Primitive of Reciprocal of p by Sine of a x plus q by Cosine of a x

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Theorem

$\ds \int \frac {\d x} {p \sin a x + q \cos a x} = \frac 1 {a \sqrt {p^2 + q^2} } \ln \tan \size {\frac {a x + \arctan \dfrac q p} 2} + C$


Proof

Lemma

$\ds \frac 1 2 \map \arctan {\dfrac {-p} q} + \frac \pi 4 = \frac {\arctan \dfrac q p} 2$

$\Box$


\(\ds \int \frac {\d x} {p \sin a x + q \cos a x}\) \(=\) \(\ds \int \frac {\d x} {\sqrt {p^2 + q^2} \map \cos {a x + \arctan \dfrac {-p} q} }\) Multiple of Sine plus Multiple of Cosine
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {p^2 + q^2} } \int \map \sec {a x + \arctan \dfrac {-p} q} \rd x\) Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {p^2 + q^2} } \int \sec z \rd z\) Primitive of Function of $a x + b$: $z = a x + \arctan \dfrac {-p} q$
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {p^2 + q^2} } \ln \tan \size {\frac z 2 + \frac \pi 4} + C\) Primitive of $\sec a x$
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {p^2 + q^2} } \ln \tan \size {\frac 1 2 \paren {a x + \arctan \dfrac {-p} q + \frac \pi 2} } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {p^2 + q^2} } \ln \tan \size {\frac {a x + \arctan \dfrac q p} 2} + C\) Lemma

$\blacksquare$


Also see


Sources

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