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

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int \frac {\d x} {p \sin a x + q \cos a x + r} = \begin{cases} \displaystyle \frac 2 {a \sqrt {r^2 - p^2 - q^2} } \map \arctan {\frac {p + \paren {r - q} \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } } + C & : p^2 + q^2 < r^2 \\ \displaystyle \frac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {p - \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} {p + \sqrt {p^2 + q^2 - r^2} + \paren {r - q} \tan \dfrac {a x} 2} } + C & : p^2 + q^2 > r^2 \end{cases}$


Proof 1

Let $z = a x + \arctan \dfrac q p$.

Then:

\(\displaystyle \int \frac {\mathrm d x} {p \sin a x + q \cos a x + r}\) \(=\) \(\displaystyle \int \frac {\mathrm d x} {r + \sqrt {p^2 + q^2} \sin \left({a x + \arctan \dfrac q p}\right)}\) Multiple of Sine plus Multiple of Cosine
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \int \frac {\mathrm d z} {r + \sqrt {p^2 + q^2} \sin z}\) Primitive of Function of $a x + b$


Let $d = \sqrt {p^2 + q^2}$.

Then:

\(\displaystyle \) \(\) \(\displaystyle \int \frac {\mathrm d x} {p \sin a x + q \cos a x + r}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \int \frac {\mathrm d z} {r + d \sin z}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \begin{cases} \displaystyle \frac 2 {\sqrt {r^2 - d^2} } \arctan \left({\frac {r \tan \dfrac z 2 + d} {\sqrt {r^2 - d^2} } }\right) + C & : d^2 - r^2 < 0 \\ \displaystyle \frac 1 {\sqrt {d^2 - r^2} } \ln \left\vert{\frac {r \tan \dfrac z 2 + d - \sqrt {r^2 - d^2} } {r \tan \dfrac z 2 + d + \sqrt {r^2 - d^2} } }\right\vert + C & : r^2 - d^2 > 0 \\ \end{cases}\) Primitive of $\dfrac 1 {p + q \sin a x}$
\(\displaystyle \) \(=\) \(\displaystyle \begin{cases} \displaystyle \frac 2 {a \sqrt {r^2 - p^2 - q^2} } \arctan \left({\frac {r \tan \dfrac z 2 + \sqrt {p^2 + q^2} } {\sqrt {r^2 - p^2 - q^2} } }\right) + C & : r^2 > p^2 + q^2 \\ \displaystyle \frac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \left\vert{\frac {r \tan \dfrac z 2 + \sqrt {p^2 + q^2} - \sqrt {r^2 - p^2 - q^2} } {r \tan \dfrac z 2 + \sqrt {p^2 + q^2} + \sqrt {r^2 - p^2 - q^2} } }\right\vert + C & : r^2 < p^2 + q^2 \\ \end{cases}\) substituting for $d$


Then:

\(\displaystyle \tan \frac z 2\) \(=\) \(\displaystyle \tan \left({\frac {a x} 2 + \frac 1 2 \arctan \frac q p}\right)\) substituting for $z$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\tan \dfrac {a x} 2 + \tan \left({\frac 1 2 \arctan \frac q p}\right)} {1 - \tan \dfrac {a x} 2 \tan \left({\frac 1 2 \arctan \frac q p}\right)}\) Tangent of Sum


Now let:

\(\displaystyle y\) \(=\) \(\displaystyle \tan \left({\frac 1 2 \arctan \frac q p}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \arctan y\) \(=\) \(\displaystyle \frac 1 2 \arctan \frac q p\)
\(\displaystyle \implies \ \ \) \(\displaystyle 2 \arctan y\) \(=\) \(\displaystyle \arctan \frac q p\)
\(\displaystyle \implies \ \ \) \(\displaystyle \tan \left({2 \arctan y}\right)\) \(=\) \(\displaystyle \frac q p\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {2 \tan \left({\arctan y}\right)} {1 - \tan^2 \left({\arctan y}\right)}\) \(=\) \(\displaystyle \frac q p\) Double Angle Formula for Tangent
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {2 y} {1 - y^2}\) \(=\) \(\displaystyle \frac q p\) Definition of Arctangent
\(\displaystyle \implies \ \ \) \(\displaystyle q y^2 + 2 p y - q\) \(=\) \(\displaystyle 0\) rearranging
\(\displaystyle \implies \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \frac {-2 p \pm \sqrt{4 p^2 + 4 q ^2} } {2 q}\) Quadratic Formula
\(\displaystyle \) \(=\) \(\displaystyle \frac {-p \pm \sqrt{p^2 + q^2} } q\) simplifying


