Primitive of Reciprocal of p squared plus square of 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 of $\ds \int \frac {\d x} {p^2 + q^2 \sin^2 a x}$ is:
- $\ds \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {p^2 \paren {u^2}^2 + \paren {2 p + 4 q^2} u^2 + p}$
Proof
\(\ds \int \frac {\d x} {p^2 + q^2 \sin^2 a x}\) | \(=\) | \(\ds \frac 1 a \int \frac {\d z} {p^2 + q^2 \sin^2 z}\) | Primitive of Function of Constant Multiple: $z = a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac 1 {p^2 + q^2 \paren {\frac {2 u} {u^2 + 1} }^2} \frac {2 \rd u} {u^2 + 1}\) | Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {2 \rd u} {\paren {u^2 + 1} \frac {p^2 \paren {u^2 + 1}^2 + 4 q^2 u^2} {\paren {u^2 + 1}^2} }\) | common denominator | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 a \int \frac {\paren {u^2 + 1} \rd u} {p^2 \paren {u^2}^2 + \paren {2 p + 4 q^2} u^2 + p}\) | simplifying |
$\blacksquare$