Using the property \int_{a}^{b}{f(x)dx} = \int_{a}^{b}{f(a+b -x)dx}
we have
I = \int_{0}^{\pi}{\frac{dx}{1+5^{cosx}}} = \int_{0}^{\pi}{\frac{dx}{1+5^{-cosx}}} = \int_{0}^{\pi}{\frac{5^{cosx}dx}{1+5^{cosx}}}
so now 2I = \int_{0}^{\pi}{\frac{1+5^{cosx}}{1+5^{cosx}}}dx = \int_{0}^{\pi}{dx} = \pi
so I = π/2