# Attain a two-term approximation for the following integral as m goes to 1 from below:

Attain a two-term approximation for the following integral as $m$ goes to $1$ from below:
$I={\int }_{0}^{\pi /2}\frac{\mathrm{d}\theta }{\sqrt{1-\left({m}^{2}\right)\cdot \mathrm{sin}\left(\theta {\right)}^{2}}}.$
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Mekjulleymg
This is just the complete elliptic intagral of the first kind, thus
$I=K\left(m\right)=-\frac{1}{2\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathrm{\Gamma }}^{2}\phantom{\rule{negativethinmathspace}{0ex}}\left(n+\frac{1}{2}\right)}{{\mathrm{\Gamma }}^{2}\left(n+1\right)}\left(1-{m}^{2}{\right)}^{n}\left(\mathrm{log}\left(1-{m}^{2}\right)+{a}_{n}\right)$
as $m\to 1-0$, where ${a}_{0}=2\mathrm{log}2$ and
${a}_{n+1}={a}_{n}+\frac{4}{\left(2n+1\right)\left(2n+2\right)}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$