Primitive of Reciprocal of square of p plus q by Sine of a x/Proof 2

Theorem

$\ds \int \frac {\d x} {\paren {p + q \sin a x}^2} = \frac {q \cos a x} {a \paren {p^2 - q^2} \paren {p + q \sin a x} } + \frac p {p^2 - q^2} \int \frac {\d x} {p + q \sin a x}$

Proof

First a pair of lemmata:

Lemma

Let $u = \tan \dfrac \theta 2$.

Then:

$\dfrac 1 {p + q \sin a x} = \dfrac {u^2 + 1} {p u^2 + 2 q u + p}$

$\Box$

Weierstrass Substitution

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

$\Box$

\(\ds \int \frac {\d x} {\paren {p + q \sin a x}^2}\)

\(=\)

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

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

\(\ds \)

\(=\)

\(\ds \frac 2 a \int \frac {u^2 \rd u} {\paren {p u^2 + 2 q u + p}^2} + \frac 2 a \int \frac {\d u} {\paren {p u^2 + 2 q u + p}^2}\)

Linear Combination of Primitives

\(\ds \)

\(=\)

\(\ds \frac 2 a \paren {\frac {\paren {4 q^2 - 2 p^2} u + 2 p q} {p \paren {4 p^2 - 4 q^2} \paren {p u^2 + 2 q u + p} } + \frac {2 p} {4 p^2 - 4 q^2} \int \frac {\d u} {p u^2 + 2 q u + p} } + C\)

Primitive of $\dfrac {x^2} {\paren {a x^2 + b x + c}^2}$

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \frac 2 a \paren {\frac {2 p u + 2 q} {\paren {4 p^2 - 4 q^2} \paren {p u^2 + 2 q u + p} } + \frac {2 p} {4 p^2 - 4 q^2} \int \frac {\d u} {p u^2 + 2 q u + p} } + C\)

Primitive of $\dfrac 1 {\paren {a x^2 + b x + c}^2}$

\(\ds \)

\(=\)

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

simplifying ...

\(\ds \)

\(\)

\(\, \ds + \, \)

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

... and creating common denominators

\(\ds \)

\(=\)

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

summing

\(\ds \)

\(=\)

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

simplifying, and strategic rearrangement

\(\ds \)

\(=\)

\(\ds \frac {2 q \paren {q u + p} } {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

Primitive of $\dfrac 1 {p + q \sin a x}$: Weierstrass Substitution

Then we note that:

\(\ds \dfrac {2 q \paren {q u + p} } p - \dfrac {q \paren {p u^2 + 2 q u + p} } p\)

\(=\)

\(\ds \dfrac {2 q^2 u + 2 p q - p q u^2 - 2 q^2 u - p q} p\)

\(\ds \)

\(=\)

\(\ds \dfrac {p q - p q u^2} p\)

\(\ds \)

\(=\)

\(\ds q \paren {1 - u^2}\)

Hence:

\(\ds \int \frac {\d x} {\paren {p + q \sin a x}^2}\)

\(=\)

\(\ds \frac {2 q \paren {q u + p} } {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

from above

\(\ds \)

\(=\)

\(\ds \frac {q \paren {1 - u^2} + \frac {q \paren {p u^2 + 2 q u + p} } p} {a \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

\(\ds \)

\(=\)

\(\ds \frac {q \paren {1 - u^2} } {a \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac {q \paren {p u^2 + 2 q u + p} } {a p \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

\(\ds \)

\(=\)

\(\ds \frac {q \paren {1 - u^2} } {a \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac q {a p \paren {p^2 - q^2} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

\(\ds \)

\(=\)

\(\ds \frac {q \paren {1 - u^2} } {a \paren {p^2 - q^2} \paren {p u^2 + 2 q u + p} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

subsuming $\dfrac q {a p \paren {p^2 - q^2} }$ into the constant

\(\ds \)

\(=\)

\(\ds \frac {q \paren {1 - u^2} } {a \paren {p^2 - q^2} \paren {1 + u^2} \paren {p + q \sin a x} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

Lemma

\(\ds \)

\(=\)

\(\ds \frac {q \cos a x} {a \paren {p^2 - q^2} \paren {p + q \sin a x} } + \frac p {\paren {p^2 - q^2} } \int \frac {\d x} {p + q \sin a x} + C\)

Tangent Half-Angle Substitution for Cosine

$\blacksquare$