Question

Evaluate the integral: $$\int [\sin(\log x) + \cos(\log x)]\ dx$$

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

Hint:

You can also verify this instantly using the derivative product rule on the options!
Differentiating option (D) via the product rule:
$$\frac{d}{dx}[x\sin(\log x)] = 1 \cdot \sin(\log x) + x \cdot \cos(\log x) \cdot \frac{1}{x} = \sin(\log x) + \cos(\log x)$$ Since its derivative matches the integrand, option (D) must be the correct answer.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the indefinite integral of the sum of the sine and cosine functions of $\log x$.

Step 2: Key Formula or Approach:
We can solve this efficiently by substituting $\log x = t$. This transforms the integral into a standard exponential form.
The relevant standard integration identity for exponential forms is:
$$\int e^t [f(t) + f'(t)]\ dt = e^t f(t) + c$$

Step 3: Detailed Explanation:
Let the integral be $I = \int [\sin(\log x) + \cos(\log x)]\ dx$.
Substitute $\log x = t \implies x = e^t$.
Differentiating both sides gives:
$$dx = e^t\ dt$$ Now, substitute these expressions back into the integral:
$$I = \int [\sin t + \cos t] e^t\ dt = \int e^t (\sin t + \cos t)\ dt$$ Let's assign $f(t) = \sin t$. Differentiating $f(t)$ gives:
$$f'(t) = \cos t$$ This directly matches our standard integration form $\int e^t [f(t) + f'(t)]\ dt$.
Applying the identity yields:
$$I = e^t \sin t + c$$ Now, substitute back $t = \log x$ and $e^t = x$:
$$I = x\sin(\log x) + c$$

Step 4: Final Answer:
The result of the integration is $x\sin(\log x) + c$, which matches option (D).