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: Understanding the Question:
The problem requires computing the value of the definite integral of a logarithmic trigonometric quotient evaluated over the interval from $0$ to $\frac{\pi}{2}$.

Step 2: Key Formula or Approach:
We use King's Property of definite integration, which states that an integral remains unchanged when the variable $x$ is replaced by the sum of its boundaries minus $x$: $$I = \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$$ For our boundaries, this means replacing $x$ with $\left(\frac{\pi}{2} - x\right)$. Recall the co-function identities: $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$.

Step 3: Detailed Explanation:
Let our initial integral equation be designated as $I$: $$I = \int_{0}^{\pi / 2} \log\left(\frac{4+3\sin x}{4+3\cos x}\right) dx \quad \text{---- (1)}$$ Apply King's Property by replacing $x$ with $\left(\frac{\pi}{2} - x\right)$: $$I = \int_{0}^{\pi / 2} \log\left(\frac{4+3\sin\left(\frac{\pi}{2}-x\right)}{4+3\cos\left(\frac{\pi}{2}-x\right)}\right) dx$$ Simplify the expression inside the logarithm using our trigonometric co-function identities: $$I = \int_{0}^{\pi / 2} \log\left(\frac{4+3\cos x}{4+3\sin x}\right) dx \quad \text{---- (2)}$$ Add equations (1) and (2) together to combine the integrals: $$2I = \int_{0}^{\pi / 2} \left[ \log\left(\frac{4+3\sin x}{4+3\cos x}\right) + \log\left(\frac{4+3\cos x}{4+3\sin x}\right) \right] dx$$ Using the logarithmic addition rule $\log A + \log B = \log(A \cdot B)$, multiply the two fractions together: $$2I = \int_{0}^{\pi / 2} \log\left(\frac{4+3\sin x}{4+3\cos x} \times \frac{4+3\cos x}{4+3\sin x}\right) dx$$ Since the terms in the numerator and denominator cancel each other out completely, the product inside reduces to exactly 1: $$2I = \int_{0}^{\pi / 2} \log(1) \, dx$$ We know that $\log(1) = 0$, so the entire definite integral evaluates to zero: $$2I = \int_{0}^{\pi / 2} 0 \, dx = 0 \implies I = 0$$ This matches option (A).

Step 4: Final Answer:
The value of the definite integral is $0$, which corresponds to option (A).