Question

The value of the integral \( \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx \) is:

• A. \( 2 \)
• B. \( \frac{3}{4} \)
• C. \( 0 \)
• D. \( -2 \)

Hint:

Definite integrals with symmetric limits or involving trigonometric functions often simplify using appropriate properties of integrals and trigonometric identities.

Answer 1

The Correct Option is C

Step 1: Define the integral.
\nLet \( I = \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx \).\n\n
Step 2: Apply the definite integral property.
\nUsing \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \) with \( a = \pi/2 \), we get:\n\[\nI = \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin(\frac{\pi}{2} - x)}{4 + 3\cos(\frac{\pi}{2} - x)}\right) dx\n\]\nUsing trigonometric identities:\n\[\nI = \int_{0}^{\pi/2} \log\left(\frac{4 + 3\cos x}{4 + 3\sin x}\right) dx\n\]\n\n
Step 3: Use the logarithm property.
\n\[\nI = \int_{0}^{\pi/2} \log\left(\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right)^{-1}\right) dx\n\]\n\[\nI = \int_{0}^{\pi/2} -\log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx\n\]\n\[\nI = - \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx\n\]\n\n
Step 4: Relate to the original integral.
\nThe integral from Step 3 is \( -I \), thus:\n\[\nI = -I\n\]\n\n
Step 5: Solve for \( I \).
\nAdding \( I \) to both sides:\n\[\n2I = 0\n\]\n\[\nI = 0\n\]\n\nTherefore, the integral's value is \( 0 \).