The value of ${\int }_0^{\pi } \frac{1}{5+3\cos x}dx,$ is 

Let $t={\int }_0^{\pi } \frac{1}{5+3\cos x}dx$………………(i)

Then 

            $I={\int }_0^{\pi } \frac{1}{5-3\cos x}dx\dots ..................\text{ (ii) }$                $\left[\because {\int }_0^a f\left(x\right)dx={\int }_0^a f(a-x)dx\right]$

Adding (i) and (ii), we get

$\begin{array}{l}2I=10{\int }_0^{\pi } \frac{1}{25-9{\cos }^{2}x}dx\\ \Rightarrow I=5\times 2{\int }_0^{\pi /2} \frac{1}{25-9{\cos }^{2}x}dx\\ \Rightarrow I=10{\int }_0^{\pi /2} \frac{{\sec }^{2}x}{16+25{\tan }^{2}x}dx\\ \Rightarrow I=10\times \frac{1}{\left(4\right).\left(5\right)}{\left[{\tan }^{-1}\left(\frac{5\tan x}{4}\right)\right]}_0^{\pi /2}=\frac{\pi }{4}\end{array}$