I am trying to attain a two-term approximation for the following integral as $m$ goes to $1$ from below: $$I=\int_{0}^{\pi /2}\frac{\mathrm d\theta }{\sqrt{1-(m^2)\cdot\sin(\theta)^2 }}.$$ So far I am struggling to get the approximation because I have not been able to put my hand on a proper change of variable.
I would appreciate any comments.
This is just the complete elliptic intagral of the first kind, thus $$ I = K(m) = - \frac{1}{{2\pi }}\sum\limits_{n = 0}^\infty {\frac{{\Gamma ^2 \!\left( {n + \frac{1}{2}} \right)}}{{\Gamma ^2 (n + 1)}}(1 - m^2 )^n (\log (1 - m^2 ) + a_n )} $$ as $m \to 1-0$, where $a_0 =2\log 2$ and $$ a_{n + 1} = a_n + \frac{4}{{(2n + 1)(2n + 2)}} \quad (n\geq 0). $$