How can I integrate cos(3x)cos(2x) without using tables?

[Naeem]

Develop the argument:

Integral cos(3x)cos(2x)dx , without using integral tables.

I know,

cos (A+B) = cosA.cosB-sinA.sinB

Any ideas:

 

[dextercioby]

Okay.

\cos (A+B) =\cos A \cos B -\sin A \sin B

\cos (A-B) =\cos A \cos B +\sin A \sin B

Add the 2 expressions and c what u get...

Daniel.

 

[Naeem]

Yes, I added them, and
I came up with:

cos(A+B) + cos (A-B) = 2cosAcosB
= 2cos(3x)cos(2x) if we let A = 3x and B = 2x
But how to get rid of that extra 2.

if we write 3x = x + 2x, and then write the cos (A+B) of the LHS as:

cos (x+2x) and expand, similarly expand cos (A-B)

Equate the LHS and the RHS, and do some cancellations, but how to get that integral sign... that's another thing to worry about.

 

[whozum]

just put a 1/2 outside the integral and put a 2 inside. Same result.

As for the integral, you ahvent posed an argument. You've just made a statement.

 

[dextercioby]

\cos A \cos B =\frac{1}{2} \left[\cos (A+B) + \cos (A-B) \right]

Agree?

Then set

\left\{\begin{array}{c} A=3x\\B=2x \end{array}\right

And apply the formula.The integrations will be simple,involving almost trivial substitutions.

Daniel.

 

[Naeem]

Ok, did this:

Using cosA.cosB = 1/2[cos(A+B) + cos (A-B)]

A = 3x, B = 2x

cos3x.cos2x = 1/2 [ cos (5x) + cos (x) ]

Integrating both sides now:

Integral ( cos3x.cos2x) = Integral ( 1/2 [ cos (5x) + cos(x) ]

= 1/2 Integral cos 5x + 1/2 Integral cos x
= 1/2 sin5x /5 + 1/2 sinx

= 1/10 ( sin 5x) + 1/2 sinx /2

Integral cos 2x. dx = cos3x = 1/2 [ sin5x/5 +sinx ]

Is the above correct!

 

[Naeem]

I mean:
Integral cos 2x.cos 3x dx = 1/2 [ sin5x/5 +sinx ]

 

[dextercioby]

Don't forget the integration constant.You might consider learning to edit formulas in Latex..

Daniel.