Question

The value of \[ \int \sin(\log x) \, dx + \int \cos(\log x) \, dx \] is equal to
 

• A. \( \sin(\log x) + C \)
• B. \( \cos(\log x) + C \)
• C. \( x \sin(\log x) + C \)
• D. \( x \cos(\log x) + C \)

Hint:

Use integration by parts to solve integrals of the form \( \int \sin(\log x) \, dx \) and \( \int \cos(\log x) \, dx \).

Answer 1

The Correct Option is C

Step 1: Applying integration by parts.
To solve the integral \( \int \sin(\log x) \, dx \), we use integration by parts. Let: \[ u = \sin(\log x), \quad dv = dx \] Then: \[ du = \cos(\log x) \cdot \frac{1}{x} \, dx, \quad v = x \] Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get: \[ \int \sin(\log x) \, dx = x \sin(\log x) - \int x \cdot \cos(\log x) \cdot \frac{1}{x} \, dx = x \sin(\log x) - \int \cos(\log x) \, dx \]
Step 2: Conclusion.
Thus, the value of \( \int \sin(\log x) \, dx + \int \cos(\log x) \, dx \) simplifies to: \[ x \sin(\log x) + C \] Therefore, the correct answer is (3) \( x \sin(\log x) + C \).