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 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 |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$: $14.361$