Primitive of Reciprocal of square of p plus q by Sine of a x/Proof 1
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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
| \(\ds \map {\dfrac \d {\d x} } {\dfrac {\cos a x} {p + q \sin a x} }\) | \(=\) | \(\ds \dfrac {\paren {p + q \sin a x} \map {\frac \d {\d x} } {\cos a x} - \cos a x \map {\frac \d {\d x} } {p + q \sin a x} } {\paren {p + q \sin a x}^2}\) | 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.
$\blacksquare$