Primitive of Reciprocal of p plus q by Tangent of a x

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 1

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 $\dfrac {\cos a x} {\map \sin {a x + \phi} }$

This needs considerable tedious hard slog to complete it.
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Proof 2

We have:

$\dfrac \d {\d x} \paren {q \sin a x + p \cos a x} = a q \cos a x - a p \sin a x$

Thus:

$\ds \int \frac {\d x} {p + q \tan a x}$

$=$

$\ds \int \frac {\d x} {p + q \dfrac {\sin a x} {\cos a x} }$

Tangent is Sine divided by Cosine

$\ds $

$=$

$\ds \int \frac {\cos a x \rd x} {p \cos a x + q \sin a x}$

$\ds $

$=$

$\ds \frac 1 {p^2 + q^2} \int \frac {\paren {p^2 + q^2} \cos a x \rd x} {p \cos a x + q \sin a x}$

$\ds $

$=$

$\ds \frac 1 {p^2 + q^2} \int \frac {p^2 \cos a x + p q \sin a x - p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x$

$\ds $

$=$

$\ds \frac 1 {p^2 + q^2} \paren {\int \frac {p^2 \cos a x + p q \sin a x} {p \cos a x + q \sin a x} \rd x + \int \frac {-p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x}$

$\ds $

$=$

$\ds \frac 1 {p^2 + q^2} \paren {\int p \rd x + \frac q a \int \frac {\map \d {p \cos a x + q \sin a x} } {p \cos a x + q \sin a x} }$

$\ds $

$=$

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

Primitive of Constant and Primitive of Reciprocal

$\blacksquare$