Primitive of Reciprocal of square of p plus q by Sine of a x/Lemma

Lemma for Primitive of $\dfrac 1 {\paren {p + q \sin a x}^2}$

Let $u = \tan \dfrac \theta 2$.

Then:

$\dfrac 1 {p + q \sin a x} = \dfrac {u^2 + 1} {p u^2 + 2 q u + p}$

$\ds u$

$=$

$\ds \tan \dfrac \theta 2$

$\ds \leadsto \ \ $

$\ds \sin \theta$

$=$

$\ds \dfrac {2 u} {u^2 + 1}$

$\ds \leadsto \ \ $

$\ds \dfrac 1 {p + q \sin a x}$

$=$

$\ds \dfrac 1 {p + q \paren {\dfrac {2 u} {u^2 + 1} } }$

$\ds $

$=$

$\ds \dfrac 1 {\frac {p \paren {u^2 + 1} + 2 q u} {u^2 + 1} }$

$\ds $

$=$

$\ds \dfrac {u^2 + 1} {p u^2 + 2 q u + p}$

$\blacksquare$