\[\int^{ \pi /2}_0 \frac{1}{ \sqrt{cos y}} dy \]
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\[\begin{equation} \begin{aligned} \int^{ \pi /2}_0 \frac{1}{ \sqrt{cos y}} dy &= 2 \int^{\pi /2}_0 \frac{sin \theta cos \theta}{cos \theta \sqrt{1-cos^4 \theta }} d \theta \\ &= 2 \int^{\pi /2}_0 \frac{\sqrt{1-cos^2 \theta }}{\sqrt{1-cos^2 \theta } \sqrt{1+cos^2 \theta }} d \theta \\ &= 2 \int^{\pi /2}_0 \frac{1}{\sqrt{1+1- sin^2 \theta}} d \theta \\ &= \sqrt{2}\int^{\pi /2}_0 \frac{1}{\sqrt{1- \frac{1}{2} sin^2 \theta}} d \theta \end{aligned} \end{equation}\]
.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
\[\sqrt{2} \frac{\sqrt{2 \pi}}{2} = \sqrt{ \pi }\]
.