Approximate $\int_0^{\pi /2} \frac{ds}{\sqrt{1-x\sin^2s}}$ I am trying to approximate the following integral 
$$K(x)=\int\limits_0^{\pi /2} \frac{ds}{\sqrt{1-x\sin^2s}}$$
with $0<x<1$.
I need to show that for x close to one that $K(x)\sim -\frac{1}{2}\ln(1-x)$.
My first attempt was to Taylor expand the function $f(x)= (1-x\sin^2s)^{-1/2}$ about $x=1$. I did this using $f(x)\approx f(1)+f'(1)(x-1)$. I integrated the result and found that $K(x)\sim \left[\frac{5-x}{4}\ln|\sec s + \tan s|\right]_0^{\pi/2} - \left[\frac{1-x}{4}\sec s \tan s \right]_0^{\pi/2}$, which goes to infinity when $s=\pi/2$. 
I can't think of another way to approach this, so any advice would be helpful. 
 A: The integral as written is the 
complete elliptic integral of the first kind and it can be expressed in term of hypergeometric function:
$$K(x) = \frac{\pi}{2}\,_2F_1(\frac12,\frac12;1;x)\tag{*1}$$
We know when $\gamma = \alpha + \beta$, $\alpha$ and $\beta \ne 0, -1, -2, \ldots$, $|\arg(1-x)| < \pi$, $|1-x| < 1$, the hypergeometric function has following 
expansion near $x = 1$:
$$\begin{align}
\,_2F_1(\alpha,\beta;\gamma;x) = 
& \frac{-\Gamma(\gamma)}{\Gamma(\alpha)\Gamma(\beta)}
\sum_{n=0}^{\infty}\frac{(\alpha)_n(\beta)_n}{n!^2}(1-x)^n \times\\
& \left\{
\psi(\alpha+n)+\psi(\beta+n) -2\psi(1+n) + \log(1-x)
\right\}
\end{align}$$
where $(t)_n$ is the rising Pochhammer symbol and $\psi(t)$ is the 
digamma function.
Substitute this back into $(*1)$ for $x$ near $1$, we have up to $O((1-x)\log(1-x))$,
$$K(x) \sim -\frac12 \left\{ 2\psi(\frac{1}{2}) - 2\psi(1) + \log(1-x)\right\}
          = \log 4 - \frac12\log(1-x)$$
Actually, we can derive this leading term by elementary means.
Let $\delta = \sqrt{\frac{1-x}{x}}$, $u = \cos s = \delta \sinh\theta$, we have
$$\begin{align}
K(x) 
= & \frac{1}{\sqrt{x}}\int_0^{\frac{\pi}{2}} \frac{ds}{\sqrt{\delta^2 + \cos^2\!s}}\\
= & \frac{1}{\sqrt{x}}\int_0^1 \frac{du}{\sqrt{\delta^2 + u^2}\sqrt{1-u^2}}\\
= & \frac{1}{\sqrt{x}}\left\{
\int_0^1  \frac{du}{\sqrt{\delta^2 + u^2}} + 
\int_0^1 \frac{1-\sqrt{1-u^2}}{\sqrt{\delta^2+u^2}} \frac{du}{\sqrt{1 - u^2}}
\right\}\\
= & \frac{1}{\sqrt{x}}\left\{
\int_0^1 \frac{du}{\sqrt{\delta^2 + u^2}} +
\int_0^{\frac{\pi}{2}}\frac{1-\sin s}{\sqrt{\delta^2 + \cos^2\!s}} ds
\right\}\\
= & \frac{1}{\sqrt{x}}\left\{
\int_0^{\sinh^{-1}\frac{1}{\delta}} d\theta +
\int_0^{\frac{\pi}{2}}\frac{1-\sin s}{\sqrt{\delta^2 + \cos^2\!s}} ds
\right\}\\
\end{align}$$
The $2^{nd}$ integral in the RHS has a finite limit as $\delta \to 0$. 
If we replace it by its limit, we will introduce an error that won't affect
our determination of the leading term. Up to $o(1)$$\color{blue}{^{[1]}}$, we find:
$$\begin{align}
K(x) 
\sim & \frac{1}{\sqrt{x}}\left\{\sinh^{-1}\frac{1}{\delta} + \int_0^{\frac{\pi}{2}}\frac{1-\sin s}{\cos s} ds
\right\}\\
= & \frac{1}{\sqrt{x}}\left\{ \log\left(\frac{1+\sqrt{1+\delta^2}}{\delta}\right) + \log 2\right\}\\
= & \frac{1}{\sqrt{x}}\left\{ \log\left(\frac{\sqrt{x}+1}{\sqrt{1-x}}\right) + \log 2\right\}\\
\sim & (1 + O(1-x))
\left\{\log\left(\frac{1 + O(1-x) + 1}{\sqrt{1-x}}\right) + \log 2\right\}\\
\sim & \log 4 - \frac12\log(1-x) + O((1-x)\log(1-x))
\end{align}$$
recovering the leading term as expected.
Notes


*

*$\color{blue}{[1]}$ A more careful analysis shows the error introduced in this step is of the order $O(\delta^2\log\delta) = O((1-x)\log(1-x))$.

