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

Theorem

$\ds \int \frac {\rd x} {p + q \cos a x} = \begin {cases}

\dfrac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\sqrt {\dfrac {p - q} {p + q} } \tan \dfrac {a x} 2} + C & : p^2 > q^2 \\ \dfrac 1 {a \sqrt {q^2 - p^2} } \ln \size {\dfrac {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } } + C & : p^2 < q^2 \\ \end {cases}$ for $p \ne q$.

Proof

\(\ds \int \frac {\d x} {p + q \cos a x}\)

\(=\)

\(\ds \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + \dfrac {p + q} {p - q} }\)

Weierstrass Substitution: $u = \tan \dfrac {a x} 2$

Let $p^2 > q^2$.

Then, by Sign of Quotient of Factors of Difference of Squares:

$\dfrac {p + q} {p - q} > 0$

Thus, let $\dfrac {p + q} {p - q} = d^2$.

Then:

\(\ds \int \frac {\d x} {p + q \cos a x}\)

\(=\)

\(\ds \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + d^2}\)

\(\ds \)

\(=\)

\(\ds \frac 2 {a \paren {p - q} } \frac 1 d \arctan \frac u d + C\)

Primitive of $\dfrac 1 {x^2 + a^2}$

\(\ds \)

\(=\)

\(\ds \frac 2 {a \paren {p - q} } \frac 1 {\sqrt {\dfrac {p + q} {p - q} } } \map \arctan {\frac {\tan \dfrac {a x} 2} {\sqrt {\dfrac {p + q} {p - q} } } } + C\)

substituting for $u$ and $d$

\(\ds \)

\(=\)

\(\ds \frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\sqrt {\frac {p - q} {p + q} } \tan \dfrac {a x} 2} + C\)

simplifying

$\Box$

Now let $p^2 < q^2$.

Then, by Sign of Quotient of Factors of Difference of Squares:

$\dfrac {p + q} {p - q} < 0$

Thus, let:

$-\dfrac {p + q} {p - q} = d^2$

or:

$\dfrac {q + p} {q - p} = d^2$

Then:

\(\ds \int \frac {\d x} {p + q \cos a x}\)

\(=\)

\(\ds \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 - d^2}\)

\(\ds \)

\(=\)

\(\ds \frac 2 {a \paren {p - q} } \frac 1 {2 d} \ln \size {\frac {u - d} {u + d} } + C\)

Primitive of $\dfrac 1 {x^2 - a^2}$

\(\ds \)

\(=\)

\(\ds \frac 1 {a \paren {p - q} } \frac 1 {\sqrt {\dfrac {q + p} {q - p} } } \ln \size {\frac {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } } + C\)

substituting for $u$ and $d$

\(\ds \)

\(=\)

\(\ds \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\frac {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } } + C\)

simplifying

$\blacksquare$