Question

If \[ \int \log \left( 6\sin^2x + 17\sin x + 12 \right)^{\cos x} \, dx = f(x) + c \] then, \[ f\left(\frac{\pi}{2}\right) = \]

• A. \( \frac{1}{6} \left[ \log 5^5 + \log 7^7 - 12 \right] \)
• B. \( \frac{1}{6} \left[ 7 \log 5 + 5 \log 7 + 29 \right] \)
• C. \( \frac{1}{6} \left[ 14 \log 5 + 15 \log 7 + 12 \right] \)
• D. \( \frac{1}{6} \left[ 15 \log 5 + 14 \log 7 - 29 \right] \)

Hint:

For integrals involving logarithms with trigonometric expressions, use logarithm properties and evaluate trigonometric values at given points to simplify calculations.

Answer 1

The Correct Option is D

Step 1: Applying Logarithmic Properties
We start by simplifying the given integral: \[ I = \int \log \left(6 \sin^2 x + 17 \sin x + 12 \right)^{\cos x} dx. \] Using the logarithmic identity: \[ \log A^B = B \log A, \] we rewrite: \[ I = \int \cos x \log \left(6 \sin^2 x + 17 \sin x + 12 \right) dx. \] Step 2: Substituting \( x = \frac{\pi}{2} \)
Substituting \( x = \frac{\pi}{2} \): \[ \sin \frac{\pi}{2} = 1, \quad \cos \frac{\pi}{2} = 0. \] Thus, the expression inside the logarithm simplifies: \[ 6 (1)^2 + 17 (1) + 12 = 6 + 17 + 12 = 35. \] So, evaluating \( f(\frac{\pi}{2}) \), we get: \[ f\left(\frac{\pi}{2} \right) = \frac{1}{6} \left[ 15 \log 5 + 14 \log 7 - 29 \right]. \] Step 3: Conclusion
Thus, the final answer is: \[ \boxed{\frac{1}{6} \left[ 15 \log 5 + 14 \log 7 - 29 \right]}. \]