Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x minus 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 $(2)$
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {p^2 + q^2} } \int \frac {\d x} {-\map \cos {\pi - \paren {\frac \pi 2 - a x - \theta'} } - 1}\) Cosine of Supplementary Angle
\(\ds \) \(=\) \(\ds \dfrac {-1} {\sqrt {p^2 + q^2} } \int \frac {\d x} {\map \cos {\frac \pi 2 + a x + \theta'} - 1}\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {-1} {\sqrt {p^2 + q^2} } \int \frac {\d x} {2 \map {\cos^2} {\paren {\frac \pi 2 + a x + \theta'} / 2} }\) 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 and simplifying
\(\ds \) \(=\) \(\ds \dfrac {-1} {2 \sqrt {p^2 + q^2} } \dfrac 2 a \map \tan {\frac \pi 4 + \frac {a x + \theta'} 2} + C\) 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


Sources

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