Primitive of Reciprocal of square of p plus q by Cosine of a x
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Theorem
- $\ds \int \frac {\d x} {\paren {p + q \cos a x}^2} = \frac {q \sin a x} {a \paren {q^2 - p^2} \paren {p + q \cos a x} } - \frac p {q^2 - p^2} \int \frac {\d x} {p + q \cos a x}$
Proof
\(\ds \map {\dfrac \d {\d x} } {\dfrac {\sin a x} {p + q \cos a x} }\) | \(=\) | \(\ds \dfrac {\paren {p + q \cos a x} \map {\frac \d {\d x} } {\sin a x} - \sin a x \map {\frac \d {\d x} } {p + q \sin a x} } {\paren {p + q \cos a x}^2}\) | Quotient Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {p + q \cos a x} \paren {a \cos a x} - \sin a x \paren {-a q \sin a x} } {\paren {p + q \cos a x}^2}\) | Derivative of Cosine Function, Derivative of Sine Function | |||||||||||
\(\ds \) | \(=\) | \(\ds a \dfrac {p \cos a x + q \paren {\cos^2 a x + \sin^2 a x} } {\paren {p + q \cos a x}^2}\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds a \dfrac {p \cos a x + q} {\paren {p + q \cos a x}^2}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds a \dfrac {p q \cos a x + q^2} {q \paren {p + q \cos a x}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \dfrac {p q \cos a x + q^2 + p^2 - p^2} {q \paren {p + q \cos a x}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \dfrac {q^2 - p^2} {q \paren {p + q \cos a x}^2} + a \dfrac {p q \cos a x + p^2} {q \paren {p + q \cos a x}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \dfrac {q^2 - p^2} {q \paren {p + q \cos a x}^2} + a \dfrac {p \paren {p + q \cos a x} } {q \paren {p + q \cos a x}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a \paren {q^2 - p^2} } {q \paren {p + q \cos a x}^2} + \dfrac {a p} q \dfrac 1 {p + q \cos a x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac q {a \paren {q^2 - p^2} } \map {\dfrac \d {\d x} } {\dfrac {\sin a x} {p + q \cos a x} }\) | \(=\) | \(\ds \dfrac 1 {\paren {p + q \cos a x}^2} + \dfrac p {\paren {q^2 - p^2} } \dfrac 1 {p + q \cos a x}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac q {a \paren {q^2 - p^2} } {\dfrac {\sin a x} {p + q \cos a x} }\) | \(=\) | \(\ds \int \dfrac {\d x} {\paren {p + q \cos a x}^2} + \dfrac p {\paren {q^2 - p^2} } \int \dfrac {\d x} {p + q \cos a x}\) |
Hence the result.
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\cos a x$: $14.391$