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.

Since
$$
(\sqrt{1-2\sin x})'=-\frac{\cos x}{\sqrt{1-2\sin x}},
$$
using integration by parts we have
\begin{eqnarray}
\int_0^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}\,dx&=&-\int_0^{\pi/6}\frac{x}{\cos x}(\sqrt{1-2\sin x})'\,dx\\
&=&-\frac{x}{\cos x}\sqrt{1-2\sin x}\Big|_0^{\pi/6}+\int_0^{\pi/6}\sqrt{1-2\sin x}\left(\frac{x}{\cos x}\right)'\,dx\\
&=&\int_0^{\pi/6}\sqrt{1-2\sin x}\frac{\cos x+x\sin x}{\cos^2x}\,dx
\end{eqnarray}
It is clear that the function $\displaystyle x\mapsto f(x):=\sqrt{1-2\sin x}\frac{\cos x+x\sin x}{\cos^2x}$ is continuous on $[0,\pi/6]$, and therefore its integral over $[0,\pi/6]$ exists and is finite. It follows that the integral $\displaystyle \int_0^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}\,dx$ is convergent.
