## Elliptic Integral of the First Kind Example

To evaluate
$\int^{ \pi /2}_0 \frac{1}{ \sqrt{cos y}} dy$
substitute
$cosy=cos^2 \theta$
then
$-sin y dy=-2cos \theta sin \theta d \theta$
then
$dy= \frac{2cos \theta sin \theta}{sin \theta} d \theta$
so
\begin{aligned} \int^{ \pi /2}_0 \frac{1}{ \sqrt{cos y}} dy &= 2 \int^{\pi /2}_0 \frac{sin \theta cos \theta}{cos \theta \sqrt{1-cos^4 \theta }} d \theta \\ &= 2 \int^{\pi /2}_0 \frac{\sqrt{1-cos^2 \theta }}{\sqrt{1-cos^2 \theta } \sqrt{1+cos^2 \theta }} d \theta \\ &= 2 \int^{\pi /2}_0 \frac{1}{\sqrt{1+1- sin^2 \theta}} d \theta \\ &= \sqrt{2}\int^{\pi /2}_0 \frac{1}{\sqrt{1- \frac{1}{2} sin^2 \theta}} d \theta \end{aligned}
.
This integral is called an elliptic integral of the first kind and is denoted by
$K( \frac{1}{ \sqrt{2}}$
, being of the form
$\int^{\pi /2}_0 \frac{1}{\sqrt{1-k^2 sin^2 \theta}} d \theta$
.
Hence
$\int^{ \pi /2}_0 \frac{1}{ \sqrt{cos y}} dy = \sqrt{2} K( \frac{\sqrt{2}}{2} )$
.
The value of the integral is
$\sqrt{2} \frac{\sqrt{2 \pi}}{2} = \sqrt{ \pi }$
.