$A:\int \frac{1}{5+4\sin x}dx=\frac{2}{3}Ta{n}^{-1}\left(\frac{{\displaystyle 4+5\tan (x/2)}}{{\displaystyle 3}}\right)+c$ 
$R:Ifa>0,a>b,then$$\int \frac{dx}{a+b\sin x}=$$\frac{2}{\sqrt{a^2-b^2}}Ta{n}^{-1}\left[\frac{b+a \tan \left(x/2\right)}{\sqrt{a^2-b^2}}\right]+c$

$\begin{array}{l}I=\int \frac{dx}{5+4\sin x}\\ =\int \frac{dx}{5+4\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}\\ =\int \frac{\left(1+{\tan }^2\frac{x}{2}\right)dx}{5+5{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}}\\ =\int \frac{\left({\sec }^2\frac{x}{2}\right)dx}{5+5{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}}\end{array}$
$$\begin{gathered} \begin{array}{l}\text{ Let }t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx\\ =\int \frac{2dt}{5t^2+8t+5}\\ =2\int \frac{dt}{5\left(t^2+\frac{8}{5}t+1\right)}\\ =\frac{2}{5}\int \frac{dt}{{\left(t+{\displaystyle \frac{4}{5}}\right)}^2+{\left({\displaystyle \frac{3}{5}}\right)}^2} \\ =\frac{2}{5}\frac{5}{3}{\tan }^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)+C \\ =\frac{2}{3}{\tan }^{-1}\left(\frac{5t+4}{3}\right)+C \\ =\frac{2}{3}{\tan }^{-1}\left(\frac{5\tan \left(x/2\right)+4}{3}\right)+C\\ =c\end{array} \end{gathered}$$