Primitive of Reciprocal of p plus q by Tangent of a x/Proof 1

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Theorem

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


Proof

First, let $\arctan \dfrac p q = \phi$.

Let $z = a x + \phi$.

\(\ds z\) \(=\) \(\ds \map \sin {a x + \phi}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds a \map \cos {a x + \phi}\) Derivative of $\sin a x$ etc.
\(\ds \) \(=\) \(\ds a \cos z\)


Then:

\(\ds \int \frac {\d x} {p + q \tan a x}\) \(=\) \(\ds \int \frac {\d x} {p + q \dfrac {\sin a x} {\cos a x} }\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \int \frac {\cos a x \rd x} {p \cos a x + q \sin a x}\) multiplying top and bottom by $\cos a x$
\(\ds \) \(=\) \(\ds \int \frac {\cos a x \rd x} {\sqrt {p^2 + q^2} \map \sin {a x + \phi} }\) Multiple of Sine plus Multiple of Cosine: Sine Form
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {p^2 + q^2} } \int \frac {\cos a x \rd x} {\map \sin {a x + \phi} }\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {p^2 + q^2} } \paren {\frac {\ln \size {\map \sin {a x + \phi} } } {a \cos \phi} + \tan \phi \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} } + C}\) Primitive of $\dfrac {\cos a x} {\map \sin {a x + \phi} }$