Question

3.
\\( \int x \sin ( 2 x ) d x = \\)
(a) \\( - \frac { x } { 2 } \cos ( 2 x ) + \frac { 1 } { 4 } \sin ( 2 x ) + c \\)
(b) \\( - \frac { x } { 2 } \cos ( 2 x ) - \frac { 1 } { 4 } \sin ( 2 x ) + c \\)
(c) \\( \frac { x } { 2 } \cos ( 2 x ) - \frac { 1 } { 4 } \sin ( 2 x ) + c \\)
(d) \\( - 2 x \cos ( 2 x ) + \sin ( 2 x ) + c \\)
(e) \\( - 2 x \cos ( 2 x ) - 4 \sin ( 2 x ) + c \\)

Explanation:

Step1: Apply integration by parts

Let \(u = x\), \(dv=\sin(2x)dx\). Then \(du = dx\), \(v=-\frac{1}{2}\cos(2x)\)

$$ \int x\sin(2x)dx=uv-\int vdu=-\frac{x}{2}\cos(2x)+\frac{1}{2}\int\cos(2x)dx $$

Step2: Integrate \(\int\cos(2x)dx\)

\(\int\cos(2x)dx=\frac{1}{2}\sin(2x)+C\)

$$ -\frac{x}{2}\cos(2x)+\frac{1}{2}\times\frac{1}{2}\sin(2x)+C=-\frac{x}{2}\cos(2x)+\frac{1}{4}\sin(2x)+C $$

Answer:

A. \(-\frac{x}{2}\cos(2x)+\frac{1}{4}\sin(2x)+C\)