Math Help - Integrate cos(at)(cos(bt) ... Can't figure it out

Well, I've been looking at this problem for a while, and need some help. I'm sure some smart folks on this forum can help me, perhaps? This is somewhat urgent unfortunately...

{ cos(at)cos(bt) where a and b are positive integers and not equal to zero.

There is a little more to the problem, but I just need to know how to integrate these. I can probably do the rest myself.

Originally Posted by taichu

Well, I've been looking at this problem for a while, and need some help. I'm sure some smart folks on this forum can help me, perhaps? This is somewhat urgent unfortunately...

{ cos(at)cos(bt) where a and b are positive integers and not equal to zero.

There is a little more to the problem, but I just need to know how to integrate these. I can probably do the rest myself.

May I assume your whole integrand is cos(at)*cos(bt) *dt?
Because that last part, the "dt", is very important.

INT.[cos(at)cos(bt)]dt ---------------(1)

As it is, the integrand is the product of two functions. To integrate that, we can do it by parts. INT[u]dv = uv -INT.[v]du. But that won't do here because there is no sin(at) or sin(bt) involved.
So we have to separate/decompose the product into addition or subtraction.

There is this trig identity
cosAcosB = (1/2)[cos(A-B) +cos(A+B)] ---------(i)
that we can use to decompose the original integrand.

cos(at)cos(bt)
= (1/2)[cos(at -bt) +cos(at +bt)]
= (1/2)[cos(t(a-b)) +cos(t(a+b))]

So, (1) becomes
= (1/2)INT.[cos((a-b)t) +cos((a+b)t)]dt -----------(2)

It is easy now.

= (1/2)INT.[cos((a-b)t)]dt +1/2INT.[cos((a+b)t)]dt

= (1/2)INT.[cos((a-b)t)]*[(a-b)/(a-b)]dt +(1/2)INT.[cos((a+bt))]*[(a+b)/(a+b)]dt

= [(1/2)(1/(a-b))]INT.[cos((a-b)t)]*(a-b)dt +[(1/2)(1/(a+b))]INT.[cos((a+b)t)]*(a+b)dt

= [1 / (2(a-b))]sin((a-b)t) +[1 / (2(a+b))]sin((a+b)t) +C

That is it.

ticbol forget to mention one thing, the equation fails when a=b.

In that case the formula is not true.
Rather you have,
INT sin (at) cos (bt) dt = (1/2)*INT 2sin(at)*cos(at) dt=(1/2)*INT sin(2at) dt=-(1/4a) cos(2at)+C

Originally Posted by ThePerfectHacker

ticbol forget to mention one thing, the equation fails when a=b.

In that case the formula is not true.
Rather you have,
INT sin (at) cos (bt) dt = (1/2)*INT 2sin(at)*cos(at) dt=(1/2)*INT sin(2at) dt=-(1/4a) cos(2at)+C

Yeah?

If a=b,
then cos(at) cos(bt) becomes cos^2(at) or cos^(bt).

So what's the use of asking for Integral of cos(at)cos(bt) dt then? )-:

So what's the use of asking for Integral of cos(at)cos(bt) dt then? )-:

There is none.
But as a mathematician you need to consider all cases, even the simpler ones. Otherwise that leads to errors.

Originally Posted by ThePerfectHacker

There is none.
But as a mathematician you need to consider all cases, even the simpler ones. Otherwise that leads to errors.

In this case it is useless. Again, if a=b, .....blah, blah, blah.

Originally Posted by ticbol

In this case it is useless. Again, if a=b, .....blah, blah, blah.

But your solution is wrong unless your write a not = b.

For example, solve the differencial equation,
y'=2xy
Divide by "y",
y'/y=2x
Variables seperable,
ln |y| = x^2 +C
Thus,
y=C*e^(x^2) where C>0.
But that cannot be right!
Because y=0 is also a solution.
The error occured after the division, that case was not considered.
Like here.