The value of $\frac{\pi^2}{\ln 3} \int\limits_{7 / 6}^{5 / 6} \sec (\pi x) d x$, is

$\pi$

$\frac{\pi^2}{\ln 3} \int\limits_{7 / 6}^{5 / 6} \sec (\pi x) d x$

$=\frac{\pi^2}{\ln 3} \times \frac{1}{\pi}[\log (\sec \pi x+\tan \pi x)]_{7 / 6}^{5 / 6}$ sub nothing

$=\frac{\pi}{\ln 3}\left[\log \left(\sec \frac{5 \pi}{6}+\tan \frac{5 \pi}{6}\right)-\log \left(\sec \frac{7 \pi}{6}+\tan \frac{7 \pi}{6}\right)\right]$

$=\frac{\pi}{\ln 3}\left[\log \left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)-\log \left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right)\right]$

$=\frac{\pi}{\ln 3}\left[\log \sqrt{3}-\log \left(\frac{1}{\sqrt{3}}\right)\right]=\frac{\pi}{\ln 3}(\log 3)=\pi$