Convergence of the Integral $ \int_0^1 \frac{1}{\sqrt{\sin x}} \, dx$ Here is an old question from my real analysis exam. It has been bugging me for the good part of a year. Does the following integral converge?
$$ \int_0^1 \frac{1}{\sqrt{\sin x}} \, dx$$
I'm pretty sure the comparison test is the way to go. Any insight would be greatly appreciated. Thanks.
 A: For $ 0 \leq x \leq 1,$ we get
$$    \frac{x}{2} \leq \; \sin x \; \leq x        $$
A: HINT: 
You're right: the comparison test is the way to go. In the domain of integration, i.e., the interval $[0,1]$, the only point that causes concern is $0$. As $x \to 0$, note that the integrand grows unbounded. Therefore, to decide the convergence or divergence of the integral, we need to bound the growth of the integrand near $0$. This is the idea behind the (limit) comparison test. 
To implement the above idea, we could use the standard fact $\sin x \sim x$ for $x$ close to $0$ (i.e., as $x \to 0$). Therefore, our integral $\int_0^1 \frac{1}{\sqrt{\sin x}} ~\mathrm dx$ converges if and only if $\int_0^1 \frac{1}{\sqrt{x}} ~\mathrm dx$ (the integral of the test function) converges. Do you know how to establish the convergence (or divergence) of the latter integral?

Convergence of the test integral: The integral $\int_0^1 \frac{1}{\sqrt{x}} ~\mathrm dx$  in fact converges (and so does our original integral). To see this, note that
$$
\int_{\delta}^{1} \frac{1}{\sqrt{x}} ~\mathrm dx = \left. 2 \sqrt{x} \right|_{\delta}^{1} = 2 - 2 \sqrt{\delta} \to 2,
$$
as $\delta \to 0$. 
In fact, one could similarly see that the integral $\int_0^1 x^p ~\mathrm dx$ converges if and only if $p \gt -1$.
A: $\newcommand{\+}{^{\dagger}}%
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With $\ds{t \equiv \sin\pars{x}\quad\iff x = \arcsin\pars{t}}$:
\begin{align}
\color{#00f}{\large\int_{0}^{1}{\dd x \over \root{\sin\pars{x}}}}&=
\int_{0}^{\sin\pars{1}}t^{-1/2}\,{\dd t \over \root{1 - t^{2}}}
=
\int_{0}^{\root{\sin\pars{1}}}t^{-1/4}\pars{1 - t}^{-1/2}\half\,t^{-1/2}\,\dd t
\\[3mm]&=
\half\int_{0}^{\root{\sin\pars{1}}}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t
=\color{#00f}{\large\half\,{\rm B}_{\sin^{1/2}\pars{1}}\pars{{1 \over 4},\half}}
\\[3mm]&=\color{#00f}{\large2\sin^{1/8}\pars{1}\  _{2}{\rm F}_{1}\pars{{1 \over 4},\half;{3 \over 4};\sin^{1/8}\pars{1}}} \approx 2.3283
\end{align}
${\rm B}_{x}\pars{p,q}$ and $_{2}{\rm F}_{1}$ are the Incomplete Beta and the Hypergeometric functions, respectively. 
A: A possible solution only with the comparison test:
It suffices to show that $\displaystyle{\int_{0}^{\pi /2}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}}$ converges, since the integrand is positive, therefore $\displaystyle{\int_{0}^{1}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}} \le \int_{0}^{\pi /2}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}$.
Set $x = \arcsin y$ and $\mathrm{d}x = \dfrac{\mathrm{d}y}{\sqrt{1-y^2}}$. Then,
$$
\begin{aligned}
\int_{0}^{\pi /2}\dfrac{\mathrm{d}x}{\sqrt{\sin x}}&=\int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{y}\sqrt{1-y^2}} \\
&=\int_{0}^{1} \dfrac{\mathrm{d}y}{\sqrt{y}\sqrt{1-y}\sqrt{1+y}} \\
&\le \int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{y}\sqrt{1-y}} \\
&=2\int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{1-y^2}}\\
&\le2\int_{0}^{1}\dfrac{\mathrm{d}y}{\sqrt{1-y}} \\
&=4.
\end{aligned}
$$
So, the integral is convergent and
$$ \int_{0}^{1}\dfrac{\mathrm{d}x}{\sqrt{\sin x}} \le 4.$$
