Key Concepts and Formulas

• Trigonometric Identity: $\cos^2\theta + \sin^2\theta = 1$
• Partial Fraction Decomposition: $\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$ for some constants A and B.
• Integral of $\frac{1}{x}$: $\int \frac{1}{x} dx = \ln|x| + C$

Step-by-Step Solution

Step 1: Rewrite the integrand using the trigonometric identity.

• We are given the integral $\int \frac{\cos \theta}{5 + 7\sin \theta - 2\cos^2 \theta} d\theta$. To simplify the denominator, we replace $\cos^2 \theta$ with $1 - \sin^2 \theta$. $\int \frac{\cos \theta}{5 + 7\sin \theta - 2(1 - \sin^2 \theta)} d\theta$

Step 2: Simplify the denominator.

• Expanding and simplifying the denominator, we get: $\int \frac{\cos \theta}{5 + 7\sin \theta - 2 + 2\sin^2 \theta} d\theta = \int \frac{\cos \theta}{3 + 7\sin \theta + 2\sin^2 \theta} d\theta$

Step 3: Perform a u-substitution.

• Let $t = \sin \theta$. Then, $dt = \cos \theta \, d\theta$. Substituting these into the integral, we obtain: $\int \frac{dt}{3 + 7t + 2t^2}$

Step 4: Factor the quadratic in the denominator.

• The quadratic expression $2t^2 + 7t + 3$ can be factored as $(2t + 1)(t + 3)$. Therefore, the integral becomes: $\int \frac{dt}{(2t + 1)(t + 3)}$

Step 5: Perform partial fraction decomposition.

• We decompose the fraction $\frac{1}{(2t + 1)(t + 3)}$ into partial fractions: $\frac{1}{(2t + 1)(t + 3)} = \frac{A}{2t + 1} + \frac{B}{t + 3}$
• Multiplying both sides by $(2t + 1)(t + 3)$, we get: $1 = A(t + 3) + B(2t + 1)$
• To find $A$, we set $2t + 1 = 0$, which means $t = -\frac{1}{2}$. Substituting this into the equation, we have: $1 = A\left(-\frac{1}{2} + 3\right) + B(0) \Rightarrow 1 = A\left(\frac{5}{2}\right) \Rightarrow A = \frac{2}{5}$
• To find $B$, we set $t + 3 = 0$, which means $t = -3$. Substituting this into the equation, we have: $1 = A(0) + B(2(-3) + 1) \Rightarrow 1 = B(-5) \Rightarrow B = -\frac{1}{5}$
• Therefore, the partial fraction decomposition is: $\frac{1}{(2t + 1)(t + 3)} = \frac{2/5}{2t + 1} - \frac{1/5}{t + 3} = \frac{1}{5}\left(\frac{2}{2t + 1} - \frac{1}{t + 3}\right)$

Step 6: Integrate the partial fractions.

• Substituting the partial fraction decomposition back into the integral, we get: $\int \frac{1}{5}\left(\frac{2}{2t + 1} - \frac{1}{t + 3}\right) dt = \frac{1}{5} \int \left(\frac{2}{2t + 1} - \frac{1}{t + 3}\right) dt$ $= \frac{1}{5} \left(\int \frac{2}{2t + 1} dt - \int \frac{1}{t + 3} dt\right) = \frac{1}{5} \left(\ln|2t + 1| - \ln|t + 3|\right) + C$

Step 7: Simplify using logarithm properties.

• Using the property $\ln a - \ln b = \ln \frac{a}{b}$, we have: $\frac{1}{5} \ln\left|\frac{2t + 1}{t + 3}\right| + C$

Step 8: Substitute back for $t$.

• Since $t = \sin \theta$, we substitute back to get: $\frac{1}{5} \ln\left|\frac{2\sin \theta + 1}{\sin \theta + 3}\right| + C$

Step 9: Identify A and B($\theta$).

• Comparing this to the given form $A \log_e |B(\theta)| + C$, we identify $A = \frac{1}{5}$ and $B(\theta) = \frac{2\sin \theta + 1}{\sin \theta + 3}$.

Step 10: Calculate $\frac{B(\theta)}{A}$.

• We have $\frac{B(\theta)}{A} = \frac{\frac{2\sin \theta + 1}{\sin \theta + 3}}{\frac{1}{5}} = 5\left(\frac{2\sin \theta + 1}{\sin \theta + 3}\right) = \frac{5(2\sin \theta + 1)}{\sin \theta + 3}$.

Common Mistakes & Tips

• Remember to use the trigonometric identity correctly when simplifying the integrand.
• Be careful with the signs when performing partial fraction decomposition.
• Don't forget to substitute back to the original variable after integration.
• When integrating $\int \frac{2}{2t+1} dt$, remember the chain rule in reverse: the integral is $\ln |2t+1|$, not $2\ln|2t+1|$.

Summary

We started by simplifying the integrand using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$. Then, we performed a u-substitution with $t = \sin \theta$ to transform the integral into a rational function. We factored the denominator and used partial fraction decomposition to split the rational function into simpler terms. We integrated these terms and then substituted back to express the result in terms of $\theta$. Finally, we compared the result with the given form to find $A$ and $B(\theta)$ and calculated $\frac{B(\theta)}{A}$.

The final answer is $\boxed{\frac{5(2\sin \theta + 1)}{\sin \theta + 3}}$, which corresponds to option (C).