The value of \(\int \frac{d x}{5+4 \sin x}\) is equal to

\(\begin{array}{ll} & \text { Let } I=\int \frac{\mathrm{d} x}{5+4 \sin x} \\ & \text { Put } \tan \left(\frac{x}{2}\right)=\mathrm{t} \\ \therefore \quad & x=2 \tan ^{-1} \mathrm{t} \\ \therefore \quad & \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2} \text { and } \sin x=\frac{2 \mathrm{t}}{1+\mathrm{t}^2} \\ \therefore \quad & \mathrm{I}=\int \frac{2 \mathrm{dt}}{5+4\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^2}\right)}\end{array}\)
\(\begin{aligned} & =2 \int \frac{d t}{5+5 t^2+8 t} \\ & =\frac{2}{5} \int \frac{d t}{t^2+\frac{8 t}{5}+1} \\ & =\frac{2}{5} \int \frac{d t}{t^2+\frac{8}{5} t+1-\frac{16}{25}+\frac{16}{25}} \\ & =\frac{2}{5} \int \frac{d t}{\left(t+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}\end{aligned}\)
\(\begin{aligned} & =\frac{2}{5} \cdot \frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{\mathrm{t}+\frac{4}{5}}{\frac{3}{5}}\right)+\mathrm{c} \\ & =\frac{2}{3} \tan ^{-1}\left[\frac{5 \tan \left(\frac{x}{2}\right)+4}{3}\right]+\mathrm{c}\end{aligned}\)