Which of the following is NOT true for all possible non-zero choices of integers m, n; m ≠ n, or all possible non-zero choices of real numbers p, q; p ≠ q, as applicable?

The correct options are:
1. $\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\sin m\theta \sin n\theta d\theta =0$
2. $\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\sin p\theta \cos q\theta d\mathrm{��}=0$
3. $\underset{\alpha \to \infty }{lim}\frac{1}{2\alpha }\underset{-\alpha }{\overset{\alpha }{\int }}\sin p\theta \sin q\theta d\theta =0$

Step-by-step Analysis of the Statements:

1. Analysis of the first statement:
Consider the integral:
$I=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\sin m\theta \sin n\theta d\theta$
Using the product-to-sum trigonometric identity:
$\sin A\sin B=\frac{1}{2}\left[\cos \right(A-B)-\cos (A+B\left)\right]$
We rewrite the integrand:
$I=\frac{1}{2\pi }\underset{0}{\overset{\pi }{\int }}\left[\cos \right(m-n)\theta -\cos (m+n\left)\theta \right]d\theta$
If we choose non-zero integers such that $m=-n$ (where $m\ne n$), then:
$I=\frac{1}{2\pi }\underset{0}{\overset{\pi }{\int }}\left[\cos \right(2m\theta )-1]d\theta =\frac{1}{2\pi }{[\frac{\sin \left(2m\theta \right)}{2m}-\theta ]}_0^{\pi }=-\frac{1}{2}\ne 0$
Thus, the first statement is NOT true for all possible non-zero choices of integers $m,n$.

2. Analysis of the second statement:
Consider the integral:
$J=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\sin p\theta \cos q\theta d\theta$
For general real numbers $p,q$, the integrand $f\left(\theta \right)=\sin p\theta \cos q\theta$ is an odd function because:
$f(-\mathrm{��})=\sin (-p\theta )\cos (-q\theta )=-\sin p\theta \cos q\theta =-f\left(\theta \right)$
The integration limits are symmetric from $-\pi$ to $\pi$. Therefore, the integral of any odd function over this symmetric interval is identically zero.

3. Analysis of the third statement:
Consider the limit:
$K=\underset{\alpha \to \infty }{lim}\frac{1}{2\alpha }\underset{-\alpha }{\overset{\alpha }{\int }}\sin p\theta \sin q\theta d\theta$
Using the product-to-sum identity:
$K=\underset{\alpha \to \infty }{lim}\frac{1}{4\alpha }$
If we choose real numbers such that $p=-q$ (where $p\ne q$ since both are non-zero), we get:
$K=\underset{\alpha \to \infty }{lim}\frac{1}{4\alpha }$
Evaluating the limits:
$K=\underset{\alpha \to \infty }{lim}\frac{1}{4\alpha }[\frac{\sin \left(2p\alpha \right)}{p}-2\alpha ]=\underset{\alpha \to \infty }{lim}[\frac{\sin \left(2p\alpha \right)}{4p\alpha }-\frac{1}{2}]=-\frac{1}{2}\ne 0$
Thus, the third statement is NOT true for all possible non-zero choices of real numbers $p,q$.