What is ∫

Question

What is

\int \frac{8}{9} ﻿ x \cos (3 x) d x

?

Answer 1

Simplify.

$\int \frac{8 x \cos (3 x)}{9} d x$

Since $\frac{8}{9}$ is constant with respect to $x$ , move $\frac{8}{9}$ out of the integral.

$\frac{8}{9} \int x \cos (3 x) d x$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \cos (3 x)$ .

$\frac{8}{9} (x (\frac{1}{3} \sin (3 x)) - \int \frac{1}{3} \sin (3 x) d x)$

Simplify.

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - \int \frac{\sin (3 x)}{3} d x)$

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - (\frac{1}{3} \int \sin (3 x) d x))$

Let $u = 3 x$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - \frac{1}{3} \int \sin (u) \frac{1}{3} d u)$

Combine $\sin (u)$ and $\frac{1}{3}$ .

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - \frac{1}{3} \int \frac{\sin (u)}{3} d u)$

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - \frac{1}{3} (\frac{1}{3} \int \sin (u) d u))$

Simplify.

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - \frac{1}{9} \int \sin (u) d u)$

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

$\frac{8}{9} (\frac{x \sin (3 x)}{3} - \frac{1}{9} (- \cos (u) + C))$

Rewrite $\frac{8}{9} (\frac{x \sin (3 x)}{3} - \frac{1}{9} (- \cos (u) + C))$ as $\frac{8}{9} (\frac{x \sin (3 x)}{3} + \frac{\cos (u)}{9}) + C$ .

$\frac{8}{9} (\frac{x \sin (3 x)}{3} + \frac{\cos (u)}{9}) + C$

Replace all occurrences of $u$ with $3 x$ .

$\frac{8}{9} (\frac{x \sin (3 x)}{3} + \frac{\cos (3 x)}{9}) + C$

Simplify.

$\frac{8 x \sin (3 x)}{27} + \frac{8 \cos (3 x)}{81} + C$

Reorder terms.

$\frac{8}{27} x \sin (3 x) + \frac{8}{81} \cos (3 x) + C$

Answers (1)