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: Check by differentiation:
Differentiate $F = \frac x2[\sin(\log x) - \cos(\log x)]$.

Step 2: Result:
$F' = \frac12[\sin(\log x) - \cos(\log x)] + \frac x2\left[\frac{\cos(\log x)}{x} + \frac{\sin(\log x)}{x}\right] = \sin(\log x)$.
So option A is the antiderivative. Options B and C have wrong signs, and D has a wrong factor of $\frac14$.

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