Does the following improper integral converges? $$\int _ {0}^{1} \frac{1}{\sqrt{(1-x) \sin{x}}} dx $$ I have tried some approaches but I'm not sure whether it was correct or not.
First I split the integral since there are 2 critical points at $x=0$ and $x=1$:
$$\int_{0}^{1}\frac{1}{\sqrt{(1-x) \sin{x}}}dx = \int_{0}^{a} \frac{1}{\sqrt{(1-x) \sin{x}}}dx + \int_{a}^{1} \frac{1}{\sqrt{(1-x) \sin{x}}}dx$$ For the first part, I notice that $$\lim_{x\rightarrow0}\frac{1}{\sqrt{(1-x) \sin{x}}} = \frac{1}{\sqrt{\sin{x}}} $$ So I consider the limit $$\lim_{x\rightarrow0}\frac{\frac{1}{\sqrt{(1-x) \sin{x}}}}{\frac{1}{\sqrt{\sin{x}}}} = 1 $$
Now we know that $\int_{0}^{a}{\frac{1}{\sqrt{\sin{x}}}}dx$ coverges (the reason is that we know that from $x=0$ to $a$ the following inequality holds: $ \frac{\sin{a}}{a} x < \sin{x}\rightarrow \int_{0}^{a}{\frac{1}{\sqrt{\sin{x}}}}dx < \int_{0}^{a}\sqrt{\frac{a}{\sin{a}}}{\frac{1}{\sqrt{x}}}dx$, and the right hand side term converges since it's a p-series with $p=0.5$). Consequently, by LCT, the $\int_{0}^{a} \frac{1}{\sqrt{(1-x) \sin{x}}}dx$ converges.
For the second part, similarly. At $x=1$,
$$\lim_{x\rightarrow1}\frac{1}{\sqrt{(1-x) \sin{x}}} = \frac{1}{\sqrt{(1-x) \sin{1}}}$$ So I consider the limit $$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{(1-x) \sin{x}}}}{\frac{1}{\sqrt{(1-x) }}} = \frac{1}{\sqrt{\sin{1}}} $$
Now we know that $\int_{a}^{1}{\frac{1}{\sqrt{1-x}}}dx$ coverges (by Integral Test). Consequently, by LCT, the $\int_{a}^{1} \frac{1}{\sqrt{(1-x) \sin{x}}}dx$ converges. For these reasons, the improper integral converges.
If it is correct, I still don't understand how it really works. I just follow the approach in this link (at the end of the page, problem 5).
So my question is:
Is this idea correct? Can I to compare the improper integral by using LCT in this way? Any help will be greatly appreciated.
Hint: $$\int _ {0}^{1} \frac{dx}{\sqrt{(1-x) \sin{x}}} =\\\int _ {0}^{\epsilon} \frac{dx}{\sqrt{(1-x) \sin{x}}} + \int _ {\epsilon}^{1-\epsilon} \frac{dx}{\sqrt{(1-x) \sin{x}}} + \int _ {1-\epsilon}^{1} \frac{dx}{\sqrt{(1-x) \sin{x}}} \approx\\ \int _ {0}^{\epsilon} \frac{dx}{\sqrt{x}} + \int _ {\epsilon}^{1-\epsilon} \frac{dx}{\sqrt{(1-x) \sin{x}}} + \frac1{\sqrt{\sin(1)}}\int _ {1-\epsilon}^{1} \frac{dx}{\sqrt{1-x}}.$$
All three integrals are finite.