I need to show that the following tricky integral: $$\int_0^\pi{12\cos x\ \mathrm{sech}\left(\frac {\pi}2\tan\frac x2\right)}\mathrm{d}x$$ is equal to exactly $\pi^2$. I have no idea how to start. I tried the substitution $u=\tan\frac x2$ and ended up with:
$$\int_0^\infty24\ \mathrm{sech}\frac{\pi u}{2}\frac{1-u^2}{(1+u^2)^2}\mathrm{d}u$$
I might have to use residue theorem or Fourier transformation here, but I'm lost. Thank you!
Continue with \begin{align} &\int_0^\infty \operatorname{sech}\frac{\pi x}{2}\frac{1-x^2}{(1+x^2)^2}dx\\ =& \int_0^\infty \operatorname{sech}\frac{\pi x}{2}\bigg(\int_0^\infty e^{-y} y \cos(x y) dy\bigg) dx\\ =& \int_0^\infty e^{-y}y \int_0^\infty \frac{\cos (xy)}{\cosh\frac{\pi x}2}dx \ dy =\int_0^\infty \frac{ e^{- y}y} {\cosh y} \overset{t=e^{-2y}}{dy}\\ =& - \frac12\int_0^1 \frac{\ln t}{1+t}dt\overset{ibp}= \frac12\int_0^1 \frac{\ln (1+t)}{t}dt=\frac{\pi^2}{24}\\ \end{align} where $\int_0^\infty \frac{\cos (xy)}{\cosh\frac{\pi x}2}dx=\text{sech}\ y$ and $ \int_0^1 \frac{\ln (1+t)}{t}dt =\frac{\pi^2}{12}$.
COMMENT.-By definitions $$I=\int_0^\pi{12\cos(x)\ \mathrm{sech}\left(\frac 12\pi\tan\left(\frac x2\right)\right)}\mathrm{d}x$$ $$I=\int_0^\pi6\cos(x)( e^A-e^{-A}) dx\text { where } A={\frac 12\pi\tan(\frac x2)}$$ or$$I=\int_0^\pi6\cos(x)(e^B-e^{-B})dx\text { where } B=\frac12\pi\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}$$ where, obviously $$A=B=\tan(\frac x2)=\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}$$ What results with changes of possible variables is always non-elementary. My opinion is that there is no clever trick possible as you believe.
