Determine whether $\int^{1/2}_0\frac{1}{\sin x\cdot \ln x}dx$ is convergence, absolute convergence or divergent. I'm trying to determine whether $\int^{1/2}_0\frac{1}{\sin x\cdot \ln x}dx$ is convergence, absolute convergence or divergent. 
Let $f(x) = \frac{1}{\sin x\cdot \ln x}$, $f(x)< 0$ for $(0,0.5]$.
Therefore I assume I need to work with $-f(x) \ge 0$ in this question.
I'm trying to use the comparing rule without success to solve this.
I thought of using the function $h(x) = \frac{-1}{\ln x}$ which holds $f(x)>g(x)$ in the question's interval, although that didn't get me anywhere. 
What are my options here?

$\sin{x}<x$ for $0<x<\pi/2$ (draw a picture to see this). So
$$ \frac{1}{\sin{x} \cdot (-\log{x})} >\frac{1}{x \cdot (-\log{x})}, $$
and this has antiderivative $-\log{(-\log{x})}$. This is finite when $x=1/2$ and diverges for $x \to 0$, so the original integral diverges.
