Answer: 12

Small Hint:

Set $t=x\sin x,$ which is positive on the given interval

Big Hint:

Apply AM-GM to $9t+\frac4t$ and verify that equality is attainable

Solution:

Let $t=x\sin x>0.$ The expression becomes $$9t+\frac4t\geq2\sqrt{9t\cdot\frac4t}=12,$$ with equality when $9t=\frac{4}{t},$ or $t=\frac{2}{3}.$ The continuous function $x\sin x$ tends to $0$ as $x$ tends to $0$ and equals $\frac{\pi}{2}>\frac{2}{3}$ at $x=\frac{\pi}{2}.$ Thus it takes the value $\frac{2}{3}$ somewhere in $(0,\frac{\pi}{2}),$ so the lower bound $12$ is attained.