New Year's Countdown Day 26: Day After Christmas Integral

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac 1{12+26\cos x}dx & \small \color{#3D99F6} \text{Let }t = \tan \frac x 2 \implies dt = \frac 12 \sec^2 \frac x 2 \ dx \\ & = \int_0^1 \frac 1{12 + 26\left(\frac {1-t^2}{1+t^2}\right)} \cdot \frac {2\ dt}{1+t^2} & \small \color{#3D99F6} \text{and }\cos x = \frac {1-t^2}{1+t^2} \\ & = \int_0^1 \frac {dt}{19-7t^2} \\ & = \int_0^1 \frac {dt}{(\sqrt{19}+\sqrt 7t)(\sqrt{19}-\sqrt 7t)} \\ & = \frac 1{2\sqrt{19}} \int_0^1 \left(\frac 1{\sqrt{19}+\sqrt 7t} + \frac 1{\sqrt{19}-\sqrt 7t} \right) dt \\ & = \frac 1{2\sqrt{19}\cdot \sqrt 7} \bigg[\ln \left(\sqrt{19}+\sqrt 7t\right) - \ln \left(\sqrt{19}-\sqrt 7t\right) \bigg]_0^1 \\ & = \frac 1{2\sqrt{133}} \ln \left(\frac {\sqrt{19}+\sqrt 7}{\sqrt{19}-\sqrt 7}\right) \\ & = \frac 1{2\sqrt{133}} \ln \left(\frac {\left(\sqrt{19}+\sqrt 7\right)^2}{19-7}\right) \\ & = \frac 1{2\sqrt{133}} \ln \left(\frac {26 + 2\sqrt{133}}{12}\right) \\ & = \frac 1{\sqrt{133}} \ln \sqrt {\frac {13 + \sqrt{133}}6} \end{aligned}$

Therefore, $a+b-c = 133+13-6 = \boxed{140}$ .