If $\int\frac{dx}{5+4\cos x}=k\tan^{-1}(m\tan\frac{x}{2})+C$ then:

$k=\frac{2}{3},m=\frac{1}{3}$

$I=\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5(\sin^2\frac{x}{2}+\cos^2\frac{x}{2})+4(\cos^2\frac{x}{2}-\sin^2\frac{x}{2})}=\int\frac{dx}{9\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}=\int\frac{\sec^2\frac{x}{2}}{9+\tan^2\frac{x}{2}}dx$

Let $I = \tan\frac{x}{2}⇒2\,dt=\sec^2\frac{x}{2}dx$

$⇒I=\int\frac{2dt}{9+t^2}=\frac{2}{3}\tan^{-1}(\frac{t}{3})+C=\frac{2}{3}\tan^{-1}(\frac{\tan(\frac{x}{2})}{3})+C⇒k=\frac{2}{3},m=\frac{1}{3}$