<s>[INST] 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.
 [/INST] Using  the right hand side of the well known Jordan inequality that states $\sin x < x$ for all $x \in \left[{0,\frac{\pi}{2}}\right]$ we have $1-2\sin x > 1-2x$ so that $\frac{1}{1-2\sin x} < \frac{1}{1-2x}$ so that for all $0\le x\le \frac{\pi}{6}$ we have  $\frac{x}{\sqrt{1-2\sin x}} < \frac{x}{\sqrt{1-2x}}$. 
Thus, 
\begin{align}
\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx} < \int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2x}}dx}
\end{align}
since  $0\le x\le \frac{\pi}{6}$ we may consider the one-to-one substitution $u^2=1-2x$, so that $x= \frac{1-u^2}{2}$ and $dx=-u du$. Also, when $x=0$ say $u=u_0$ and $x=\frac{\pi}{6}$ say $u=u_1$, therefore,
\begin{align}
\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx} < \int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2x}}dx}&= \int_{u_0}^{u_1}{\frac{\frac{1-u^2}{2}}{u}(-u)du} 
\\
&= \int_{u_0}^{u_1}{\frac{u^2-1}{2}du} = \frac{u^3_1-u^3_0}{6}-u_1 +u_0<\infty
\end{align}
which means that $\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx}$ converges.
</s>