Question

The integral \[ \int \cot x \cdot \log \left[ \log (\sin x) \right] \, dx = \]

• A. \( \log (\sin x) \left[ \log (\sin x) \right] + 1 + c \)
• B. \( \log (\sin x) \left[ \log (\sin x) \right] + 1 + c \)
• C. \( \log (\sin x) \left[ \log (\log (\sin x)) \right] - 1 + c \)
• D. \( \log (\sin x) \left[ \log (\sin x) \right] - 1 + c \)

Hint:

When solving integrals involving logarithms, use integration by parts and simplify the terms to obtain the result.

Answer 1

The Correct Option is C

Step 1: Substituting and simplifying.
We are given the integral \( \int \cot x \cdot \log \left[ \log (\sin x) \right] \, dx \). The most straightforward method to solve this is to apply integration by parts. Let: \[ u = \log \left[ \log (\sin x) \right] \quad \text{and} \quad dv = \cot x \, dx \]
Step 2: Applying integration by parts.
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we calculate the result: \[ \int \cot x \cdot \log \left[ \log (\sin x) \right] \, dx = \log (\sin x) \left[ \log (\log (\sin x)) \right] - 1 + c \]
Step 3: Conclusion.
Thus, the correct answer is \( \log (\sin x) \left[ \log (\log (\sin x)) \right] - 1 + c \), which makes option (C) the correct answer.