Question

\(\int sin(logx)\,dx =\)

• A. \(\frac{x}{2}[sin(logx)-cos(logx)]+c\)
• B. \(\frac{x}{2}[sin(logx)+cos(logx)]+c\)
• C. \(\frac{x}{2}[cos(logx)-sin(logx)]+c\)
• D. \(\frac{x}{4}[cos(logx)-sin(logx)]+c\)

Hint:

Substitute x = e^t, then use integration by parts twice.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Put \(\log x = t\), so \(x = e^t\) and \(dx = e^t\,dt\). The integral becomes \(\int e^t\sin t\,dt\), a standard form.

Step 2: Key Formula or Approach:
\(\int e^t\sin t\,dt = \frac{e^t}{2}(\sin t - \cos t) + c\), found by applying integration by parts twice.

Step 3: Detailed Explanation:
Let \(I = \int e^t\sin t\,dt\). By parts: \(I = e^t\sin t - \int e^t\cos t\,dt\).
Apply parts again: \(\int e^t\cos t\,dt = e^t\cos t + \int e^t\sin t\,dt = e^t\cos t + I\).
So \(I = e^t\sin t - e^t\cos t - I\), hence \(2I = e^t(\sin t - \cos t)\).
\[ I = \frac{x}{2}[\sin(\log x) - \cos(\log x)] + c \]

Final Answer:
The integral is \(\frac{x}{2}[\sin(\log x) - \cos(\log x)] + c\), option (A). \[ \boxed{\frac{x}{2}[\sin(\log x)-\cos(\log x)]+c} \]