convergence of improper integral, sin in denominator

Question

Let $f: (0,\infty) \times (0,1) \rightarrow \mathbb{R} $ where $f(x,t) = \frac{t^x}{\sin{\sqrt{t(1-t)}}}$ . I want to check if $g(x) = \int_0^1{f(x,t)dt}$ is convergent for $x\in(0,\infty)$. Have tried different ways, for example integration by parts or substituting $t$ as $\sin^2{\theta}$, none of which worked. The biggest problem I'm facing is that $\frac{1}{\sin{\sqrt{t(1-t)}}} \rightarrow \infty$ as $t \rightarrow 0$, thus I'm finding it hard to bound $f(x,t)$, namely to find $g(t)$ such that $\lvert f(x,t) \rvert \leq g(t)$, which is needed to say $g(x)$ converges.