Question

If $\int (x^3-2x^2+3x-1)cos2xdx=\frac{sin2x}{4}u\left(x\right)+\frac{cos2x}{8}v\left(x\right)+c$, then

• A. $u\left(x\right)=x^3-4x^2+3x$
• B. $u\left(x\right)=2x^3-4x^2+3x$
• C. $v\left(x\right)=3x^2-4x+3$
• D. $v\left(x\right)=k6x^2-8x$

Answer 1

Answer: (B)

$u\left(x\right)=2x^3-4x^2+3x$

$\int (x^2-2x^2+3x-1)\cos 2xdx=\int x^2\cos 2xdx-2\int x^2\cos 2xdx+\int 3x\cos 2xdx-\int \cos 2xdx=\frac{1}{2}{x}^3\sin 2x-\frac{3}{2}\int x^2\sin 2xdx-2\int x^2\cos 2xdx+3\int x\cos 2xd-\int \cos 2xdx$
$=\frac{3}{4}{x}^2\cos 2x+\frac{1}{2}{x}^3\sin 2x+\frac{3}{2}\int x\cos 2xdx-2\int x^2\cos 2xdx-\int \cos 2xdx$
$=x^3\sin x\cos x-x^2\sin 2x+\frac{3}{4}{x}^2\cos 2x\frac{3}{4}x\sin 2x-x\cos 2x+\frac{3}{8}\cos 2x$
$=\frac{1}{4}({2x}^3-{4x}^2+3x)\sin 2x+\frac{1}{8}({12x}^3-{16x}^2+6x)\cos 2x$