Integral of the day III

Corollary of the integral representation of the $sec(x)$ function

Today we show the proof of this beautiful integral posted by @gureponnn on Twitter. \[ \int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x + \sin^{2n} x}dx = \frac{\pi^2}{8n}\sum_{j=1}^{n}\binom{n-1}{j-1} \csc \left(\frac{\pi(2j-1)}{2n}\right)\] The proof of this result rely on the well-known integral representation of the $sec(x)$ function

Proof

\begin{align*} I=\int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x + \sin^{2n} x}dx =& \int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x (1+\tan^{2n} x)}dx \\ =& \int_{0}^{\infty} \frac{\arctan(w)(w^2+1)^{n-1}}{1+w^{2n}}dw \quad (w \mapsto \tan x)\\ =& \int_{0}^{\infty} \frac{\arctan\left(\frac{1}{s}\right)(s^2+1)^{n-1}}{s^{2n}+1}ds \quad \left(s \mapsto \frac{1}{w}\right)\\ =& \int_{0}^{\infty} \frac{\left(-\arctan(s)+\frac{\pi}{2}\right)(s^2+1)^{n-1}}{s^{2n}+1}ds \quad \left(\arctan \left(\frac{1}{a}\right) = -\arctan(a)+\frac{\pi}{2}\right)\\ =& -\int_{0}^{\infty}\frac{\arctan(s)(s^2+1)^{n-1}}{s^{2n}+1}ds + \frac{\pi}{2} \int_{0}^{\infty}\frac{(s^2+1)^{n-1}}{s^{2n}+1}ds \end{align*} Hence \[ 2I = \frac{\pi}{2} \int_{0}^{\infty}\frac{(s^2+1)^{n-1}}{s^{2n}+1}ds \] Therefore \begin{align*} I = \frac{\pi}{4} \int_{0}^{\infty}\frac{(s^2+1)^{n-1}}{s^{2n}+1}ds = &\frac{\pi}{4n} \int_{0}^{\infty}\frac{(r^{\frac{2}{n}}+1)^{n-1}}{(r^2+1)r^{\frac{n-1}{n}}}dr \quad (r^2 \mapsto s^{2n})\\ =& \frac{\pi}{4n} \int_{0}^{\infty}\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{r^{\frac{2k}{n}}}{(r^2+1)r^{\frac{n-1}{n}}}dr \quad (\textrm{binomial theorem})\\ =& \frac{\pi}{4n}\sum_{k=0}^{n-1}\binom{n-1}{k} \int_{0}^{\infty}\frac{r^{\frac{2k+1-n}{n}}}{r^2+1}dr \end{align*} Recall the integral representation of the secant function: \[\boxed{\sec z = \frac{2}{\pi} \int_{0}^{\infty} \frac{t^{\frac{2z}{\pi}}}{t^2+1}dt \quad |\operatorname{Re}(z)|<\frac{\pi}{2}}\] If $\displaystyle z = \frac{\pi(2k+1-n)}{2n}$ \[ \int_{0}^{\infty} \frac{t^{\frac{2k+1-n}{n}}}{t^2+1}dt = \frac{\pi}{2} \sec \left(\frac{\pi(2k+1)}{2n} - \frac{\pi}{2} \right) = \frac{\pi}{2} \csc \left(\frac{\pi(2k+1)}{2n} \right)\]

Hence
\[ I = \frac{\pi}{4n}\sum_{k=0}^{n-1}\binom{n-1}{k} \int_{0}^{\infty}\frac{r^{\frac{k-1+n}{n}}}{r^2+1}dr = \frac{\pi^2}{8n}\sum_{k=0}^{n-1}\binom{n-1}{k} \csc \left(\frac{\pi(2k+1)}{2n} \right)\] If $\displaystyle k = j-1$ \[ I = \frac{\pi^2}{8n}\sum_{j=1}^{n}\binom{n-1}{j-1} \csc \left(\frac{\pi(2j-1)}{2n}\right)\] \[\boxed{\int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x + \sin^{2n} x}dx = \frac{\pi^2}{8n}\sum_{j=1}^{n}\binom{n-1}{j-1} \csc \left(\frac{\pi(2j-1)}{2n}\right)}\]