Integrating Sin^2: Solving for n=m & m≠n

[Sophia_AI]

Hi guys,


I have a problem with the following integrals( the integral is between [-L,L]) :

1) ∫sin(( nπ/L )x).sin(( mπ/L )x) dx = L when m = n

and :

2) ∫sin(( nπ/L )x).sin(( mπ/L )x) dx= 0 when m ≠ n

I know that in the first equation we have :

∫sin^2(ax) dx = (x/2) - (sin2ax/4a)

but it doesn't work like that, could you please help me with it?

and for the second equation I think when m≠n then m or n should be even and one of the sin will be 0 so the integral will be 0, as in second equation. Am I right about it?

Thanks in advance.
Sophia

 

[clamtrox]

I know that in the first equation we have :

∫sin^2(ax) dx = (x/2) - (sin2ax/4a)

but it doesn't work like that, could you please help me with it?

What do you mean it doesn't work like that? Isn't that completely correct?

and for the second equation I think when m≠n then m or n should be even and one of the sin will be 0 so the integral will be 0, as in second equation. Am I right about it?

It should work for any m and n -- perhaps integrating in parts might help.

 

[epenguin]

Yes, i was going to say, it does work like that.

This exercise is meant to introduce you to a very major mathematical theme.

Apart from calculations, for the case m ≠ n do a sketch of sin(( nπ/L )x) and sin(( mπ/L )x) on same fig. over nm cycles*. e.g. n = 1, m = 2 or n = 2, m = 3. Look at any value of sin(( nπ/L )x) - there are general several. Look what the value of sin(( mπ/L )x) is at that x. For every value some you will find that at some of your x points sin(( mπ/L )x) is exactly minus what it is at other such points. If you think about it you may realize this has to be. For every point. Result: this product, and hence your definite integral averages out as 0. On the other hand when n = m you have a square which is never negative so the integral is not 0 in this one case.

Very useful conclusion you will hear no end of later.*or lcm of n, m