Let's do some calculus! (10)

$\begin{aligned} I & = \frac {\pi^2}{\ln 3} \int_\frac 76 ^\frac 56 \sec (\pi x) \ dx \\ & = \frac {\pi^2}{\ln 3} \int_\frac 76 ^\frac 56 \frac 1{\cos (\pi x)} \ dx & \small \color{#3D99F6}{\text{Let }t = \tan \frac {\pi x}2, \ \cos (\pi x) = \frac {1-t^2}{1+t^2}, \ dx = \frac 2{\pi (1+t^2)} dt } \\ & = \frac {2 \pi}{\ln 3} \int_{\color{#3D99F6}{\tan \frac {7\pi}{12}}} ^{\tan \frac {5\pi}{12}} \frac 1{1-t^2} \ dt & \small \color{#3D99F6}{\text{Note that } \tan \frac {7\pi}{12} = - \tan \frac {5\pi}{12}} \\ & = \frac {2 \pi}{\ln 3} \int_{\color{#3D99F6}{- \tan \frac {5\pi}{12}}} ^{\tan \frac {5\pi}{12}} \frac 1{(1-t)(1+t)} \ dt \\ & = \frac {\pi}{\ln 3} \int_{-\tan \frac {5\pi}{12}} ^{\tan \frac {5\pi}{12}} \left( \frac 1{1-t} + \frac 1{1+t} \right) dt \\ & = \frac {\pi}{\ln 3} \left( -\ln |1-t| + \ln |1+t| \right)\bigg|_{-\tan \frac {5\pi}{12}} ^{\tan \frac {5\pi}{12}} \\ & = \frac {\pi}{\ln 3} \ln \left( \frac {\tan \frac {5\pi}{12}+1}{\tan \frac {5\pi}{12}-1} \right)^2 & \small \color{#3D99F6}{\text{Note that } \tan \frac {5\pi}{12} = \frac {\sqrt 3+1}{\sqrt 3-1}} \\ & = \frac {\pi}{\ln 3} \ln (\sqrt 3)^2 = \frac {\pi}{\ln 3} \ln 3 = \boxed{\pi} \end{aligned}$

Let

$\begin{aligned} I & = \dfrac{{\pi}^{2}}{\ln 3} \displaystyle \int_{\frac{7}{6}}^{\frac{5}{6}}{\sec(\pi x)} \,dx \\ & = \dfrac{{\pi}^{2}}{\ln 3} \left(\dfrac{\ln {\left| \sec \pi x + \tan \pi x \right|}_{{7}/{6}}^{{5}/{6}}}{\pi}\right) \\ & = \dfrac{\pi}{\ln 3} \left(\ln {\left| \sec \dfrac{5\pi}{6} + \tan \dfrac{5\pi}{6} \right|} - \ln {\left| \sec \dfrac{7\pi}{6} + \tan \dfrac{7\pi}{6} \right|} \right) \\ & = \dfrac{\pi}{\ln 3} \left(\ln 3\right) \\ & = \boxed{\pi} \end{aligned}$