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 we are only interested in whether the improper integral converges or diverges, let us write out the integrand in terms of their power series and see what we get:
$$  2 \sin x = 2 \cdot \sum_{n=0}^{\infty} (-1)^n \frac{x^{(2n+1)}}{(2n+1)!} = 2x - \frac{x^3}{3}+\frac{x^5}{60}+ O(x^7)$$
Then we have: 
$$ 1 - 2 \sin x = 1 -2 \cdot \sum_{n=0}^{\infty} (-1)^n \frac{x^{(2n+1)}}{(2n+1)!} = 1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7) $$
So now we have (and we rationalize the integral):
$$\int_0^\frac{\pi}{6}\frac{x}{ \sqrt{1-2 \sin x}} dx= \int_0^\frac{\pi}{6} \frac{x}{\sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)}} \cdot \frac {\sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)}} {\sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)}} dx= \int_0^ \frac{\pi}{6}  \frac{x}{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} \cdot \sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} dx$$
Then by long division:
$$\frac{x}{1-2 \sin x} = \frac{x}{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} = x + 2x^2 +4x^3+ \frac{23}{x}x^4+\frac{44}{3}x^5+ O(x^6) $$
So the integral simplifies to:
$$ \int_0^\frac{\pi}{6} \left( x + 2x^2 +4x^3+ \frac{23}{x}x^4+\frac{44}{3}x^5+ O(x^6) \right )  \cdot \sqrt{1 - 2x + \frac{x^3}{3} - \frac{x^5}{60} + O(x^7)} dx$$ 
Then by multiplication we get:
$$ \int_0^\frac{\pi}{6} \left (x+ x^2 + \frac{3x^3} {2} + \frac{7x^4}{3}+\frac{31x^5}{8}+O(x^6) \right )dx $$
Finally when we integrate piecewise and evaluate at the upper and lower endpoints:
$$ \int_0^\frac{\pi}{6} \left (x+ x^2 + \frac{3x^3} {2} + \frac{7x^4}{3}+\frac{31x^5}{8}+O(x^6) \right )dx = \left ( \frac{x^2}{2}+ \frac{x^3}{3} + \frac{3x^4}{8} + \frac{7x^5}{15} + O(x^6) \right)_0^{\frac{\pi}{6}} \approx 0.231477911 + 0 \approx 0.2315$$
Thus we conclude that the improper integral must converge. 
