Find $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$

(N/A) Let $y = \sin \phi$.
Then $dy = \cos \phi \, d\phi$.
Substituting these into the integral,we get:
$\int \frac{(3y-2) dy}{5 - (1 - y^2) - 4y} = \int \frac{3y-2}{y^2 - 4y + 4} dy = \int \frac{3y-2}{(y-2)^2} dy$.
Using partial fractions,let $\frac{3y-2}{(y-2)^2} = \frac{A}{y-2} + \frac{B}{(y-2)^2}$.
Then $3y-2 = A(y-2) + B$.
Comparing coefficients,$A = 3$ and $-2A + B = -2 \implies B = 4$.
Thus,the integral becomes $\int \left( \frac{3}{y-2} + \frac{4}{(y-2)^2} \right) dy$.
$= 3 \ln |y-2| - \frac{4}{y-2} + C$.
Substituting $y = \sin \phi$ back,we get $3 \ln |\sin \phi - 2| - \frac{4}{\sin \phi - 2} + C$.
Since $\sin \phi - 2$ is always negative,$|\sin \phi - 2| = 2 - \sin \phi$.
$= 3 \ln (2 - \sin \phi) + \frac{4}{2 - \sin \phi} + C$.