The Student Room Group
Reply 1
zoidberg
which trig identity/method should i use for this?

int cos(4x)cos(5x)

ta


I think you use the sum to product (or is it product to sum??) formula - then take it from ther!!!

I don't know them off the top of my head so i can't tell you which one to use!!!

Hope this helps :smile:
Reply 2
zoidberg
which trig identity/method should i use for this?

int cos(4x)cos(5x)

ta


Parts or the sum and product formulae. :wink:
Reply 3
zoidberg
which trig identity/method should i use for this?

int cos(4x)cos(5x)

ta


Parts:
u = cos4x
du/dx = -4sin4x
dv/dx = cos5x
v = -sin5x/5

I = INT cos4xcos5x = -cos4x(sin5x/5) - INT (-sin5x/5)(-4sin4x)
= -1/5(cos4x)(sin5x) - 4/5 INT sin5xsin4x

Parts again:
u = sin5x
du/dx = 5cos5x
dv/dx = sin4x
v = -cos4x/4

INT sin5xsin4x = -(sin5x)(cos4x/4) - INT (-cos4x/4)(5cos5x)
= -1/4(sin5x)(cos4x) + 5/4 INT cos4xcos5x

Note that the bold is equal to I, substitute for I and make I the subject.
Reply 4
The easiest way to solve anything in this form is to use the identity cos(A+B)+cos(A-B).
Reply 5
I went the long way round then. No formula book to hand at the moment is my lame excuse.
Reply 6
Ralfskini
The easiest way to solve anything in this form is to use the identity cos(A+B)+cos(A-B).


sorry, could you just do a quick solution using that identity? was never actually taught this in p2 and kinda hoped i wouldnt need these identities for p3 :]
Reply 7
zoidberg
sorry, could you just do a quick solution using that identity? was never actually taught this in p2 and kinda hoped i wouldnt need these identities for p3 :]


cosA+cosB = 2cos(A+B/2)cos(A-B/2)

Equate to cos4xcos5x and you get

4x = A+B/2, 5x = A-B/2 and cos4xcos5x = 1/2(cosA+cosB)

Solve for A and B, substitute and integrate.
Reply 8
ZJuwelH
cosA+cosB = 2cos(A+B/2)cos(A-B/2)

Equate to cos4xcos5x and you get

4x = A+B/2, 5x = A-B/2 and cos4xcos5x = 1/2(cosA+cosB)

Solve for A and B, substitute and integrate.


To complete:

8x = A+B, 10x = A-B
A=9x
B=-x

INT cos4xcos5x = INT 1/2 cos9x+cos(-x) = 1/2 INT cos9x+cos(-x)
= 1/2 (-sin9x)/9 + sinx + c

I think.
Reply 9
Using 2cos(A) cos(B) = cos(A + B) + cos(A - B),

(int) cos(4x)cos(5x) dx
= (1/2) (int) [cos(9x) + cos(x)] dx
= (1/18) sin(9x) + (1/2) sin(x) + c.
Reply 10
Will product-to-sum and sum-to-product formulae be provided in the exam?
Reply 11
They're in the AQA formula book at least (which is given in the exam). Not sure about the other specifications.
Reply 12
ZJuwelH
They're in the AQA formula book at least (which is given in the exam). Not sure about the other specifications.


They're in the Edexcel formula book as well.
Reply 13
Nylex
They're in the Edexcel formula book as well.


Is it this?
Reply 14
keisiuho
Is it this?

yep
Reply 15
Jonny W
Using 2cos(A) cos(B) = cos(A + B) + cos(A - B),

(int) cos(4x)cos(5x) dx
= (1/2) (int) [cos(9x) + cos(x)] dx
= (1/18) sin(9x) + (1/2) sin(x) + c.


this is the answer i got, slightly different to ZJuwelH's, thanks everyone