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

  • #1
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
20,145
4,214

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
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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
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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
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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.
 
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  • #6
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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
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
20,145
4,214
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
 

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