Question

$ \int\left[sin \left(log\,x\right)+cos\left(log\,x\right)\right]dx $ is equal to

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

Answer 1

The Correct Option is C

$ \int[\sin (\log x)+\cos (\log x)] d x $ $=\int \frac{d}{d x}\{x \sin (\log x)\} d x $ $=x \sin (\log x)+c$

Answer 2

Given:
\(\int [\sin(\log x) + \cos(\log x)] \, dx\)

To integrate \(\sin(\log x) + \cos(\log x)\), we can recognize that its derivative to x is \(\frac{d}{dx}[x \sin(\log x)]\):

\(\frac{d}{dx}[x \sin(\log x)] = \sin(\log x) + x \cdot \frac{d}{dx}[\sin(\log x)]\)

Now, differentiate \(\sin(\log x)\):
\(\frac{d}{dx}[\sin(\log x)] = \cos(\log x) \cdot \frac{1}{x}\)

So,
\(\frac{d}{dx}[x \sin(\log x)] = \sin(\log x) + x \cos(\log x) \cdot \frac{1}{x} = \sin(\log x) + \cos(\log x)\)

Therefore,
\(\int [\sin(\log x) + \cos(\log x)] \, dx = x \sin(\log x) + C\)

So, the correct option is (C): \(x \sin(\log x) + C\)