integration - Check convergence of $ int_{0}^{1}frac{dx}{|sin{x}|^{1/2}} $

Check convergence of integral: $$ \int_{0}^{1}
\frac{dx}{|\sin{x}|^{1/2}} $$

My attempt

Dirichlet's test would't help there so I am going to use comparison test:

$$ x \ge \sin{x} \\
\frac{1}{x} \le \frac{1}{\sin{x}} \\
\frac{1}{|x|} \le \frac{1}{|\sin{x}|} \\
\frac{1}{|x|^{1/2}} \le \frac{1}{|\sin{x}|^{1/2}}$$
So

$$ \int_{0}^{1} \frac{dx}{|\sin{x}|^{1/2}} \ge \int_{0}^{1} \frac{1}{|x|^{1/2}} $$
But $\int_{0}^{1} \frac{1}{|x|^{1/2}}$ converges so it doesn't help me.

Answer

Instead of using the upper bound $\sin x\le x$, use the to lower bound $\sin x\ge\frac{2x}{\pi}$, which follows from $\sin x$ being concave on $[0,\,\pi/2]$.