Solve 3 Integrals with Fourier Integral - Cyrus

[Cyrus]

Can someone please show me how to solve these three integrals, or proivde a link that does. I can't think of a way to solve it right now and I don't have any free time to do so. Thank you!


$$\int cos(nx)cos(mx)dx$$

$$\int sin(nx)sin(mx)dx$$

$$\int sin(nx)cos(mx) dx$$

The thing that makes it tricky is the different coefficients of m and n in each trig term, so you can't make them cos^2(x) or sin^2(x) or use trig identities. (at least I can't see how). Anywho, thanks once more.


Cyrus.

 

[HallsofIvy]

Trig identities.

sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)
sin(a- b)= sin(a)cos(-b)+ cos(a)sin(-b) but cos(-b)= cos(b) sin(-b)= -sin(b)
= sin(a)cos(b) - cos(a)sin(b)

Adding, sin(a+b)+ sin(a- b)= 2sin(a)cos(b).

To integrate sin(nx)cos(mx) let a= nx, b= mx to write
sin(nx)cos(mx)= (1/2)sin((m+n)x)+ (1/2)sin((m-n)x).

cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)
cos(a- b)= cos(a)cos(b)+ sin(a)sin(b)

Adding, cos(a+b)+ cos(a-b)= 2cos(a)cos(b)
To integrate cos(nx)cos(mx), let a= nx, b= mx to write
cos(nx)cos(mx)= (1/2)cos((m+n)x)+ (1/2)cos((m-n)x)

Subtracting, cos(a+b)- cos(a-b)= 2sin(a)sin(b)
To integrate sin(nx)sin(mx), let a= nx, b= mx to write
sin(nx)sin(mx)= (1/2)cos((m+n)x)- (1/2)cos((m-n)x)

 

[Cyrus]

cos(-b)= cos(b) sin(-b)= -sin(b)

Im not following how this is true, pluging in a value for b and none of them are comming out equal?

Ahhh, I figure your just missing a comma inbetween there. You meant cos(-b) = cos(b) and sin(-b)= - sin(b).

Thats a VERY clever trick to solve it. I would never have spotted it! Thanks!

 

[Cyrus]

A follow up, can you show me how the series is reduced, its been a while since i touched a series, a long while.

$$cos(n \pi) = (-1) ^n$$

My book says $$cos(x) = (-1)^n x^{2n} / (2n!)$$

what happened to the fractional part?

The n is different for each equation, poor notation sorry.

Thanks twice.

Cyrus

 

[quasar987]

Isn't it more like

$$cos(x) = \sum_{n=0}^{\infty}(-1)^n x^{2n} / (2n!)$$.

Your question is strangely worded. What do you mean by reduced?

 

[Cyrus]

oops your right.
My question is how did they arrive at this:

$$cos(n \pi) = (-1) ^n$$

oh, I see now, because its just positive of negative 1 depending on the coefficent. Crap. never mind......I got it now. :-)