Try Solving this (3)

$\large \displaystyle I = \int_0^{\pi} \frac{dx}{7+5 \cos x}\\ \large \displaystyle I = \int_0^{\pi} \frac{dx}{7 + 5 \cos (\pi - x)}\\ \large \displaystyle I = \int_0^{\pi} \frac{dx}{7 - 5 \cos x}$

By Adding, We get

$\begin{aligned} \large \displaystyle 2I = \int_0^{\pi} \frac{14}{49 - 25 \cos ^2x} .dx = 2 \times 14 \int_0^{\pi/2} \frac{dx}{49-25 \cos ^2x}\\ \large \displaystyle = 28 \int_0^{\pi/2} \frac{\sec ^2x . dx}{49(1+\tan ^2x) - 25}\\ \large \displaystyle = 28 \int_0^{\pi/2} \frac{\sec ^2x . dx}{24 + 49 \tan ^2x}\\ \large \displaystyle = \frac{28}{49} \int_0^{\pi/2} \frac{\sec ^2x}{\frac{24}{49} + \tan ^2x} .dx\\ \large \displaystyle = \frac{28}{49} \times \left[\frac{1}{\sqrt{\frac{24}{49}}} \tan ^{-1} \frac{7 \tan x}{\sqrt{24}} \right]_0^{\pi/2}\\ \large \displaystyle = \frac{2}{\sqrt{6}} \left (\frac{\pi}{2} - 0 \right) = \frac{\pi}{\sqrt{6}} = \frac{22}{7 \times \sqrt6} \end{aligned}$

$\large \displaystyle 2I = \frac{22}{7 \times \sqrt6}\\ \large \displaystyle \implies I = \frac{22}{7 \times 2 \times \sqrt6} = \boxed{0.64153}$

$I = \displaystyle \int_{0}^{\pi} \dfrac{dx}{7+5\cos(x)}$
$\tan\left(\dfrac{x}{2}\right) = t \rightarrow dx = \dfrac{2dt}{1+t^{2}}$
$I = \displaystyle \int_{0}^{\infty} \dfrac{\frac{2dt}{1+t^{2}}}{7+5\frac{1-t^{2}}{1+t^{2}}} = \int_{0}^{\infty} \dfrac{dt}{t^{2}+6} = \dfrac{1}{2\sqrt{6}}\left[\arctan\left(\dfrac{t}{\sqrt{6}}\right)\right]_{0}^{\infty} = \dfrac{\pi}{2\sqrt{6}}$