integral of cosine product
Author
ive been looking at some different papers and every now and than a difficult integral comes up, most of which is somehow manage.. however i have no idea how to solve integrals of functions like this... i cant remember where i read it...
f(x) = cos(ax)cos(bx)
F(x) = ???
where a and b are some numbers
Integration by parts maybe? I vaguely recall seeing something like that where if you take it out a few steps eventually you can get something to cancel or combine or something. It's been a while since I've done messy integrals by hand, and I'm not at home so I don't have access to my books right now...
The lazy way: go to this page enter cos(ax)cos(bx). Read the result.
Slightly less lazy: go to a table of integral and find a table of trigonometric functions. cos(ax)cos(bx) is almost guaranteed to be on the list.
By hand: I believe this one can be solved with integration by parts.
Slightly less lazy: go to a table of integral and find a table of trigonometric functions. cos(ax)cos(bx) is almost guaranteed to be on the list.
By hand: I believe this one can be solved with integration by parts.
Author
cant se how to simplify this function since it just goes between cos and sin... i cant simplify it since the innerfunctions differ... its not in any table ive looked at
If I remember correctly my highschool math, it's
cos(ax)cos(bx) = (cos((a+b)x)+cos((a-b)x))/2
That's very easy to integrate.
EDIT: They beat me to it.
cos(ax)cos(bx) = (cos((a+b)x)+cos((a-b)x))/2
That's very easy to integrate.
EDIT: They beat me to it.
Author
Quote:Original post by jjd
Remember the identity
cos( ax ) cos( bx ) = ( cos( ( a + b ) x ) + cos( ( a - b ) x ) ) / 2
ooh... that seems to work.. but i cant remeber ever seeing that.. i looked into my highschool books but didnt find it anywhere... could u please explain that? how do u get that equation based on the standard trigonometric equations?
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
cos(a + b) + cos(a - b) = 2cos(a)cos(b)
( cos(a + b) + cos(a - b) ) / 2 = cos(a)cos(b)
cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
cos(a + b) + cos(a - b) = 2cos(a)cos(b)
( cos(a + b) + cos(a - b) ) / 2 = cos(a)cos(b)
Quote:Original post by Dragon_StrikeQuote:Original post by jjd
Remember the identity
cos( ax ) cos( bx ) = ( cos( ( a + b ) x ) + cos( ( a - b ) x ) ) / 2
ooh... that seems to work.. but i cant remeber ever seeing that.. i looked into my highschool books but didnt find it anywhere... could u please explain that? how do u get that equation based on the standard trigonometric equations?
Simple.
It's based off of the trigonometric addition Identities, specifically
cos( a + b ) = cos a cos b - sin a sin b
and
cos( a - b ) = cos a cos b + sin a sin b
Working backwards, we get
( cos( ( a + b ) x ) + cos( ( a - b ) x ) ) / 2 =
( ( cos ax cos bx - sin ax sin bx ) + ( cos ax cos bx + sin ax sin bx ) ) / 2 =
( 2 cos ax cos bx ) / 2 =
cos ax cos bx
This topic is closed to new replies.
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