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

  • #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!
 

Answers and Replies

  • #2
22,071
5,032

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!

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

Add these two equations together and see what you get.
 
  • #3
ThomasMagnus
138
0
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 :(
 
  • #4
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
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 your answer deviates from the answer key's by merely a constant of integration, your original expression was correct as well.

If they do NOT equate, you have made some trivial integration error somewhere.
 
  • #5
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
687
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.
 
  • #6
ThomasMagnus
138
0
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!
 
  • #7
Pixardaki_Lamba
1
0
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
 
  • #8
22,071
5,032
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
 

Suggested for: Integrating cos(2x)*cos(6x)

  • Last Post
Replies
5
Views
3K
Replies
3
Views
1K
  • Last Post
Replies
6
Views
4K
Replies
4
Views
17K
Replies
8
Views
28K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
86K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
4K
Top