\documentclass[a4paper,12pt]{article}
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\newcommand{\pl}{\partial}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\df}{\ds\frac}
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\begin{document}


{\bf Question}

Show that the elliptic integral

$$I(m)=\ds\int_0^{\frac{\pi}{2}}d\theta\
\sqrt{1-m^2\sin^2\theta}$$

has the expansion

$$I(m)=\ds\frac{\pi}{2}\left(1-\ds\frac{1}{4}m-\ds\frac{3}{64}m^2
-\ds\frac{5}{256}m3-\ds\frac{175}{16384}m^4+\cdots\right)$$

as $m\to 0$.

$$\rm{\bf Hint:} \ds\int_0^{\frac{\pi}{2}} \sin{2n}\theta\,
\theta=\pi\ds\frac{(2n)!}{\{2^{n+1}n!\}}$$

\vspace{.5in}

{\bf Answer}

Do it the dirty way! $m$ is small so a binomial expansion may
work!

$\sqrt{1-m^2\sin^2\theta} = 1- \df{1}{2} \df{m^2\sin^2\theta}{1!}
 + \df{(\frac{1}{2})(-\frac{1}{2})}{2!} m^4\sin^4\theta +\cdots$

Need $m^2\sin^2\theta<1$ for convergence.

So if $O<\theta<\df{\pi}{2} \Rightarrow 0<\sin\theta<1$ this last
bit means $|m^2|<1$ to be safe.

General expansion is:

$\sqrt{1-m^2\sin^2\theta}=1-\sum_{n=1}^\infty
\df{m^2\sin^{2n}\theta}{n!}\ \ m^2>1$

so integrating term by term $\ds\int_0^{\frac{\pi}{2}}\,d\theta$
we get from the hint in the question

$$I(m)=\df{\pi}{2}\left(1-\df{1}{4}m-\df{3}{64}m^2-\df{5}{256}m^3-\df{175}{16384}m^4
+\cdots\right)\ \rm{etc.}$$

\end{document}