An integral limit $\lim_{x\to0^+}\int_0^x\frac1{\sqrt{\cosh(x)-\cosh(y)}}\ dy$ While solving this question, I stumbled upon a strange limit:
$$\lim_{x\to0^+}\int_0^x\frac1{\sqrt{\cosh(x)-\cosh(y)}}\ dy\approx2.22$$
I'm not really sure how to go about handling such limits within the integral.  Any suggestions?
 A: Hint. By the change of variable $u=\dfrac yx$, $dy=xdu$, one gets
$$
\int_0^x\frac1{\sqrt{\cosh(x)-\cosh(y)}}\ dy=\int_0^1\frac{x}{\sqrt{\cosh(x)-\cosh(x \cdot u)}}\ du
$$ then one may observe that, for $0<u<1$, by applying a Taylor series expansion, as $x \to 0^+$, 
$$
\lim_{x\to0^+}\frac{x}{\sqrt{\cosh(x)-\cosh(x \cdot u)}}=\frac{\sqrt{2}}{\sqrt{1-u^2}}
$$ giving, by using the Dominated Convergence Theorem,
$$
\lim_{x\to0^+}\int_0^x\frac1{\sqrt{\cosh(x)-\cosh(y)}}\ dy=\int_0^1\frac{\sqrt{2}}{\sqrt{1-u^2}}du=\frac{\pi}{\sqrt{2}}\approx \color{red}{2.2214\cdots}.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\lim_{x \to 0^{+}}\int_{0}^{x}
{\dd y \over \root{\cosh\pars{x} - \cosh\pars{y}}} =
\root{2}\lim_{x \to 0^{+}}\int_{0}^{x}
{\dd y \over \root{\sinh^{2}\pars{x} - \sinh^{2}\pars{y}}}\label{1}\tag{1}
\end{align}
Note that, by simplicity, I replaced $\ds{x/2 \mapsto x}$ which doesn't matter in the $\ds{x \to 0^{+}}$ limit.

Lets $\ds{\sinh\pars{y} = \sinh\pars{x}\sin\pars{t}}$.

Then,
\begin{align}
&\lim_{x \to 0^{+}}\int_{0}^{x}
{\dd y \over \root{\cosh\pars{x} - \cosh\pars{y}}} =
\root{2}\lim_{x \to 0^{+}}\int_{0}^{\pi/2}
{\dd t \over \root{1 + x\sin^{2}\pars{t}}}\label{2}\tag{2}
\\[5mm] = &\
\root{2}\int_{0}^{\pi/2}\dd t = \bbx{\ds{\root{2}\pi \over 2}}
\approx 2.2214
\end{align}
In the line \eqref{2} squared root, I replaced
$\ds{\sinh^{2}\pars{x} \mapsto x}$ which doesn't matter in the
$\ds{x \to 0^{+}}$ limit.

Note that $\ds{\pars{~\mbox{with}\ x\ >\ 0~}}$

\begin{align}
&0 < \verts{\int_{0}^{\pi/2}
{\dd t \over \root{1 + x\sin^{2}\pars{t}}} - \int_{0}^{\pi/2}\dd t} <
\int_{0}^{\pi/2}
{\root{1 + x\sin^{2}\pars{t}} - 1 \over \root{1 + x\sin^{2}\pars{t}}}\,\dd t
\\[5mm] = &\
x\int_{0}^{\pi/2}
{\sin^{2}\pars{t} \over
\root{1 + x\sin^{2}\pars{t}}\bracks{\root{1 + x\sin^{2}\pars{t}} + 1}}\,\dd t <
x\int_{0}^{\pi/2}{1 \over 1 \times 2}\,\dd t
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0^{+}}{\to}\,\,\, {\large 0}
\end{align}
