<s>[INST] Calculate integral $\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$ I recently saw the integral problem
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$$
and tried to solve it. Below is what I did.
Interesting to look at other easier solutions.
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=4\int_{0}^{\pi /2}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}\\\overset{t=\operatorname{tg} x}{=}\int\limits_{0}^{\infty }\frac{1+t^2}{\left ( 1+\left ( 1+n^2 \right )t^2 \right )^2}dt\\ 
\overset{t=\frac{y}{\sqrt{1+n^2}}}{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int\limits_{0}^{\infty }\frac{1+n^2+y^2}{\left ( 1+y^2 \right )^2}dy\\ \overset{y=\operatorname{tg} \theta }{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int_{0}^{\pi /2}\left ( 1+n^2\cos^2 \theta  \right )d\theta \\
=\frac{\pi \left ( 2+n^2 \right )}{\left ( 1+n^2 \right )\sqrt{1+n^2}}$$
 [/INST] Alternatively
$$I(n)= \int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=\frac4{n^4}\int_{0}^{\pi /2}\frac{dx}{\left ( a-\cos^2 x \right )^2},\>\>\>\>\>a= 1+\frac1{n^2}$$
Note that
$$J(a)= \int_{0}^{\pi/2}\frac{dx}{ a- \cos^2 x }
= \int_{0}^{\pi/2}\frac{d(\tan x)}{ a\tan^2x  +(a-1)}dx= \frac\pi{2\sqrt{a(a-1)}}
$$
Then
$$I(n)= -\frac4{n^4} \frac{dJ(a)}{da}\bigg|_{a= 1+\frac1{n^4}}=\frac{\pi(2+n^2)}{ (1+n^2)^{3/2}}
$$
</s>