Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus Root of p squared plus q squared
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Theorem
- $\ds \int \frac {\d x} {p \sin a x + q \cos a x + \sqrt {p^2 + q^2} } = \frac {-1} {a \sqrt {p^2 + q^2} } \map \tan {\frac \pi 4 - \frac {a x + \arctan \frac q p} 2} + C$
Proof
Let $\theta = \arctan \dfrac p q$.
Then by the definitions of sine, cosine and tangent:
\(\text {(1)}: \quad\) | \(\ds \cos \theta\) | \(=\) | \(\ds \frac q {\sqrt {p^2 + q^2} }\) | |||||||||||
\(\ds \sin \theta\) | \(=\) | \(\ds \frac p {\sqrt {p^2 + q^2} }\) |
Now consider:
\(\ds \map \cos {\frac \pi 2 - a x - \theta}\) | \(=\) | \(\ds \map \cos {\frac \pi 2 - a x} \cos \theta + \map \sin {\frac \pi 2 - a x} \sin \theta\) | Cosine of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin a x \cos \theta + \cos a x \sin \theta\) | Sine of Complement equals Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \sin {a x + \theta}\) | Sine of Sum |
Then let $\theta' = \arctan \dfrac q p$.
From Arctangent of Reciprocal equals Arccotangent:
- $\theta' = \arccot \dfrac p q$.
Hence:
\(\ds \map \tan {\frac \pi 2 - \theta}\) | \(=\) | \(\ds \cot \theta\) | Tangent of Complement equals Cotangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \tan \theta'\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \theta\) | \(=\) | \(\ds \frac \pi 2 - \theta'\) |
Thus:
\(\ds \int \frac {\d x} {p \sin a x + q \cos a x + \sqrt {p^2 + q^2} }\) | \(=\) | \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \int \frac {\d x} {\dfrac p {\sqrt {p^2 + q^2} } \sin a x + \dfrac q {\sqrt {p^2 + q^2} } \cos a x + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \int \frac {\d x} {\sin \theta \sin a x + \cos \theta \cos a x + 1}\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \int \frac {\d x} {\map \cos {\theta - a x} + 1}\) | Cosine of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \int \frac {\d x} {\map \cos {\frac \pi 2 - a x - \theta'} + 1}\) | from $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \int \frac {\d x} {2 \map {\cos^2} { {\paren {\frac \pi 2 - a x - \theta'} / 2} } }\) | Square of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2 \sqrt {p^2 + q^2} } \int \map {\sec^2} {\frac \pi 4 - \frac {a x + \theta'} 2} \rd x\) | Definition of Real Secant Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2 \sqrt {p^2 + q^2} } \dfrac {-2} a \map \tan {\frac \pi 4 - \frac {a x + \theta'} 2} + C\) | Primitive of $\sec^2 a x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac {-1} {a \sqrt {p^2 + q^2} } \map \tan {\frac \pi 4 - \frac {a x + \arctan \frac q p} 2} + C\) | substituting for $\theta'$ and simplifying |
$\blacksquare$
Also see
- Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x plus r: the general case for where $r^2 \ne p^2 + q^2$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.422$