$\int \frac{1}{7+5\cos x}dx=$

$\int \frac{1}{7+5\cos x}dx=$

$Put\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}\frac{1}{2}dx=dt$

$(1+t^2)dx=2dt\Rightarrow dx=\frac{2dt}{1+t^2}$

$=\int \frac{1}{\left[7+5\left(\frac{1-t^2}{1+t^2}\right)\right]}\frac{2dt}{(1+t^2)}$

$=\int \frac{2dt}{7+7t^2+(5-5t^2)}$

$=2\int \frac{2dt}{2t^2+12}$

$=\int \frac{dt}{t^2+6}=\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\frac{{\displaystyle t}}{{\displaystyle \sqrt{6}}}\right)+c$

$=\frac{1}{\sqrt{6}}{\tan }^{-1}\left(\frac{{\displaystyle \tan x/2}}{{\displaystyle \sqrt{6}}}\right)+c$