Convergence of improper integral $\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx$ I'm trying to determine whether the following improper integral is convergent or divergent.
$$
\int_{0}^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}dx
$$
At first, I substituted $t=\dfrac{\pi}{2} - x $
and then I used $1-\dfrac{1}{2}x^2  \le \cos x$.
But I couldn't determine. :-(
$$$$
Second attempt, I used $\sin x\le x $ on $[ 0, \frac{\pi}{6} ]$.
But I couldn't determine. :-[
Could you give me some advice? 
Thanks in advance.

Thm: Suppose $0<p<1.$ Let $f$ be continuous on $[a,b],$ with $f(b)=0$ and $f>0$ on $[a,b).$ If $f'(b)$ exists and is nonzero, then
$$\int_a^b \frac{1}{f(x)^p}\, dx < \infty.$$
In our problem, $f(x) = 1-2\sin x, p = 1/2,$ and $[a,b]=[0,\pi/6].$ We have $f'(\pi/6)=-\sqrt 3,$ so the integral in this problem converges.
Sketch of proof of theorem: Because $f(b)=0, f'(b)\ne 0,$ there exists $\delta > 0$ such that
$$|f(x)| \ge (|f'(b)|/2)|x-b|, \ x\in[b-\delta,b].$$
This falls right out of the difference quotients whose limit equals $f'(b).$ Thus
$$\int_{b-\delta}^b \frac{1}{f(x)^p}\,dx \le \int_{b-\delta}^b\frac{2^p}{|f'(b)|^p|x-b|^p} \,dx.$$
Because $p<1,$ that integral converges. By the positivity of $f, \int_a^{b-\delta}(1/f^p)$ is no problem. This proves the theorem.
