Asymptotic expansion of integral with hyperbolic functions Consider the integral given by 
$$f(r)=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}d\sigma,$$
where $a>0$. I am wondering how one can derive the asymptotic power series for $r\downarrow 0$?
I have tried the following
$$\arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)=\pi/2-\arcsin\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\\
=\pi/2+\sum_{k\geq 0}\binom{2k}{k}\frac{\sigma^{2k+1}}{(2k+1)4^k \sinh(r)^{2k+1}\cdot (1-\sigma^2)^{k+1/2}}.$$
Hence we get intuitively (?)
$$f(r)=\frac{\pi}{2}\left( \log\left( \sqrt{a^2+\tanh(r)^2}+\tanh(r)\right)-\log(a)\right)+ \sum_{k\geq 0}\binom{2k}{k}\frac{1}{(2k+1)4^k \sinh(r)^{2k+1}}\cdot \int_{0}^{\tanh(r)}\frac{\sigma^{2k+1}}{ (1-\sigma^2)^{k+1/2}\sqrt{\sigma^2+a^2}}d\sigma.$$
But the expressions become very complicated. Maybe I have not seen a simplification which could be made at the beginning. Does anyone know how one can solve this problem?
Best wishes
 A: Consider $\;\sigma:=\tanh(u)\,$ then :
\begin{align}
f(r)&:=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}\,d\sigma\\
&=\int_{0}^{r} \arccos\left(\frac{\sinh(u)}{\sinh(r)}\right) \frac{1}{\sqrt{\tanh(u)^2+a^2}}\,\frac{du}{\cosh(u)^2}\\
\end{align}
Set $\;t:=\dfrac{\sinh(u)}{\sinh(r)}\;$ (i.e. $\;u:=\operatorname{argsinh}(\sinh(r)\;t)\,$) then :
\begin{align}
&\sinh(u)=t\,\sinh(r),\\
&\cosh(u)^2=1+(t\,\sinh(r))^2\\
\\
&\text{and}\\
f(r)&=\int_{0}^{1} \arccos\left(t\right) \frac{\sqrt{1+t^2\sinh(r)^2}}{\sqrt{t^2\sinh(r)^2+a^2(1+t^2\sinh(r)^2)}}\,\frac{\sinh(r)\;dt}{\sqrt{1+t^2\sinh(r)^2}^3}\\
&=\sinh(r)\int_{0}^{1} \frac{\arccos\left(t\right) }{\sqrt{a^2+(1+a^2)t^2\sinh(r)^2}}\,\frac{dt}{1+t^2\sinh(r)^2}\\
\end{align}
At this point you may get an asymptotic expansion as $r\to 0$ especially by  considering the expansion in powers of $s:=\sinh(r)$ and observing that :
$$a_n:=\int_{0}^{1} \arccos\left(t\right) t^{2n}\,dt=\frac{4^n}{(2n+1)^2\binom{2n}{n}}$$
(the indefinite integral of $\arccos\left(t\right) t^{2n}$ admits a closed form) 
For $\,s=\sinh(r)\,$ and using the expansion of $\dfrac 1{\sqrt{1-x}}$ in central binomial terms I get :
$$f(r)=\sum_{n=0}^\infty \frac{(-4)^n}{(2n+1)^2\binom{2n}{n}}\left(\frac sa\right)^{2n+1}\sum_{k=0}^n a^{2n-2k}(a^2+1)^k\frac{\binom{2k}{k}}{4^k}$$
The first terms of this expansion are : 
$$f(r)=\frac sa-\frac 29\left(\frac sa\right)^3\frac{3a^2+1}{2}+\frac 8{75}\left(\frac sa\right)^5\frac{15a^4+10a^2+3}{8}-+\frac {16}{245}\left(\frac sa\right)^7\frac{35a^6+35a^4+21a^2+5}{16}+O\left(s^9\right) $$
(as confirmed by numerical evaluation)
Using the Maclaurin series for $\sinh(r)$ this becomes (with an error of $\;O\left(r^9\right)\;$) :
$$\frac ra-\left(\frac ra\right)^3\frac{3a^2+2}{18}+\left(\frac ra\right)^5\frac{75a^4+140a^2+72}{1800}-\left(\frac ra\right)^7\frac{1281a^6+4634a^4+5544a^2+2160}{105840}$$
