So I was bored, and decided to evaluate some integrals because why not. After a while, I came up with this monstrosity:$$\int_0^1\dfrac{4\pi\tan^4(2x)}{\cos^2(x)}dx$$Which I decided to try and evaluate. Here is my attempt at evaluating the integral:$$\begin{align}\int_0^1\dfrac{4\pi\tan^4(2x)}{\cos^2(x)}dx=&4\pi\int_0^1\dfrac{\tan^2(2x)\tan^2(2x)}{\dfrac{1+\cos(2x)}{2}}dx\\=&8\pi\int_0^1\dfrac{\left(\dfrac{1-\cos(2x)}{1+\cos(2x)}\right)^2}{1+\cos(2x)}dx\\=&8\pi\int_0^1\dfrac{(1-\cos(2x))^2}{(1+\cos(2x))^3}dx\\=&8\pi\int_0^1\dfrac{(2-2\cos^2(x))^2}{(2\cos^2(x))^3}dx\\=&8\pi\int_0^1\dfrac{4\sin^4(x)}{8\cos^6(x)}dx\\=&4\pi\int_0^1\tan^4(x)\sec^2(x)dx\\=&\dfrac{4\pi}5\left.\tan^5(x)\right]_0^1=\dfrac{4\pi}5\tan^5(1)\end{align}$$
My question
Did I evaluate the monstrosity of an integral correctly, or what could I do to integrate it correctly?
Best Answer
For any $\;x\in\left[\dfrac{\pi}8, \dfrac\pi4\right[,\;$ it results that
$\begin{align}\dfrac{4\pi\tan^4(2x)}{\cos^2(x)}&\geqslant\dfrac{4\pi\tan^4(2x)}{\cos^2(\frac\pi8)}>4\pi\tan^4(2x) =\dfrac{4\pi\sin^4(2x)}{\cos^4(2x)}\geqslant\\[3pt]&\geqslant\dfrac{4\pi\sin^4(\frac{\pi}4)}{\sin^4\!\left(\frac{\pi}2\!-\!2x\right)}=\dfrac{\pi}{\sin^4\!\left(\frac{\pi}2\!-\!2x\right)}>\dfrac{\pi}{\left(\frac{\pi}2\!-\!2x\right)^4}\,.\end{align}$
Moreover ,
$\begin{align}\!\!\!\!\!\!\displaystyle\int_0^1\!\dfrac{4\pi\tan^4(2x)}{\cos^2(x)}\,\mathrm dx&>\!\int_{\frac\pi8}^{\frac\pi4}\!\dfrac{4\pi\tan^4(2x)}{\cos^2(x)}\,\mathrm dx>\!\int_{\frac\pi8}^{\frac\pi4}\!\!\dfrac{\pi}{\left(\frac{\pi}2\!-\!2x\right)^4}\,\mathrm dx=\\[3pt]&=\lim\limits_{\varepsilon\to0^+}\,\left[\dfrac{\pi}{6\left(\frac\pi2\!-\!2x\right)^3}\right]_{\frac\pi8}^{\frac\pi4-\varepsilon}\!\!\!\!=+\infty\;.\end{align}$