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?
On
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 $2.000,000,100,966,347,331$.
We have $$\int_{-\frac\pi 2}^{\frac \pi 2} \sqrt{1-(1-a^2) \sin(k)^2} dk =2 E\left(1-a^2\right)$$ provided that $a\in \mathbb{R}\lor \Re(a)\neq 0$.
Expanded as series around $a=0^+$, we have $$2 E\left(1-a^2\right)=2-\sum_{n=1}^\infty \Bigg(\alpha_n+\beta_n\,\log \left(\frac{\sqrt{a}}{2}\right)\Bigg)\,a^{2n}$$ The first $\alpha_n$ are $$\left\{\frac{1}{2},\frac{13}{32},\frac{9}{32},\frac{5255}{24576}, \frac{11291}{65536},\frac{189021}{1310720},\frac{2600191}{20971520 },\cdots\right\} $$ The first $\beta_n$ are $$\left\{2,\frac{3}{4},\frac{15}{32},\frac{175}{512},\frac{2205}{8192 },\frac{14553}{65536},\frac{99099}{524288},\cdots\right\}$$