Chapter 7.3, Problem 24E

To determine

To determine the integral ${\displaystyle \int x^3\cos 2xdx}$

Answer to Problem 24E

The value of the integral ${\displaystyle \int x^3\cos 2xdx}$ is $\sin 2x\left[\frac{x^3}{2}-\frac{6x}{8}\right]+\cos 2x\left[\frac{3x^2}{4}-\frac{6}{16}\right]$

Explanation of Solution

Given information: ${\displaystyle \int x^3\cos 2xdx}$

Formula used: ${\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \left(u^{\prime }{\displaystyle \int vdx}\right)}}}dx$

Calculation :

Differentiation of $u(x)$	Signs	Integration of   $v(x)$
$x^3$	(+)	$\cos 2x$
$3x^2$	(-)	$\frac{\sin 2x}{2}$
$6x$	(+)	$-\frac{\cos 2x}{4}$
$6$	(-)	$-\frac{\sin 2x}{8}$
0	(+)	$\frac{\cos 2x}{16}$

  ${\displaystyle \int x^3\cos 2xdx}=(x^3)\left(\frac{\sin 2x}{2}\right)-(3x^2)\left(-\frac{\cos 2x}{4}\right)+(6x)\left(-\frac{\sin 2x}{8}\right)-(6)\left(\frac{\cos 2x}{16}\right)$

Hence, ${\displaystyle \int x^3\cos 2xdx}=\sin 2x\left[\frac{x^3}{2}-\frac{6x}{8}\right]+\cos 2x\left[\frac{3x^2}{4}-\frac{6}{16}\right]$