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

Question

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 1

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 $[0,\,\pi/2]$.

Answer 2

Near $0$, $|\sin x|\sim |x|$, so $\sqrt{|\sin x|}\sim \sqrt{|x|}$ and $$\frac 1{\sqrt{\sin x}}\sim \frac 1{\sqrt x}\quad \text{ for } x>0.$$

As $\displaystyle\int_0^1\frac{\mathrm d x}{\sqrt x}$ converges, $\;\displaystyle\int_0^1\frac{\mathrm d x}{\sqrt{\sin x}}$ converges too.