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: Denote the integral by \( I \).
Let \( I = \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx \).
Step 2: Use the property of definite integrals \( \int_{0}^{a} f(x) dx = \int_{0^{a} f(a - x) dx \).}
Applying this property with \( a = \pi/2 \), we have: \[ I = \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin(\frac{\pi}{2} - x)}{4 + 3\cos(\frac{\pi}{2} - x)}\right) dx \] Using the trigonometric identities \( \sin(\frac{\pi}{2} - x) = \cos x \) and \( \cos(\frac{\pi}{2} - x) = \sin x \), we get: \[ I = \int_{0}^{\pi/2} \log\left(\frac{4 + 3\cos x}{4 + 3\sin x}\right) dx \]
Step 3: Use the property of logarithms \( \log\left(\frac{a}{b}\right) = -\log\left(\frac{b}{a}\right) \).
\[ I = \int_{0}^{\pi/2} \log\left(\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right)^{-1}\right) dx \] \[ I = \int_{0}^{\pi/2} -\log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx \] \[ I = - \int_{0}^{\pi/2} \log\left(\frac{4 + 3\sin x}{4 + 3\cos x}\right) dx \]
Step 4: Relate the new integral to the original integral \( I \).
We see that the integral obtained in Step 3 is \( -I \). Therefore, we have: \[ I = -I \]
Step 5: Solve for \( I \).
Adding \( I \) to both sides of the equation \( I = -I \), we get: \[ 2I = 0 \] \[ I = 0 \] Thus, the value of the integral is \( 0 \).