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: Simplify:
\(e^{\log(\sin x)} = \sin x\), so the integral is \(\int x(\sin x + \cos x)\,dx\).

Step 2: Integration by parts:
Take \(u = x\) and \(dv = (\sin x + \cos x)dx\). Then \(v = -\cos x + \sin x\).
\[ \int u\,dv = uv - \int v\,du = x(\sin x - \cos x) - \int(\sin x - \cos x)dx \]

Step 3: Finish:
\(\int(\sin x - \cos x)dx = -\cos x - \sin x\). So
\[ x(\sin x - \cos x) + \cos x + \sin x + c = x(\sin x - \cos x) + (\sin x + \cos x) + c \]

Step 4: Why the other options are wrong.
Options (A) and (C) have \(x(\sin x + \cos x)\) as the first term, which uses the integrand itself instead of its antiderivative. Option (B) has the wrong sign on the last bracket because the integral of \(\sin x - \cos x\) was not subtracted correctly.

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