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zoidberg

which trig identity/method should i use for this?

int cos(4x)cos(5x)

ta

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

zoidberg

which trig identity/method should i use for this?

int cos(4x)cos(5x)

ta

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.

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.

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.

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.

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