Convergence of Type II Improper Integral

State whether the following integral is convergent or divergent:$$\int_{0}^{1} \frac{\sin(x)}{x^{1.5}} \ dx$$

The answer says that it converges due to a comparison with $\frac{1}{\sqrt{x}}$. I don't see how this works, as $\frac{1}{\sqrt{x}}$ is < $\frac{\sin(x)}{x^{1.5}}$ for $0 < x < 1$.

Any help will be greatly appreciated, thanks in advance.

Note that for $x\in [0,1]$, $0\le \sin(x)\le x$. Hence, we have

$$0\le \frac{\sin(x)}{x^{3/2}}\le \frac{1}{\sqrt x}$$

Can you finish now?