Does the following integral converge or diverge? $$ \int_0^{1/2}\frac{1}{\sin x\ln x}\, dx $$
When $x \in [0,1/2] , \sin x>0$ and $\ln x<0$. Therefore $-1/(\sin x\ln x)>0$, which means I can use the comparision test.
$$ \int_0^{1/2} \frac{1}{\sin x\ln x}dx < \int_0^{1/2} \frac{1}{\ln x}dx $$
I don't know what to do next in order to prove convergence (which I think is the answer). Hints will be appreciated.
Hint: Note that $\frac2\pi x\le\sin(x)\le x$ for $0\le x\le\frac\pi2$. Therefore, $$ \frac{\pi}{2}\int_0^{1/2}\frac1{x\log(x)}\,\mathrm{d}x\le\int_0^{1/2}\frac1{\sin(x)\log(x)}\,\mathrm{d}x\le\int_0^{1/2}\frac1{x\log(x)}\,\mathrm{d}x. $$