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

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

$\blacksquare$