Question

Evaluate the integral: \[ I = \int_{\frac{-3\pi}{4}}^{\frac{\pi}{6}} \log(\sin(4x+3)) \,dx \]

• A. \( -\frac{\pi}{2} \log 2 \)
• B. \( -\frac{\pi}{8} \log 2 \)
• C. \( -\frac{\pi}{14} \log 2 \)
• D. \( -\frac{\pi}{28} \log 2 \)

Hint:

Use the integral transformation property \( f(a+b-x) \) and standard definite integral results for logarithm functions involving sine and cosine.

Answer 1

The Correct Option is B

Step 1: Consider the standard transformation property. Using the transformation property of definite integrals: \[ \int_{a}^{b} f(x) \,dx = \int_{a}^{b} f(a+b-x) \,dx \] Let: \[ J = \int_{\frac{-3\pi}{4}}^{\frac{\pi}{6}} \log(\sin(4x+3)) \,dx \] Using the transformation \( x \to a+b-x \), we replace \( x \) with \( -x \): \[ J = \int_{\frac{-3\pi}{4}}^{\frac{\pi}{6}} \log(\sin(4(-x)+3)) \,dx \] Using the identity: \[ \sin(\pi - \theta) = \sin\theta \] we get: \[ \log(\sin(4x+3)) + \log(\sin(4(-x)+3)) = \log(\sin(4x+3)) + \log(\cos(4x+3)) \] Thus, we use the integral identity: \[ I + I = \int_{\frac{-3\pi}{4}}^{\frac{\pi}{6}} \log(\sin(4x+3)) \log(\cos(4x+3)) \,dx \] Using the result: \[ \int_{0}^{\pi} \log \sin x \,dx = -\frac{\pi}{2} \log 2 \] We get: \[ J = -\frac{\pi}{8} \log 2 \] Thus, the correct answer is: \[ \boxed{-\frac{\pi}{8} \log 2} \]