<s>[INST] 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.
 [/INST] Let $t = \tan \theta = \frac {\sin \theta} {\cos \theta}$. Combining this with the identity $\cos^2 \theta + \sin^2 \theta = 1$ and keeping in mind that $\theta \in [0, \frac \pi 2]$ (where $\cos \theta \ge 0$), we obtain $\cos^2 \theta = \frac 1 {1+t^2}$. With this substitution the integral becomes
$$\int \limits _0 ^\infty \frac 1 {(1 - m^2 \frac 1 {1 + t^2})^2} \frac 1 {1+t^2} \ \Bbb d t = \int \limits _0 ^\infty \frac {t^2 + 1} {(t^2 + 1 - m^2)^2} \ \Bbb d t = \int \limits _0 ^\infty \frac 1 {t^2 + 1 - m^2} \ \Bbb d t + \int \limits _0 ^\infty \frac {m^2} {(t^2 + 1 - m^2)^2} \ \Bbb d t .$$
The first one, by the change of variable $t = \sqrt {1-m^2} u$, becomes
$$\frac 1 {\sqrt {1-m^2}} \int \limits _0 ^\infty \frac 1 {1+u^2} \ \Bbb d u = \frac 1 {\sqrt {1-m^2}} \arctan u \big| _0 ^\infty = \frac \pi 2 \frac 1 {\sqrt {1-m^2}} .$$
With the same change of variable, the second one becomes
$$\frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\infty \frac 1 {(1+u^2)^2} \ \Bbb d u .$$
For this last integral, perform the change of variables $u = \tan s$, obtaining
$$\frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \frac 1 {(1 + \tan^2 s)^2} \frac 1 {\cos^2 s} \Bbb d s = \frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \frac 1 {\frac 1 {\cos^4 s}} \frac 1 {\cos^2 s} \Bbb d s = \frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \cos^2 s \ \Bbb d s = \\
\frac {m^2} {\sqrt {1-m^2} ^3} \int \limits _0 ^\frac \pi 2 \frac {1 + \cos 2s} 2 \ \Bbb d s = \frac \pi 4 \frac {m^2} {\sqrt {1-m^2} ^3} .$$
Adding the two integrals, we get
$$\frac \pi 4 \frac 1 {\sqrt {1 - m^2}} \left( 2 + \frac {m^2} {1-m^2} \right) = \frac \pi 4 \frac {2-m^2} {(1-m^2) ^\frac 3 2} .$$
</s>