\[I= \int^{\pi /2}_0 \frac{sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta\]
consider\[\begin{equation} \begin{aligned} \int^{\pi /2}_0 \frac{sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{k^2sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta \\ &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-1+k^2sin^2 \theta}{\sqrt{1-k^2 sin^2 \theta} } d \theta \\ &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-(1-k^2sin^2 \theta )}{\sqrt{1-k^2 sin^2 \theta} } d \theta \\ &= \frac{1}{k^2} \int^{\pi /2}_0 \frac{1}{\sqrt{1-k^2 sin^2 \theta} } d \theta - \frac{1}{k^2} \int^{\pi /2}_0 \frac{1-k^2sin^2 \theta }{\sqrt{1-k^2 sin^2 \theta} } d \theta \end{aligned} \end{equation}\]
.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..