Just a little difference...exchange of 6 and 5 !!!

Using Weierstrass Substitution

$\begin{aligned} \text{let I} &=\int_{0}^{2\pi}{\dfrac{dx}{31+6cosx+5sinx}} =\int_{0}^{\pi}{\dfrac{dx}{31+6cosx+5sinx}} +\int_{\pi}^{2\pi}{\dfrac{dx}{31+6cosx+5sinx}} \\ & \color{#D61F06} {\text{let }t =\tan(\frac{x}{2}) \Rightarrow \cos(x) =\frac{1-t^2}{1+t^2} \text{ and } \sin(x) =\frac{2t}{1+t^2} \text{ and } dx =\frac{2dt}{1+t^2} }\\ I & = \int_{0}^{\infty}{\dfrac{1}{31+6\frac{1-t^2}{1+t^2}+5\frac{2t}{1+t^2}}}\frac{2dt}{1+t^2} +\int_{-\infty}^{0}{\dfrac{1}{31+6\frac{1-t^2}{1+t^2}+5\frac{2t}{1+t^2}}}\frac{2dt}{1+t^2} = \int_{-\infty}^{\infty}{\dfrac{2dt}{31+31t^2+6-6t^2+10t}}\\ I &=2\int_{-\infty}^{\infty}{\dfrac{dt}{25t^2+10t+37}}=\frac{2}{25}\int_{-\infty}^{\infty}{\dfrac{dt}{\left (\frac{5t+1}{5} \right )^2+\frac{36}{25}}}=\frac{2}{36}\int_{-\infty}^{\infty}{\dfrac{dt}{\left (\frac{5t+1}{6} \right )^2+1}}=\frac{2}{36}\left .\left (\frac{6}{5} \tan^{-1}\left ( \frac{5t+1}{6} \right )\right |_{-\infty}^{ \infty} \right )\\ I &= \frac{\pi}{15} \end{aligned} \\ a=15 \\ \frac{a^2}{2}=\frac{225}{2}=112.5 \\ \left \lceil 112.5 \right \rceil=113$