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

$I={\int}_{0}^{\pi /2}\frac{\mathrm{d}\theta}{\sqrt{1-({m}^{2})\cdot \mathrm{sin}(\theta {)}^{2}}}.$

$I={\int}_{0}^{\pi /2}\frac{\mathrm{d}\theta}{\sqrt{1-({m}^{2})\cdot \mathrm{sin}(\theta {)}^{2}}}.$