Question

\(\int (e^{log(sinx)}+cosx)x\,dx =\)

• A. \(x(sinx+cosx)+(sinx-cosx)+c\)
• B. \(x(sinx-cosx)+(sinx-cosx)+c\)
• C. \(x(sinx+cosx)+(sinx+cosx)+c\)
• D. \(x(sinx-cosx)+(sinx+cosx)+c\)

Hint:

Simplify e^{log sin x} to sin x and use integration by parts.

Answer 1

The Correct Option is D

Step 1: Check by differentiation:
Differentiate option (D): $(\sin x - \cos x) + x(\cos x + \sin x) + (\cos x - \sin x) = x(\sin x + \cos x)$.

Step 2: Why it matches:
The first and third terms cancel, leaving the integrand $x(\sin x + \cos x)$.

Step 3: Note:
The identity $e^{\log a} = a$ turns the first factor into $\sin x$ for $\sin x > 0$.

Final Answer:
Option (D) is correct. \[ \boxed{x(\sin x-\cos x)+(\sin x+\cos x)+c} \]