Solving $\int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta $ I have the following integration to solve.
$$f(k) = \int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta,\quad0<k<1$$
assuming $\sin\theta = t$ which results $d\theta = \frac{dt}{\sqrt{1-t^2}}$
and when $\theta = 0, t=0$ and $\theta=\frac{\pi}{2},t=1$
so above equation can be rewritten as, 
$$f(k) = \int_0^1{t^2\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}}dt$$
I'm stuck in solving this further. 
Can somebody help me with some clues/solution to solve this further.

Possible hints to perfume some kind of calculations.
Leaving apart the extrema of the integra, for the moment.
$$\int\sin^2\theta \sqrt{1 - k^2 \sin^2\theta}\ \text{d}\theta$$
Using the substitution
$$k\sin\theta = \cos\phi ~~~~~~~ \sin\theta = \frac{\cos\phi}{k} ~~~ \to ~~~ \sin^2\theta = \frac{\cos^2\phi}{k^2}$$
$$\phi = \arccos(k\sin\theta)$$
$$\text{d}\phi = \frac{- k\cos\theta}{\sqrt{1 - k^2\sin^2\theta}}\ \text{d}\theta = -\frac{k\sqrt{1 - \sin^2\theta}}{\sqrt{1 - \cos^2\phi}}\ \text{d}\theta = -\frac{k\sqrt{1 - \frac{\cos^2\phi}{k^2}}}{\sin^2\phi}\ \ \text{d}\theta$$
Thence $$\text{d}\theta = -\frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\ \text{d}\phi$$ and the integral becomes
$$- \int \frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\left(\frac{\cos^2\phi}{k^2}\right)\sin^2\phi \text{d}\phi = -\frac{1}{k^2}\int \frac{\sin^4\phi\cos^2\phi}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi$$
Now we can use the trigonometric reduction formula for the numerator of the integrand:
$$\sin^4\phi\cos^2\phi = \frac{1}{32}(2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi))$$
to get
$$ 
-\frac{1}{32k^2}\int\ \frac{2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi)}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi
$$
Which might be splitter into four parts, and then.. who knows!
The solution, however, lies into Jacobi Elliptic Functions of the First an Second Kind.
More on Jacobi Elliptic Integrals
https://en.wikipedia.org/wiki/Elliptic_integral
