1. Simplify the integrand

We need to evaluate $I=\int \frac{\cos\theta}{5+7\sin\theta-2\cos^2\theta}\,d\theta.$

Use $\cos^2\theta=1-\sin^2\theta.$ So the denominator becomes $5+7\sin\theta-2(1-\sin^2\theta)=5+7\sin\theta-2+2\sin^2\theta$ $=2\sin^2\theta+7\sin\theta+3.$ Hence $I=\int \frac{\cos\theta}{2\sin^2\theta+7\sin\theta+3}\,d\theta.$

2. Substitute

Let $t=\sin\theta \implies dt=\cos\theta\,d\theta.$ Then $I=\int \frac{dt}{2t^2+7t+3}.$

Factor the quadratic: $2t^2+7t+3=(2t+1)(t+3).$ Thus $I=\int \frac{dt}{(2t+1)(t+3)}.$

3. Partial fractions

Write $\frac{1}{(2t+1)(t+3)}=\frac{P}{2t+1}+\frac{Q}{t+3}.$ Then $1=P(t+3)+Q(2t+1).$ So $1=(P+2Q)t+(3P+Q).$ Comparing coefficients: $P+2Q=0,\qquad 3P+Q=1.$ From the first, $P=-2Q$. Substituting into the second: $3(-2Q)+Q=1 \implies -5Q=1 \implies Q=-\frac15,$ $P=\frac25.$ Therefore $\frac{1}{(2t+1)(t+3)}=\frac{2/5}{2t+1}-\frac{1/5}{t+3}.$

4. Integrate

So $I=\int \left(\frac{2/5}{2t+1}-\frac{1/5}{t+3}\right)dt.$ Now, $\int \frac{2/5}{2t+1}dt=\frac15\ln|2t+1|,$ $\int \frac{-1/5}{t+3}dt=-\frac15\ln|t+3|.$ Thus $I=\frac15\ln|2t+1|-\frac15\ln|t+3|+C$ $=\frac15\ln\left|\frac{2t+1}{t+3}\right|+C.$ Substituting back $t=\sin\theta$, $I=\frac15\ln\left|\frac{2\sin\theta+1}{\sin\theta+3}\right|+C.$

5. Match with the given form

Given $I=A\log_e|B(\theta)|+C.$ So one valid choice is $A=\frac15,\qquad B(\theta)=\frac{2\sin\theta+1}{\sin\theta+3}.$ Then $\frac{B(\theta)}{A}=5\cdot \frac{2\sin\theta+1}{\sin\theta+3}.$ Hence $\frac{B(\theta)}{A}=\frac{5(2\sin\theta+1)}{\sin\theta+3}.$

6. Check options

This matches Option C.

Therefore, the correct answer is: $\boxed{\text{C}}$