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:

For symmetric trig integrals: \begin{itemize} \item Use \( x \to \frac{\pi}{2}-x \). \item Logs often cancel nicely. \end{itemize}

Answer 1

The Correct Option is C

Concept: Use symmetry: \[ \int_0^{\pi/2} f(x)\,dx = \int_0^{\pi/2} f\left(\frac{\pi}{2}-x\right) dx \] This swaps \( \sin x \leftrightarrow \cos x \). Step 1: {\color{red}Let integral = \( I \).} \[ I = \int_0^{\pi/2} \log\!\left(\frac{4+3\sin x}{4+3\cos x}\right) dx \] Substitute \( x \to \frac{\pi}{2}-x \): \[ I = \int_0^{\pi/2} \log\!\left(\frac{4+3\cos x}{4+3\sin x}\right) dx \] Step 2: {\color{red}Add both forms.} \[ 2I = \int_0^{\pi/2} \log\left( \frac{4+3\sin x}{4+3\cos x} \cdot \frac{4+3\cos x}{4+3\sin x} \right) dx \] \[ = \int_0^{\pi/2} \log(1)\,dx = 0 \] \[ I = 0 \]