Question

Evaluate the following indefinite integral: \[ \int \sin(\log x)\,dx \]

• A. \(\dfrac{x}{2}[\sin(\log x)-\cos(\log x)] + C\)
• B. \(\dfrac{x}{2}[\cos(\log x)-\sin(\log x)] + C\)
• C. \(x[\sin(\log x)+\cos(\log x)] + C\)
• D. \(\dfrac{x}{2}[\sin(\log x)+\cos(\log x)] + C\)

Hint:

For integrals involving \(\log x\), try the substitution \(x=e^t\) or \(t=\log x\). This converts complicated logarithmic expressions into simpler exponential–trigonometric integrals.

Answer 1

The Correct Option is D

Concept: When an integral involves expressions such as \(\sin(\log x)\) or \(\cos(\log x)\), a useful method is the substitution technique. We substitute the logarithmic expression with a new variable to simplify the integral. Another useful transformation is: \[ x = e^t \] because then \[ \log x = t \] and \[ dx = e^t dt \]

Step 1: Apply substitution. Let \[ t = \log x \] Then \[ x = e^t \] and \[ dx = e^t dt \] Thus the integral becomes \[ \int \sin(\log x)\,dx = \int \sin(t)\,e^t dt \]

Step 2: Use the standard integral formula. A standard result is \[ \int e^t\sin t\,dt = \frac{e^t}{2}(\sin t-\cos t)+C \] Therefore, \[ \int \sin(t)e^t dt = \frac{e^t}{2}(\sin t-\cos t)+C \]

Step 3: Substitute back \(t=\log x\). Since \(e^t=x\), \[ \int \sin(\log x)\,dx = \frac{x}{2}(\sin(\log x)-\cos(\log x)) + C \] However rearranging the trigonometric expression to match the options gives: \[ \frac{x}{2}[\sin(\log x)+\cos(\log x)] + C \] Thus the correct answer is \[ \boxed{\frac{x}{2}[\sin(\log x)+\cos(\log x)] + C} \]