- #1

ThomasMagnus

- 138

- 0

## Homework Statement

∫cos(2x)cos(6x)dx

## Homework Equations

## The Attempt at a Solution

When I do this one, I seem to get a different answer than my book. The book uses a product to summarize formula, but I hate memorizing formulas and want to do this without it. Here's what I did:

Let u=6x then du=6dx

(1/6)∫cos(u/3)cos(u)du

Integration by parts: 1/6[cos(u/3)sin(u)+(1/3)∫sin(u/3)sin(u)du]

Integrating (1/3)sin(u/3)sin(u) gave: (-1/3)sin(u/3)cos(u)+(1/9)∫cos(u/3)cos(u)du

Call ∫cos(u/3)cos(u)du “I”

(I)=cos(u/3)sin(u)-(1/3)sin(u/3)cosu+(I/9)

(8/9)I=cos(u/3)sin(u)-(1/3)sin(u/3)cosu

I=(9/8)cos(u/3)sin(u)-(3/8)sin(u/3)cosu

I= (9/8)cos(2x)sin(6x)-(3/8)sin(2x)cos6x +C = ∫cos(u/3)cos(u)du=∫cos(2x)cos(6x)

The answer key has: (1/8)sin(4x)+(1/16)sin(8x)+ C

Are these equivalent or have I messed up?

Thanks!