Integration of $\int\frac{1}{\cos x(5-4\sin x)}\, dx$ $$\int\dfrac{1}{\cos x(5-4\sin x)}\, dx$$


I have tried multiple methods like converting this into partial fractions or substitution but no positive!
Hint: One way is $\sin x=u$ then
$$\int\dfrac{1}{\cos x(5-4\sin x)}=\int\dfrac{\cos x\, dx}{(1-\sin^2x)(5-4\sin x)}=\int\dfrac{du}{(1-u^2)(5-4u)}$$
Let,
$$Y=\int\dfrac{  1}{cos(x)(5-4sin(x)}
  =\int \dfrac{  1}{\dfrac{1-tan^2(x/2)}{1+tan^2(x/2)}  ×  (5-4\dfrac{2tan(x/2)}{1+tan^2(x/2))})}$$
Now,simplify it and then put  $tan(x/2)=t$

On differentiate bhoth side with d(t)
I.e.  d(tan(x/2))/d(t)=d(t)/d(t)
      sec^2(x/2)/2   ×  d(x)/d(t)=1
      sec^2(x/2)d(x)=2d(t)