## Writing an Integral as a Sum of Elliptic Integrals

To evaluate
$I= \int^{\pi /2}_0 \frac{sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta$
consider
\begin{aligned} \int^{\pi /2}_0 \frac{sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{k^2sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta \\ &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-1+k^2sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta \\ &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-(1-k^2sin^2 \theta )}{\sqrt{1-k^2 sin^2 \theta} } d \theta \\ &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1}{\sqrt{1-k^2 sin^2 \theta} } d \theta - \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-k^2sin^2 \theta }{\sqrt{1-k^2 sin^2 \theta} } d \theta \end{aligned}
.
Let
$F(k, \theta )= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1}{\sqrt{1-k^2 sin^2 \theta} } d \theta$
and
$E(k, \theta )= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-k^2 sin^2 \theta }{\sqrt{1-k^2 sin^2 \theta} } d \theta = \frac{1}{k^2} \int^{\pi /2}_0 \sqrt{1-k^2 sin^2 \theta} d \theta$

Then
$\int^{\pi /2}_0 \frac{sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta = \frac{1}{k^2} (F(k, \theta ) - E(k. \theta )$
.
$F( k, \theta )$
and
$E(k, \theta )$
are elliptic integrals of the first and second kind respectively..