\(\int x\sin x\cos x\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{8}\sin 2x - \frac{x}{4}\cos 2x + C\)
Since \(\sin 2x = 2\sin x\cos x\),
\( \displaystyle \int x\sin x\cos x\,dx = \frac{1}{2}\int x\sin 2x\,dx \)
Now use integration by parts with \(u = x\) and \(dv = \sin 2x\,dx\):
\(du = dx\)
\(v = -\frac{1}{2}\cos 2x\)
\( \displaystyle \int x\sin 2x\,dx = -\frac{x}{2}\cos 2x + \frac{1}{2}\int\cos 2x\,dx \)
\( \displaystyle = -\frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x \)
Multiplying by \(\frac{1}{2}\):
\( \displaystyle \int x\sin x\cos x\,dx = \frac{1}{8}\sin 2x - \frac{x}{4}\cos 2x + C \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{4}\sin 2x - \frac{x}{2}\cos 2x + C\) — Incorrect. This is \(\int x\sin 2x\,dx\); the factor \(\frac{1}{2}\) from \(\sin x\cos x = \frac{1}{2}\sin 2x\) is missing.
Option (3) \(\displaystyle\frac{x}{4}\cos 2x - \frac{1}{8}\sin 2x + C\) — Incorrect. Every sign is reversed; its derivative is \(-x\sin x\cos x\).
Option (4) \(\displaystyle\frac{1}{8}\sin 2x + \frac{x}{4}\cos 2x + C\) — Incorrect. The sign of the \(\cos 2x\) term is wrong; its derivative is \(\frac{1}{2}\cos 2x - x\sin x\cos x\).