Primitive of Reciprocal of square of p plus q by Sine of a x/Lemma
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Lemma for Primitive of $\dfrac 1 {\paren {p + q \sin a x}^2}$
Let $u = \tan \dfrac \theta 2$.
Then:
- $\dfrac 1 {p + q \sin a x} = \dfrac {u^2 + 1} {p u^2 + 2 q u + p}$
Proof
From Tangent Half-Angle Substitution for Sine:
\(\ds u\) | \(=\) | \(\ds \tan \dfrac \theta 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin \theta\) | \(=\) | \(\ds \dfrac {2 u} {u^2 + 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 {p + q \sin a x}\) | \(=\) | \(\ds \dfrac 1 {p + q \paren {\dfrac {2 u} {u^2 + 1} } }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\frac {p \paren {u^2 + 1} + 2 q u} {u^2 + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {u^2 + 1} {p u^2 + 2 q u + p}\) |
$\blacksquare$