Find ∫ ((3 sin φ - 2) cos φ)/(5 - cos^2 φ - 4 sin φ) dφ

The correct answer is Option (1) → $3 \log (2 - \sin \phi) + \frac{4}{2 - \sin \phi} + C$

Let $y = \sin \phi$

Then

$dy = \cos \phi \, d\phi$

Therefore, $\int \frac{(3 \sin \phi - 2) \cos \phi}{5 - \cos^2 \phi - 4 \sin \phi} \, d\phi = \int \frac{(3y - 2) \, dy}{5 - (1 - y^2) - 4y}$

$= \int \frac{3y - 2}{y^2 - 4y + 4} \, dy$

$= \int \frac{3y - 2}{(y - 2)^2} \, dy = I \text{ (say)}$

Now, we write

$\frac{3y - 2}{(y - 2)^2} = \frac{A}{y - 2} + \frac{B}{(y - 2)^2} \quad [\text{by Table 7.2 (2)}] \text{}$

Therefore, $3y - 2 = A(y - 2) + B$

Comparing the coefficients of $y$ and constant term, we get $A = 3$ and $B - 2A = -2$, which gives $A = 3$ and $B = 4$.

Therefore, the required integral is given by

$I = \int \left[ \frac{3}{y - 2} + \frac{4}{(y - 2)^2} \right] \, dy = 3 \int \frac{dy}{y - 2} + 4 \int \frac{dy}{(y - 2)^2} \text{}$

$= 3 \log |y - 2| + 4 \left( -\frac{1}{y - 2} \right) + C$

$= 3 \log |\sin \phi - 2| + \frac{4}{2 - \sin \phi} + C$

$= 3 \log (2 - \sin \phi) + \frac{4}{2 - \sin \phi} + C \text{ (since, } 2 - \sin \phi \text{ is always positive)}$