Evaluate the following integral:

Let, $I=\int \frac{d x}{\cos x(5-4 \sin x)}$

Multiplying and dividing by $\cos x$

$\operatorname{Let}, I=\int \frac{\cos x d x}{\cos ^{2} x(5-4 \sin x)}$

$\Rightarrow I=\int \frac{\cos x d x}{\left(1-\sin ^{2} x\right)(5-4 \sin x)}$

Let, $\sin x=t, \cos x d x=d t$

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

Now, let $\frac{1}{\left(1-\mathrm{t}^{2}\right)(5-4 \mathrm{t})}=\frac{\mathrm{A}}{1-\mathrm{t}}+\frac{\mathrm{B}}{1+\mathrm{t}}+\frac{\mathrm{C}}{5-4 \mathrm{t}}$

$\Rightarrow 1=A(1+t)(5-4 t)+B(1-t)(5-4 t)+C\left(1-t^{2}\right)$

For $t=1, A=\frac{1}{2}$

For $t=-1, B=\frac{1}{18}$

For $t=\frac{5}{4}, C=-\frac{16}{9}$

$\therefore \mathrm{I}=\frac{1}{2} \int \frac{\mathrm{dt}}{1-\mathrm{t}}+\frac{1}{18} \int \frac{\mathrm{dt}}{1+\mathrm{t}}-\frac{16}{9} \int \frac{\mathrm{dt}}{5-4 \mathrm{t}}$

$\Rightarrow \mathrm{I}=-\frac{1}{2} \log |1-\mathrm{t}|+\frac{1}{18} \log |1+\mathrm{t}|+\frac{4}{9} \log |5-4 \mathrm{t}|+\mathrm{c}$

So, $\mathrm{I}=-\frac{1}{2} \log |1-\sin \mathrm{x}|+\frac{1}{18} \log |1+\sin \mathrm{x}|+\frac{4}{9} \log |5-4 \sin \mathrm{x}|+\mathrm{c}$