Verify that $\int_0^{\pi/2} \frac{{\rm d}\theta}{(1-m^2\cos^2{\theta})^2}= \frac{(2-m^2)\pi}{4(1-m^2)^{3/2}}$ for $0I am told that the integral
\begin{align}
I&=\int_0^{\pi/2} \frac{1}{(1-m^2\cos^2{\theta})^2}\,{\rm d}\theta\\
&= \frac{(2-m^2)\pi}{4(1-m^2)^{3/2}}
\end{align}
Where $0<m<1$. I want to verify this however I have no idea on how this is done. Can someone explain?
I was thinking possibly a substitution so I tried $x=m\cos\theta$ however this makes the problem more complicated so obviously it's wrong.

HINT:
The integral of interest $I(m)=\int_0^{\pi/2} \frac{1}{(1-m^2\cos^2(\theta))^2}\,d\theta$ can be readily simplified as 
$$\begin{align}
I&=\int_0^{\pi/2} \frac{1}{(1-m^2\cos^2(\theta))^2}\,d\theta\\\\
&=-\frac2{m^4}\left.\frac{d}{da}\left(\int_{0}^\pi\frac{1}{a-\cos(\theta)}\,d\theta\right)\right|_{a=2/m^2 -1}\\\\
\end{align}$$
Now, either simply enforce the classical Weierstrass Substitution, use contour integration, or just recall the well-known result to arrive at
$$\begin{align}
I&=-\frac2{m^4}\left.\frac{d}{da}\left(\frac{\pi}{\sqrt{a^2-1}}\right)\right|_{a=2/m^2 -1}\\\\
&=\frac{\pi(2-m^2)}{4(1-m^2)^{3/2}}
\end{align}$$
as was to be shown!
