Evaluate ∫ xsin 2x dx.

Question 82

MediumOriginal question

Evaluate $\int x\sin 2x\,dx$ .

-\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C

Correct answerYour answer

-\frac{x\cos 2x}{2} - \frac{\sin 2x}{4} + C

Correct answerYour answer

\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C

Correct answerYour answer

-x\cos 2x + \frac{\sin 2x}{2} + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

By parts with $u = x$ , $dv = \sin 2x\,dx$ : $v = -\frac{\cos 2x}{2}$ .
$-\frac{x\cos 2x}{2} + \int \frac{\cos 2x}{2}\,dx = -\frac{x\cos 2x}{2} + \frac{\sin 2x}{4} + C$ .

Worth noting

$-\frac{\sin 2x}{4}$ slips on the double minus: $-\int v\,du = +\int \frac{\cos 2x}{2}\,dx$ .

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