\sqrt{1-t^2}} dt = \frac{1}{2} (\sin^{-1}(t) - t\sqrt{1-t^2}) + C$

Step 5: Substitute back

$\int t \sin^{-1}(t) dt = \frac{t^2}{2} \sin^{-1}(t) - \frac{1}{2} \cdot \frac{1}{2} (\sin^{-1}(t) - t\sqrt{1-t^2}) + C = \frac{t^2}{2} \sin^{-1}(t) - \frac{1}{4} \sin^{-1}(t) + \frac{1}{4} t\sqrt{1-t^2} + C$

Step 6: Final Substitution

Substitute $t = x^{3/2}$ back into the expression:

$\frac{2}{3} \int t \sin^{-1}(t) dt = \frac{2}{3} \left( \frac{(x^{3/2})^2}{2} \sin^{-1}(x^{3/2}) - \frac{1}{4} \sin^{-1}(x^{3/2}) + \frac{1}{4} x^{3/2} \sqrt{1 - (x^{3/2})^2} \right) + C$

$= \frac{2}{3} \left( \frac{x^3}{2} \sin^{-1}(x^{3/2}) - \frac{1}{4} \sin^{-1}(x^{3/2}) + \frac{1}{4} x^{3/2} \sqrt{1 - x^3} \right) + C$

$= \frac{x^3}{3} \sin^{-1}(x^{3/2}) - \frac{1}{6} \sin^{-1}(x^{3/2}) + \frac{1}{6} x^{3/2} \sqrt{1 - x^3} + C$