Primitive of Reciprocal of Cosine of a x by 1 plus Sine of a x/Weierstrass Substitution

Lemma for Primitive of Reciprocal of $\cos a x \paren {1 + \sin a x}$

The Weierstrass Substitution of $\ds \int \frac {\d x} {\cos a x \paren {1 + \sin a x} }$ is:

$\ds \frac 2 a \int \frac {\d u} {\paren {1 - u} \paren {1 + u}^3}$

where $u = \tan \dfrac {a x} 2$.

Proof

\(\ds \int \frac {\d x} {\cos a x \paren {1 + \sin a x} }\)

\(=\)

\(\ds \frac 1 a \int \frac {\d z} {\cos z \paren {1 + \sin z} }\)

Primitive of Function of Constant Multiple: $z = a x$

\(\ds \)

\(=\)

\(\ds \frac 1 a \int \frac {\dfrac {2 \rd u} {1 + u^2} } {\dfrac {1 - u^2} {1 + u^2} \paren {1 + \dfrac {2 u} {1 + u^2} } }\)

Weierstrass Substitution: $u = \tan \dfrac z 2 = \tan \dfrac {a x} 2$

\(\ds \)

\(=\)

\(\ds \frac 2 a \int \frac {\d u} {\paren {1 - u^2} \paren {1 + u^2 + 2 u} }\)

multiplying top and bottom by $1 + u^2$

\(\ds \)

\(=\)

\(\ds \frac 2 a \int \frac {\d u} {\paren {1 - u^2} \paren {1 + u}^2}\)

Square of Sum

\(\ds \)

\(=\)

\(\ds \frac 2 a \int \frac {\d u} {\paren {1 - u} \paren {1 + u} \paren {1 + u}^2}\)

Difference of Two Squares

\(\ds \)

\(=\)

\(\ds \frac 2 a \int \frac {\d u} {\paren {1 - u} \paren {1 + u}^3}\)

simplifying

$\blacksquare$