Infinity and cosines

The Riemann sum for $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+4\cos^{2}\left(x\right)} dx=\displaystyle \lim_{n\to \infty} \sum_{i=0}^{n} \frac{\pi}{2n\left(1+ 4\cos^{2}\left(\frac{i\pi}{2n}\right)\right)}$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+4\cos^{2}\left(x\right)} dx =\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sec^{2} x}{\sec^{2} x +4} dx$ by factoring out $\cos^{2} x$

$\displaystyle \frac{\sec^{2} x}{\sec^{2} x +4} =\frac{\sec^{2} x}{5+\tan^{2}x}=\frac{\sec^{2} x}{5\left(1+\frac{\tan^{2}x}{5}\right)}$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+4\cos^{2}\left(x\right)} dx =\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x}{5\left(1+\frac{\tan^{2}x}{5}\right)} dx$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{5\left(1+\frac{\tan^{2}x}{5}\right)} \sec^{2} x dx$

$\sec^{2} x dx =d\tan x$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{5\left(1+\frac{\tan^{2}x}{5}\right)} d\tan x =\displaystyle \frac{\arctan\left(\frac{\tan x}{\sqrt{5}}\right)}{\sqrt{5}}\Bigg|_{0}^{\frac{\pi}{2}}=\frac{\pi}{2\sqrt{5}}$

going back to the Riemann sum:

$\displaystyle \lim_{n\to \infty} \sum_{i=0}^{n} \frac{\pi}{2n\left(1+ 4\cos^{2}\left(\frac{i\pi}{2n}\right)\right)} =\frac{\pi}{2\sqrt{5}}$

factor out $\frac{\pi}{2}$

$\displaystyle \lim_{n\to \infty} \sum_{i=0}^{n} \frac{1}{n\left(1+ 4\cos^{2}\left(\frac{i\pi}{2n}\right)\right)} =\frac{1}{\sqrt{5}}$

$\begin{aligned} L & = \lim_{n \to \infty} \sum_{k=0}^n \frac 1{n\left(1+4\cos^2 \frac {k\pi}{2n}\right)} & \small \blue {\text {Since }\cos 2\theta = 2\cos^2 \theta -1} \\ & = \lim_{n \to \infty} \sum_{k=0}^n \frac 1{n\left(3+2\cos \frac {k\pi}n \right)} & \small \blue {\text {By Riemann sums}} \\ & = \int_0^1 \frac 1{3+2\cos (x\pi)} dx & \small \blue {\text{Let }u = x\pi \implies du = \pi \ dx} \\ & = \frac 1\pi \int_0^\pi \frac 1{3+2\cos u} du & \small \blue {\text{Let }t = \tan \frac u3 \implies dt = \frac 12 \sec^2 \frac u2 \ du} \\ & = \frac 1\pi \int_0^\infty \frac 2{5+t^2} dt \\ & = \frac 2{\sqrt 5 \pi} \tan^{-1} \frac t{\sqrt 5} \bigg|_0^\infty \\ & = \frac 1{\sqrt 5} \end{aligned}$

Therefore $A = \boxed 5$ .

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Reference : Riemann sums