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 option that is NOT true for all possible non-zero choices of real numbers p, q (p ≠ q) is:

$\frac{1}{2\pi }\underset{-\pi /2}{\overset{\pi /2}{\int }}\sin p\theta \sin q\theta d\theta =0$

Step-by-Step Analysis of Each Option:

1. Analyzing the First Option:
$\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}\sin m\theta \sin n\theta d\theta =0$
For distinct non-zero integers m and n (m ≠ n), using the trigonometric identity:
$\sin m\theta \sin n\theta =\frac{1}{2}[\cos (m-n)\theta -\cos (m+n)\theta ]$
Integrating over [0, π]:
$\underset{0}{\overset{\pi }{\int }}\sin m\theta \sin n\theta d\theta =\frac{1}{2}{[\frac{\sin (m-n)\theta }{m-n}-\frac{\sin (m+n)\theta }{m+n}]}_0^{\pi }$
Since m and n are integers, sin((m - n)π) = 0 and sin((m + n)π) = 0. Thus, this integral equals 0 for all distinct non-zero integers. Therefore, this statement is TRUE.

2. Analyzing the Second Option:
$\frac{1}{2\pi }\underset{-\pi /2}{\overset{\pi /2}{\int }}\sin p\theta \sin q\theta d\theta$
Here, p and q are arbitrary real numbers (p ≠ q). Using the product identity:
$\underset{-\pi /2}{\overset{\pi /2}{\int }}\sin p\theta \sin q\theta d\theta =\frac{1}{2}\underset{-\pi /2}{\overset{\pi /2}{\int }}[\cos (p-q)\theta -\cos (p+q)\theta ]d\theta$
Evaluating the integral:
$=\frac{\sin ((p-q)\pi /2)}{p-q}-\frac{\sin ((p+q)\pi /2)}{p+q}$
For arbitrary real numbers p and q (for example, taking p = 2 and q = 4):
$\sin ((2-4)\pi /2)=\sin (-\pi )=0$
$\sin ((2+4)\pi /2)=\sin (3\pi )=0$
However, if we pick p = 1 and q = 3:
$\sin (-\pi )=0\text{and}\sin (2\pi )=0$
If we pick p = 2 and q = 1:
$\sin (\pi /2)=1\text{and}\sin (3\pi /2)=-1$
Thus, the integral evaluates to:
$\frac{1}{1}-\frac{-1}{3}=1+\frac{1}{3}=\frac{4}{3}\ne 0$
Hence, this integral is NOT zero for all possible non-zero real choices of p and q (p ≠ q). Therefore, this statement is NOT TRUE.

3. Analyzing the Third Option:
$\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\sin p\theta \cos q\theta d\theta =0$
Here, sin(pθ) is an odd function of θ and cos(qθ) is an even function of θ. The product of an odd function and an even function is always an odd function.
The integral of any odd function over a symmetric interval [-π, π] is strictly equal to 0 for all choices of real numbers p and q. Therefore, this statement is TRUE.

4. Analyzing the Fourth Option:
$\underset{\alpha \to \infty }{lim}\frac{1}{2\alpha }\underset{-\alpha }{\overset{\alpha }{\int }}\sin p\theta \sin q\theta d\theta =0$
Evaluating the inner integral:

Since sine values are bounded between -1 and 1, the numerator remains bounded as α → ∞. Dividing by 2α and taking the limit gives 0 for any distinct real numbers p ≠ q. Therefore, this statement is TRUE.

Conclusion:
The statement in Option 2 is not true for all non-zero real choices of p and q.