Approximation for elliptic integral of second kind My (physics) book gives the following approximation:
$\int_{-\pi/2}^{\pi/2} \sqrt{1-(1-a^2) \sin(k)^2} dk \approx 2 + (a_1 - b_1 \ln a^2) a^2 + O(a^2 \ln a^2)$
where a1 and b1 are "(unspecified) numerical constants."  I've been looking for either a derivation of this, or the same approximation listed elsewhere and have gotten nowhere.  Can someone help me along?
 A: This is just an outline of an answer. 
Call the integral to be calculated $I(a)$. First write 
$$I(a) = 2\int_0^{\pi/2} \sqrt{\cos^2 k + a^2 \sin^2 k} \, dk$$
by symmetry. The difficulty here is that you can't just apply Taylor's formula for $a \to 0$, because values of $k$ near $\pi/2$ make a significant contribution to the integral and $\cos k$ is small there.
Make the substitution $u = \cot k$. Then
$$I(a) = 2 \int_0^{+\infty} \frac{\sqrt{u^2 + a^2}}{(u^2 + 1)^{3/2}} \, du.$$
Now divide the integral into three parts on the intervals $[0,a]$, $[a,1]$ and $[1,+\infty)$. Then
 make the substitution $v = u/a$ on the first interval, $s = u^2$ on the second, and $w = 1/u$ on the third. We get
$$I(a) = 2a^2 \int_0^1 \sqrt{1+v^2} (1 + a^2 v^2)^{-3/2} \, dv + 2\int_0^1  (1 + w^2)^{-3/2} \sqrt{1 + a^2 w^2} \, dw \\+ \int_{a^2}^1 \frac{1}{(1+s)^{3/2}} \sqrt{1 + \frac{a^2}{s}} \, ds$$
Now the idea is to expand each integrand into a series using the Taylor series for $(1 + x)^{1/2}$ and $(1+x)^{-3/2}$, and then integrate term by term. This will be legitimate because of uniform convergence. In the first integral, expand the second factor as a function of $av$. Do the same in the second integral with respect to $aw$. The third is more complicated because $a^2$ appears as a bound, but you can expand the integrand into a double series with respect to $s$ and $a^2/s$. The resulting series is quite complicated.
However, if we only want an estimate at the level of $O(a^2)$, we can note that the square root in the last integral is $1 + a^2/2s$ to within $O(a^4/s^2)$, so the error in the integral will be at most $O(a^2)$. If we make this approximation, we find
$$I(a) = 2 - a^2 \ln a + O(a^2)$$
To evaluate the $a^2$ term is more difficult. The contribution from the first integral is $\sqrt{2} + \operatorname{arsinh}(1)$. The contribution from the second is $\operatorname{arsinh}(1) - \frac{1}{2}\sqrt{2}$. The $a^2$ term in the third integral consists of a $-a^2$ from the first term of the series, a term $a^2[-\operatorname{arsinh}(1)+ \frac{1}{\sqrt{2}} + \ln 2 - 1]$ in the second term of the series, and the remaining terms with coefficient $\sum_{n \geq 2} \frac{1}{n-1}\binom{1/2}{n}$. This last series is $f(1)$, where $f(x) = \sum_{n \geq 2} \frac{1}{n-1}\binom{1/2}{n}x^{n-1}$. We have $f(0) = 0$ and $f'(x) = \sum_{n \geq 2} \binom{1/2}{n}x^{n-2} = \frac{1}{x^2} (\sqrt{1 + x} - 1 - x/2)$, so 
$$f(1) = \int_0^1 \frac{1}{x^2} (\sqrt{1 + x} - 1 - x/2) \, dx = \frac{3}{2} - \sqrt{2} + \ln 2 - \operatorname{arsinh}(1).$$
Taking everything into account, we get
$$I(a) = 2 - a^2\ln a + a^2 (2\ln 2 - 1/2) + O(a^4 \ln a).$$
Given how simple the result is, I'll bet there's a simpler way to find it.
EDIT: For $a=0.0001$, the true value of the integral is $2.000,000,100,966,347,688$. The approximation obtained by the formula is $2.000,000,100,966,347,331$.
A: I am not sure to really answer the question here; so, please, forgive me if I am off topic.
On the lhs, we have $$\int_{-\pi/2}^{\pi/2} \sqrt{1-(1-a^2) \sin(k)^2} dk =2 E\left(1-a^2\right)$$ provided that $a\in \mathbb{R}\lor \Re(a)\neq 0$.
On the lhs, it seems that the rhs is approximated by $$2+\left(\alpha -\beta  \log \left(a^2\right)\right)a^2 $$
and the unknown parameters $\alpha,\beta$ can effectively be adjusted using nonlinear regression. Generating $100$ equally spaced data points for $0 \lt a \leq 1$, I effectively found that the fit is very good ($R^2=0.999998$) for $\alpha=1.13398$, $\beta=0.398159$.
