Primitive of Reciprocal of square of p plus q by Sine of a x/Weierstrass Substitution
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Lemma for Primitive of Reciprocal of $\paren {p + q \sin a x}^2$
The Weierstrass Substitution for $\ds \int \frac {\d x} {\paren {p + q \sin a x}^2}$ yields:
- $\ds \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}$
where $u = \tan \dfrac {a x} 2$.
Proof
\(\ds \int \frac {\d x} {\paren {p + q \sin a x}^2}\) | \(=\) | \(\ds \frac 1 a \int \frac {\d z} {\paren {p + q \sin z}^2}\) | Primitive of Function of Constant Multiple: $z = a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac 1 {\paren {p + q \frac {2 u} {u^2 + 1} }^2} \frac {2 \rd u} {u^2 + 1}\) | Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {2 \rd u} {\paren {u^2 + 1} \paren {\frac {p \paren {u^2 + 1} + 2 q u} {u^2 + 1} }^2}\) | common denominator | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {\paren {p u^2 + 2 q u + p}^2}\) | simplifying |
$\blacksquare$