# Integrating cos(2x)*cos(6x)

## Homework Statement

∫cos(2x)cos(6x)dx

## 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!

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Chestermiller
Mentor

## Homework Statement

∫cos(2x)cos(6x)dx

## 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!
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinASinB

Add these two equations together and see what you get.

cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinASinB

Add these two equations together and see what you get.
This turned out to be better than I thought. Much better than the mess above. I'm still kind of frustrated that the long way didn't work :(

arildno
Homework Helper
Gold Member
Dearly Missed
It can perfectly well be that your expressions are equivalent!

To find THAT out, simply use on your own expression the formula sin(a)cos(b)=1/2(sin(a+b)+sin(a-b)), and compare.

If they do NOT equate, you have made some trivial integration error somewhere.

D H
Staff Emeritus
Integration by parts: 1/6[cos(u/3)sin(u)+(1/3)∫sin(u/3)sin(u)du] ... Are these equivalent or have I messed up?
You messed up. You dropped that factor of 1/6.

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
Incorporating that factor of 1/6 into your answer yields ##\frac 3{16} \cos(2x)\sin(6x) - \frac 1{16} \sin(2x)\cos(6x) + C##, and that is the same as the answer key.

• 1 person
You messed up. You dropped that factor of 1/6.

Incorporating that factor of 1/6 into your answer yields ##\frac 3{16} \cos(2x)\sin(6x) - \frac 1{16} \sin(2x)\cos(6x) + C##, and that is the same as the answer key.
lol this usually happens when I do long things like this. Thanks!

Here is what I have done :

First of all I would like to remind you that cos(s)cos(t) = [cos(s+t) + cos(s-t)]/2

∫cos(8x)+cos(4x) / 2 dx = 1/2 ∫cos(8x)+cos(4x) = sin(8x)/16 + sin(4x)/8 + C

Best Regards

Chestermiller
Mentor
Here is what I have done :

First of all I would like to remind you that cos(s)cos(t) = [cos(s+t) + cos(s-t)]/2

∫cos(8x)+cos(4x) / 2 dx = 1/2 ∫cos(8x)+cos(4x) = sin(8x)/16 + sin(4x)/8 + C

Best Regards
Isn't that what I said in post #2?

By the way, you need to resist the temptation to provide the complete solution, since that is a violation of Physics Forums rules.

Chet