Question

$\int_{0}^{\pi / 2} \log\left(\frac{4+3\sin x}{4+3\cos x}\right) dx = $

• A. $0$
• B. $4\log 3$
• C. $\frac{1}{2}$
• D. $2\log 4$

Hint:

Whenever you see a definite integral from $0$ to $\frac{\pi}{2}$ containing a quotient of symmetric sine and cosine terms inside a logarithm, the answer is almost always $0$. Applying King's property flips the fraction upside down, turning it into its own reciprocal and causing the sum to cancel out to $\log(1) = 0$.

Answer 1

The Correct Option is A

Step 1: Name the integral.
Let $I = \displaystyle\int_{0}^{\pi/2} \log\left(\dfrac{4+3\sin x}{4+3\cos x}\right) dx$.
Step 2: Use the reflection property.
Replace $x$ by $\dfrac{\pi}{2}-x$. Then $\sin x \to \cos x$ and $\cos x \to \sin x$.
Step 3: Write the reflected integral.
\[ I = \int_{0}^{\pi/2} \log\left(\frac{4+3\cos x}{4+3\sin x}\right) dx \] which is the same integrand with numerator and denominator swapped.
Step 4: Add the two forms.
Adding the original and reflected versions: $2I = \displaystyle\int_{0}^{\pi/2}\left[\log\dfrac{4+3\sin x}{4+3\cos x} + \log\dfrac{4+3\cos x}{4+3\sin x}\right]dx$.
Step 5: Combine the logs.
The sum of logs is the log of the product, and the product of the two reciprocal fractions is $1$, so the integrand is $\log 1 = 0$.
Step 6: Finish.
Thus $2I = \displaystyle\int_{0}^{\pi/2} 0\, dx = 0$, giving $I = 0$. The symmetry trick makes the antiderivative unnecessary.
\[ \boxed{I = 0} \]