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
- Primitive of $\dfrac 1 {p \sin a x + q \cos a x}$ for the case where $r = 0$
- Primitive of $\dfrac 1 {p \sin a x + q \paren {1 + \cos a x} }$ for the case where $r = q$
- Primitive of $\dfrac 1 {p \sin a x + q \cos a x + \sqrt {p^2 + q^2} }$ for the case where $r = \sqrt {p^2 + q^2}$
- Primitive of $\dfrac 1 {p \sin a x + q \cos a x - \sqrt {p^2 + q^2} }$ for the case where $r = -\sqrt {p^2 + q^2}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.420$