If \(\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^2 \theta} d \theta=A \log _e|f(\theta)|+c\) (where \(\mathrm{c}\) is a constant of integration), then \(\frac{f(\theta)}{A}\) can be

\(\begin{aligned} & \text { Let } I=\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^2 \theta} d \theta \\ & =\int \frac{\cos \theta}{5+7 \sin \theta-2\left(1-\sin ^2 \theta\right)} d \theta \\ & =\int \frac{\cos \theta}{2 \sin ^2 \theta+7 \sin \theta+3} d \theta \\ & =\int \frac{\cos \theta}{(\sin \theta+3)(2 \sin \theta+1)} d \theta \\ & \text { Put } \sin \theta=t \\ & \Rightarrow \cos \theta d \theta=d t\end{aligned}\)
\(\begin{aligned} \therefore \quad I & =\int \frac{d t}{(t+3)(2 t+1)} \\ & =\int\left[\frac{2}{5(2 t+1)}-\frac{1}{5(t+3)}\right] d t \\ & =\frac{2}{5} \cdot \frac{\log |2 t+1|}{2}-\frac{1}{5} \log |t+3|+c \\ & =\frac{1}{5} \log |2 \sin \theta+1|-\frac{1}{5} \log |\sin \theta+3|+c \\ & =\frac{1}{5} \log \left|\frac{2 \sin \theta+1}{\sin \theta+3}\right|+c\end{aligned}\)
\(\begin{aligned} & \therefore \quad \mathrm{A}=\frac{1}{5}, \mathrm{f}(\theta)=\frac{2 \sin \theta+1}{\sin \theta+3} \\ & \therefore \quad \frac{\mathrm{f}(\theta)}{\mathrm{A}}=\frac{5(2 \sin \theta+1)}{\sin \theta+3}\end{aligned}\)