Let us analyze the given statements one by one:
Step 1: Bounded and continuous implies uniform continuity.
If \( f \) is bounded and continuous on \( [0, \infty) \), this does not guarantee uniform continuity. Uniform continuity requires the behavior of \( f \) to be controlled uniformly for all points in the domain, which is not guaranteed by just boundedness and continuity.
Step 2: Uniform continuity does not imply a limit at infinity.
The fact that \( f \) is uniformly continuous does not necessarily imply that \( \lim_{x \to \infty} f(x) \) exists. A function can be uniformly continuous without having a limit as \( x \to \infty \).
Step 3: Uniform continuity does not guarantee uniform continuity of \( g(x) = f(x) \sin x \).
While \( f \) is uniformly continuous, multiplying by \( \sin x \), which oscillates, can cause \( g(x) \) to fail to be uniformly continuous because the oscillations may disrupt the uniformity.
Step 4: Continuity and a finite limit at infinity imply uniform continuity.
If \( f \) is continuous on \( [0, \infty) \) and \( \lim_{x \to \infty} f(x) \) is finite, then \( f \) must be uniformly continuous because the behavior of \( f(x) \) becomes stable as \( x \) grows larger, ensuring the function remains controlled.
Final Answer: \[ \boxed{\text{(D) If } f \text{ is continuous and } \lim_{x \to \infty} f(x) \text{ is finite, then } f \text{ is uniformly continuous}}. \]
The value of \[ \int \sin(\log x) \, dx + \int \cos(\log x) \, dx \] is equal to