Primitive of Reciprocal of square of p plus q by Sine of a x

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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 1

\(\ds \map {\dfrac \d {\d x} } {\dfrac {\cos a x} {p + q \sin a x} }\) \(=\) \(\ds \dfrac {\paren {p + q \sin a x} \map {\frac \d {\d x} } {\cos a x} - \cos a x \map {\frac \d {\d x} } {p + q \sin a x} } {\paren {p + q \sin a x}^2}\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {\paren {p + q \sin a x} \paren {-a \sin a x} - \cos a x \paren {a q \cos a x} } {\paren {p + q \sin a x}^2}\) Derivative of Cosine Function, Derivative of Sine Function
\(\ds \) \(=\) \(\ds -a \dfrac {p \sin a x + q \paren {\sin^2 a x + \cos^2 a x} } {\paren {p + q \sin a x}^2}\) simplification
\(\ds \) \(=\) \(\ds -a \dfrac {p \sin a x + q} {\paren {p + q \sin a x}^2}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds -a \dfrac {p q \sin a x + q^2} {q \paren {p + q \sin a x}^2}\)
\(\ds \) \(=\) \(\ds -a \dfrac {p q \sin a x + q^2 + p^2 - p^2} {q \paren {p + q \sin a x}^2}\)
\(\ds \) \(=\) \(\ds -a \dfrac {q^2 - p^2} {q \paren {p + q \sin a x}^2} - a \dfrac {p q \sin a x + p^2} {q \paren {p + q \sin a x}^2}\)
\(\ds \) \(=\) \(\ds a \dfrac {p^2 - q^2} {q \paren {p + q \sin a x}^2} - a \dfrac {p \paren {p + q \sin a x} } {q \paren {p + q \sin a x}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {a \paren {p^2 - q^2} } {q \paren {p + q \sin a x}^2} - \dfrac {a p} q \dfrac 1 {p + q \sin a x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac q {a \paren {p^2 - q^2} } \map {\dfrac \d {\d x} } {\dfrac {\cos a x} {p + q \sin a x} }\) \(=\) \(\ds \dfrac 1 {\paren {p + q \sin a x}^2} - \dfrac p {\paren {p^2 - q^2} } \dfrac 1 {p + q \sin a x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac q {a \paren {p^2 - q^2} } {\dfrac {\cos a x} {p + q \sin a x} }\) \(=\) \(\ds \int \dfrac {\d x} {\paren {p + q \sin a x}^2} - \dfrac p {\paren {p^2 - q^2} } \int \dfrac {\d x} {p + q \sin a x}\)

Hence the result.

$\blacksquare$


Proof 2

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

$\blacksquare$


Also see


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