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