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]