How to solve the integral of sin(n*pi*y/a)*sin(n'*pi*y/a)?

[MarkNgo]

I'm trying to solve the integral below, but can't figure out how they got to that answer.

My work is shown below. What I can't understand is how they are getting 0 if n' =/= n, and a/2 if n' = n. Any help?

 

[MarkFL]

I would first look at the case where \(n'=n\):

[MATH]I=\int_0^a \sin^2\left(\frac{n\pi}{a}y\right)\,dy[/MATH]
Let:

[MATH]u=\frac{n\pi}{a}y\implies dy=\frac{n\pi}{a}\,dy[/MATH]
And we have:

[MATH]I=\frac{a}{n\pi}\int_0^{n\pi} \sin^2(u)\,du[/MATH]
Use a double-angle identity for cosine:

[MATH]I=\frac{a}{2n\pi}\int_0^{n\pi} 1-\cos(2u)\,du=\frac{a}{2n\pi}\left[u-\frac{1}{2}\sin(2u)\right]_0^{n\pi}=\frac{a}{2}[/MATH]
Now, let's look at the case where \(n'\ne n\)

Apply a product to sum identity to write:

[MATH]I=\frac{1}{2}\int_0^a \cos\left(\frac{(n-n')\pi}{a}y\right)-\cos\left(\frac{(n+n')\pi}{a}y\right)\,dy[/MATH]
[MATH]I=\frac{a}{2\pi}\left[\frac{1}{n-n'}\sin\left(\frac{(n-n')\pi}{a}y\right)-\frac{1}{n+n'}\sin\left(\frac{(n+n')\pi}{a}y\right)\right]_0^a=0[/MATH]

 

[MarkNgo]

Thank you very much! Is there a way I can get a general equation, where I can check n' =/= n or n' = n?
Or do I have to look at both cases separately every time?
Also is there a way I can mark this thread as solved?

 

[MarkFL]

I have edited the thread to change it into a question. When you create a thread, you have the option to create it either as a thread or a question. So, now you can mark it as solved if you want. You should see a green button at the top.

In my opinion, it is simpler to handle the two cases separately for this problem.