Question

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

• A. $\frac{2~\sin~\theta+1}{\sin~\theta+3}$
• B. $\frac{2~\sin~\theta+1}{5(\sin~\theta+3)}$
• C. $\frac{5(\sin~\theta+3)}{2~\sin~\theta+1}$
• D. $\frac{5(2~\sin~\theta+1)}{\sin~\theta+3}$

Hint:

Calculus Tip: When an integrand features a mix of $\sin x$ and $\cos x$, look for a way to convert all but one term (which acts as the derivative $dt$) into either all sines or all cosines.

Answer 1

The Correct Option is D

Concept: Calculus - Integration by Substitution and Partial Fractions.

Step 1: Convert the denominator into a single trigonometric function. Use the identity $\cos^2\theta = 1 - \sin^2\theta$. The denominator becomes $5 + 7\sin\theta - 2(1-\sin^2\theta) = 5 + 7\sin\theta - 2 + 2\sin^2\theta = 2\sin^2\theta + 7\sin\theta + 3$. The integral is now $\int\frac{\cos\theta}{2\sin^2\theta + 7\sin\theta + 3}d\theta$.

Step 2: Use substitution to create an algebraic fraction. Let $t = \sin\theta$. Then $dt = \cos\theta d\theta$. Substituting these into the integral transforms it entirely into terms of $t$: $I = \int\frac{dt}{2t^2 + 7t + 3}$.

Step 3: Factor the quadratic denominator. Factor $2t^2 + 7t + 3$ by splitting the middle term: $2t^2 + 6t + t + 3 = 2t(t + 3) + 1(t + 3) = (2t + 1)(t + 3)$. The integral is $\int\frac{dt}{(2t + 1)(t + 3)}$.

Step 4: Apply partial fraction decomposition and integrate. We decompose $\frac{1}{(2t + 1)(t + 3)} = \frac{2/5}{2t + 1} - \frac{1/5}{t + 3}$. Integrating this gives: $\frac{2}{5}\int\frac{dt}{2t+1} - \frac{1}{5}\int\frac{dt}{t+3} = \frac{2}{5}\cdot\frac{\ln|2t+1|}{2} - \frac{1}{5}\ln|t+3| + c$. This simplifies to $\frac{1}{5}\ln|2t+1| - \frac{1}{5}\ln|t+3| + c$. Using logarithm properties, we combine it: $\frac{1}{5}\ln\left|\frac{2t+1}{t+3}\right| + c$.

Step 5: Compare with the given expression and calculate the final value. Substitute $t = \sin\theta$ back into the integrated expression: $\frac{1}{5}\log_e\left|\frac{2\sin\theta+1}{\sin\theta+3}\right| + c$. Comparing this with $A\log_e|f(\theta)|+c$, we identify $A = \frac{1}{5}$ and $f(\theta) = \frac{2\sin\theta+1}{\sin\theta+3}$. The question asks for the ratio $\frac{f(\theta)}{A}$. Substituting our found values: $\frac{\frac{2\sin\theta+1}{\sin\theta+3}}{\frac{1}{5}} = 5\left(\frac{2\sin\theta+1}{\sin\theta+3}\right)$. $$ \therefore \text{The value of } \frac{f(\theta)}{A} \text{ is } \frac{5(2~\sin~\theta+1)}{\sin~\theta+3}. $$