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. :-(
$$$$
Second attempt, I used $\sin x\le x $ on $[ 0, \frac{\pi}{6} ]$.
But I couldn't determine. :-[
Could you give me some advice? 
Thanks in advance.
 A: Since
$$
(\sqrt{1-2\sin x})'=-\frac{\cos x}{\sqrt{1-2\sin x}},
$$
using integration by parts we have
\begin{eqnarray}
\int_0^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}\,dx&=&-\int_0^{\pi/6}\frac{x}{\cos x}(\sqrt{1-2\sin x})'\,dx\\
&=&-\frac{x}{\cos x}\sqrt{1-2\sin x}\Big|_0^{\pi/6}+\int_0^{\pi/6}\sqrt{1-2\sin x}\left(\frac{x}{\cos x}\right)'\,dx\\
&=&\int_0^{\pi/6}\sqrt{1-2\sin x}\frac{\cos x+x\sin x}{\cos^2x}\,dx
\end{eqnarray}
It is clear that the function $\displaystyle x\mapsto f(x):=\sqrt{1-2\sin x}\frac{\cos x+x\sin x}{\cos^2x}$ is continuous on $[0,\pi/6]$, and therefore its integral over $[0,\pi/6]$ exists and is finite. It follows that the integral $\displaystyle \int_0^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}\,dx$ is convergent.
A: Thm: Suppose $0<p<1.$ Let $f$ be continuous on $[a,b],$ with $f(b)=0$ and $f>0$ on $[a,b).$ If $f'(b)$ exists and is nonzero, then
$$\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.$$
Because $p<1,$ that integral converges. By the positivity of $f, \int_a^{b-\delta}(1/f^p)$ is no problem. This proves the theorem.
A: Using  the right hand side of the well known Jordan inequality that states $\sin x < x$ for all $x \in \left[{0,\frac{\pi}{2}}\right]$ we have $1-2\sin x > 1-2x$ so that $\frac{1}{1-2\sin x} < \frac{1}{1-2x}$ so that for all $0\le x\le \frac{\pi}{6}$ we have  $\frac{x}{\sqrt{1-2\sin x}} < \frac{x}{\sqrt{1-2x}}$. 
Thus, 
\begin{align}
\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx} < \int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2x}}dx}
\end{align}
since  $0\le x\le \frac{\pi}{6}$ we may consider the one-to-one substitution $u^2=1-2x$, so that $x= \frac{1-u^2}{2}$ and $dx=-u du$. Also, when $x=0$ say $u=u_0$ and $x=\frac{\pi}{6}$ say $u=u_1$, therefore,
\begin{align}
\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx} < \int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2x}}dx}&= \int_{u_0}^{u_1}{\frac{\frac{1-u^2}{2}}{u}(-u)du} 
\\
&= \int_{u_0}^{u_1}{\frac{u^2-1}{2}du} = \frac{u^3_1-u^3_0}{6}-u_1 +u_0<\infty
\end{align}
which means that $\int_{0}^{\frac{\pi}{6}}{\frac{x}{\sqrt{1-2\sin x}}dx}$ converges.
A: Since we are only interested in whether the improper integral converges or diverges, let us write out the integrand in terms of their power series and see what we get:
$$  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. 
