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

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Theorem

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

\ds \frac 2 {a \sqrt {r^2 - p^2 - q^2} } \map \arctan {\frac {p + \paren {r - q} \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } } + C & : p^2 + q^2 < r^2 \\ \ds \frac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {p - \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} {p + \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} } + C & : p^2 + q^2 > r^2 \end{cases}$


Proof

Let $u = \tan \dfrac {a x} 2$.

Then we have:

\(\ds \d x\) \(=\) \(\ds \dfrac {2 \rd u} {a \paren {1 + u^2} }\)
\(\ds \sin a x\) \(=\) \(\ds \dfrac {2 u} {1 + u^2}\)
\(\ds \cos a x\) \(=\) \(\ds \dfrac {1 - u^2} {1 + u^2}\)

Hence:

\(\ds \int \frac {\d x} {p \sin a x + q \cos a x + r}\) \(=\) \(\ds \int \dfrac {2 \rd u} {a \paren {1 + u^2} } \frac 1 {p \dfrac {2 u} {1 + u^2} + q \dfrac {1 - u^2} {1 + u^2} + r}\) Weierstrass Substitution
\(\ds \) \(=\) \(\ds \dfrac 2 a \int \frac {\d u} {2 p u + q \paren {1 - u^2} + r \paren {1 + q^2} }\)
\(\ds \) \(=\) \(\ds \dfrac 2 a \int \frac {\d u} {\paren {r - q} u^2 + 2 p u + \paren {r + q} }\)
\(\ds \) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \int \frac {\d u} {u^2 + \dfrac {2 p} {r - q} u + \dfrac {r + q} {r - q} }\)
\(\ds \) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \int \frac {\d u} {\paren {u + \dfrac p {r - q} }^2 - \dfrac {p^2} {\paren {r - q}^2} + \dfrac {r + q} {r - q} }\) completing the square
\(\ds \) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \int \frac {\d u} {\paren {u + \dfrac p {r - q} }^2 + \dfrac {r^2 - \paren {p^2 + q^2} } {\paren {r - q}^2} }\) rearranging
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \int \frac {\d z} {z^2 + \dfrac {r^2 - \paren {p^2 + q^2} } {\paren {r - q}^2} }\) Integration by Substitution: substituting $z = u + \dfrac p {r - q}$: $\d z = \d u$


There are $2$ cases: $r^2 > p^2 + q^2$ and $r^2 < p^2 + q^2$.


First suppose $r^2 > p^2 + q^2$.

Then:

$r^2 - \paren {p^2 + q^2} > 0$

and so $(1)$ may be written as:

\(\ds \int \frac {\d x} {p \sin a x + q \cos a x + r}\) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \int \frac {\d z} {z^2 + \paren {\dfrac {\sqrt {r^2 - p^2 - q^2} } {r - q} } }\)
\(\ds \) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \dfrac {r - q} {\sqrt {r^2 - p^2 - q^2} } \map \arctan {\dfrac {\paren {r - q} z} {\sqrt {r^2 - p^2 - q^2} } } + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \dfrac 2 {a \sqrt {r^2 - p^2 - q^2} } \map \arctan {\dfrac {\paren {r - q} \paren {u + \dfrac p {r - q} } } {\sqrt {r^2 - p^2 - q^2} } } + C\) substituting back for $z$
\(\ds \) \(=\) \(\ds \dfrac 2 {a \sqrt {r^2 - p^2 - q^2} } \map \arctan {\dfrac {p + \paren {r - q} \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } } + C\) simplifying and substituting back for $u$


Next suppose $r^2 < p^2 + q^2$.

Then:

$r^2 - \paren {p^2 + q^2} < 0$

and so $(1)$ may be written as:

\(\ds \int \frac {\d x} {p \sin a x + q \cos a x + r}\) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \int \frac {\d z} {z^2 - \paren {\dfrac {\sqrt {p^2 + q^2 - r^2} } {r - q} } }\)
\(\ds \) \(=\) \(\ds \dfrac 2 {a \paren {r - q} } \dfrac {r - q} {2 \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {z - \dfrac {\sqrt {p^2 + q^2 - r^2} } {r - q} } {z + \dfrac {\sqrt {p^2 + q^2 - r^2} } {r - q} } } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \dfrac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {\paren {r - q} \paren {u + \dfrac p {r - q} } - \sqrt {p^2 + q^2 - r^2} } {\paren {r - q} \paren {u + \dfrac p {r - q} } + \sqrt {p^2 + q^2 - r^2} } } + C\) substituting back for $z$
\(\ds \) \(=\) \(\ds \dfrac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {p - \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} {p + \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} } + C\) simplifying and substituting back for $u$

$\blacksquare$


Also see


Sources

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