Primitive of Reciprocal of p plus q by Cosine of a x/Weierstrass Substitution

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

$\blacksquare$