# Homework Help: Integeal of Sin^2

1. Mar 28, 2012

### 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)

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?

Sophia
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 28, 2012

### clamtrox

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

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

3. Mar 28, 2012

### 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 realise 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

Last edited: Mar 28, 2012