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.
 A: 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!
A: Instead of performing Weierstrass substitution $\theta=2\arctan\frac{t}{2}$,  it is probably faster to directly perform the substitution $\theta=\arctan t$, leading to $d\theta=\frac{dt}{1+t^2}$ and $\frac{1}{1+t^2}=\cos^2\theta$, so that
$$ I(m) = \int_{0}^{+\infty}\frac{1+t^2}{(1-m^2+t^2)^2}=\frac{1}{(1-m^2)^{3/2}}\int_{0}^{+\infty}\frac{1+(1-m^2)u^2}{(1+u^2)^2}\,du $$
and the integrals
$$ \int_{0}^{+\infty}\frac{du}{(1+u^2)^2}=\frac{\pi}{4},\qquad \int_{0}^{+\infty}\frac{u^2\,du}{(1+u^2)^2}=\frac{\pi}{4}$$
can be computed in various ways, for instance through the residue theorem. They lead to:
$$ I(m)=\frac{\pi}{4}\cdot\frac{2-m^2}{(1-m^2)^{3/2}} $$
as wanted. Another efficient trick is to recognize in the integral some multiple of the area enclosed by an ellipse, due to the polar equation of an ellipse and the formula for computing the area in polar coordinates.
Update: I would really like the downvoters explaining their downvotes. What is wrong here?
A: 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
$$\frac \pi 4 \frac 1 {\sqrt {1 - m^2}} \left( 2 + \frac {m^2} {1-m^2} \right) = \frac \pi 4 \frac {2-m^2} {(1-m^2) ^\frac 3 2} .$$
A: $\newcommand{\bbx}[1]{\bbox[8px,border:1px groove navy]{{#1}}\ }
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbx{\left.
\vphantom{\Huge A^{a}}%
\int_{0}^{\pi/2}{\dd\theta \over \bracks{1 - m^{2}\cos^{2}\pars{\theta}}^{2}}
\right\vert_{\ 0\ <\ m\ <\ 1}} =
\int_{0}^{\pi/2}{\dd\theta \over \bracks{1 - m^{2}\cos^{2}\pars{\theta}}^{2}}
\\[5mm] & \stackrel{t\ \equiv\ \tan\pars{\theta}}{=}\,\,\,
\int_{0}^{\infty}{t^{2} + 1 \over \pars{t^{2} + 1 - m^{2}}^{2}}\,\dd t =
m^{2}\int_{0}^{\infty}{\dd t \over \pars{t^{2} + 1 - m^{2}}^{2}}\,\dd t +
\int_{0}^{\infty}{\dd t \over t^{2} + 1 - m^{2}}\,\dd t
\\[5mm] = &\
\pars{{1 \over 2}\,m\,\totald{}{m} + 1}\
\underbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1 - m^{2}}}
_{\color{#f00}{\ds{{\pi \over 2}\,{1 \over \root{1 - m^{2}}}}}}\ =\
{\pi \over 2}\pars{{1 \over 2}\,m\,\totald{}{m} + 1}
{1 \over \root{1 - m^{2}}}
\\[5mm] & =
\bbx{{\pi \over 4}\,{2 - m^{2} \over \pars{1 - m^{2}}^{3/2}}}
\end{align}
A: Since we have $m\cos\left(\theta\right)<1
 $ we get $$I\left(m\right)=\int_{0}^{\pi/2}\frac{1}{\left(1-m^{2}\cos^{2}\left(\theta\right)\right)^{2}}d\theta
 $$ $$=\sum_{k\geq1}km^{2k-2}\int_{0}^{\pi/2}\cos^{2k-2}\left(\theta\right)d\theta
 $$ $$=\frac{\pi}{2}+\frac{\sqrt{\pi}}{2}\sum_{k\geq2}\frac{km^{2k-2}\Gamma\left(k-\frac{1}{2}\right)}{\left(k-1\right)!}\tag{1}
 $$ where $(1)$ follows from the definition of Beta function in the following form $$\int_{0}^{\pi/2}\sin^{2a-1}\left(\theta\right)\cos^{2b-1}\left(\theta\right)d\theta=\frac{B\left(a.b\right)}{2},a,b>0,$$ so $$I\left(m\right)=\frac{\pi}{2}+\frac{\pi}{2}\sum_{k\geq1}\frac{\left(k+1\right)m^{2k}\left(2k\right)!}{k!^{2}4^{k}}\tag{2}
 $$ where $(2)
 $ follows from this identity, hence $$I\left(m\right)=\frac{\pi}{2}+\frac{\pi}{2}\sum_{k\geq1}\dbinom{2k}{k}\left(k+1\right)\left(\frac{m}{2}\right)^{2k}
 $$ but from the generalized binomial theorem we know that $$\sum_{k\geq0}\dbinom{2k}{k}x^{k}=\frac{1}{\sqrt{1-4x^{2}}}
 $$ hence $$I\left(m\right)=\color{red}{\frac{\pi}{4}\frac{2-m^{2}}{\left(1-m^{2}\right)^{3/2}}}$$ as wanted.
