<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. :-(
$$$$
Second attempt, I used $\sin x\le x $ on $[ 0, \frac{\pi}{6} ]$.
But I couldn't determine. :-[
Could you give me some advice? 
Thanks in advance.
 [/INST] Since we are only interested in whether the improper integral converges or diverges, let us write out the integrand in terms of their power series and see what we get:
$$  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. 
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