Primitive of Reciprocal of p squared plus square of q by Sine of a x/Weierstrass Substitution

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

$\blacksquare$