Primitive of Reciprocal of p plus q by Cosine of a x/Proof 1

Proof
Let $p^2 > q^2$.

Then, by Sign of Quotient of Factors of Difference of Squares:
 * $\dfrac {p + q} {p - q} > 0$

Thus, let $\dfrac {p + q} {p - q} = d^2$.

Then:

Now let $p^2 < q^2$.

Then, by Sign of Quotient of Factors of Difference of Squares:
 * $\dfrac {p + q} {p - q} < 0$

Thus, let:
 * $-\dfrac {p + q} {p - q} = d^2$

or:
 * $\dfrac {q + p} {q - p} = d^2$

Then: