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

$\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+C$

We have,

$I=\int \frac{1}{7+5 \cos x} d x=\int \frac{1+\tan ^2 x / 2}{7\left(1+\tan ^2 \frac{x}{2}\right)+5\left(1-\tan ^2 \frac{x}{2}\right)} d x$

$\Rightarrow I=\int \frac{\sec ^2 \frac{x}{2}}{12+2 \tan ^2 \frac{x}{2}} d x=\int \frac{\frac{1}{2} \sec ^2 \frac{x}{2}}{(\sqrt{6})^2+\left(\tan \frac{x}{2}\right)^2} d x$

$\Rightarrow I=\int \frac{1}{(\sqrt{6})^2+\left(\tan \frac{x}{2}\right)^2} d\left(\tan \frac{x}{2}\right)=\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+C$