$\int \frac{d x}{5+4 \cos x}$ is equal to :

$\frac{2}{3} \tan ^{-1}\left(\frac{1}{3} \tan \frac{x}{2}\right)+c$

Let $I=\int \frac{d x}{5+4 \cos x}$

Let $t=\tan \frac{x}{2} \Rightarrow dx=\frac{2 d t}{1+t^2}$

$\Rightarrow I=\int \frac{\frac{2 d t}{1+t^2}}{5+4\left(\frac{1-t^2}{1+t^2}\right)}=2 \int \frac{d t}{t^2+9}$

$=2 . \frac{1}{3} . \tan ^{-1}\left(\frac{t}{3}\right)+c$

$\Rightarrow I=\frac{2}{3} \tan ^{-1}\left(\frac{1}{3} \tan \frac{x}{2}\right)+c$

Hence (1) is the correct answer.