Hence:

\(\displaystyle \tan \frac z 2\) \(=\) \(\displaystyle \frac {\tan \dfrac {a x} 2 + \dfrac {-p \pm \sqrt{p^2 + q^2} } q} {1 - \tan \dfrac {a x} 2 \left({\dfrac {-p \pm \sqrt{p^2 + q^2} } q}\right)}\) Tangent of Sum
\(\displaystyle \) \(=\) \(\displaystyle \frac {q \tan \dfrac {a x} 2 - p \pm \sqrt{p^2 + q^2} } {q - \tan \dfrac {a x} 2 \left({-p \pm \sqrt{p^2 + q^2} }\right)}\)


Proof 2

Let $z = a x + \arctan \dfrac {-p} q$.

Then:

\(\displaystyle \int \frac {\d x} {p \sin a x + q \cos a x + r}\) \(=\) \(\displaystyle \int \frac {\d x} {r + \sqrt {p^2 + q^2} \map \cos {a x + \arctan \dfrac {-p} q} }\) Multiple of Sine plus Multiple of Cosine
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \int \frac {\d z} {r + \sqrt {p^2 + q^2} \cos z}\) Primitive of Function of $a x + b$


Let $d = \sqrt {p^2 + q^2}$.

Then:

\(\displaystyle \) \(\) \(\displaystyle \int \frac {\d x} {p \sin a x + q \cos a x + r}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \int \frac {\d z} {r + d \cos z}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \begin{cases} \displaystyle \frac 2 {\sqrt {r^2 - d^2} } \map \arctan {\sqrt {\frac {r + d} {r - d} } \tan \dfrac z 2} + C & : r^2 > d^2 \\ \displaystyle \frac 1 {\sqrt {d^2 - r^2} } \ln \size {\frac {\tan \dfrac z 2 + \sqrt {\dfrac {d - r} {d + r} } } {\tan \dfrac z 2 - \sqrt {\dfrac {d - r} {d + r} } } } + C & : r^2 < d^2 \\ \end{cases}\) Primitive of $\dfrac 1 {p + q \cos a x}$
\(\displaystyle \) \(=\) \(\displaystyle \begin{cases} \displaystyle \frac 2 {a \sqrt {r^2 - p^2 - q^2} } \map \arctan {\sqrt {\frac {r + d} {r - d} } \tan \dfrac z 2} + C & : r^2 > p^2 + q^2 \\ \displaystyle \frac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \size {\frac {\tan \dfrac z 2 + \sqrt {\dfrac {d - r} {d + r} } } {\tan \dfrac z 2 - \sqrt {\dfrac {d - r} {d + r} } } } + C & : r^2 < p^2 + q^2 \\ \end{cases}\) partly substituting for $d$


Then:

\(\displaystyle \sqrt {\frac {r + d} {r - d} } \tan \frac z 2\) \(=\) \(\displaystyle \sqrt {\frac {r + \sqrt {p^2 + q^2} } {r - \sqrt {p^2 - q^2} } } \tan \frac z 2\) substituting for $d$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {\paren {r + \sqrt {p^2 + q^2} }^2} {\paren {r - \sqrt {p^2 + q^2} } \paren {r + \sqrt {p^2 + q^2} } } } \tan \frac z 2\) multiplying top and bottom by $r + \sqrt {p^2 + q^2}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {r + \sqrt {p^2 + q^2} } {\sqrt {r^2 - p^2 - q^2} } \tan \frac z 2\) Difference of Two Squares


Then:

\(\displaystyle \tan \frac z 2\) \(=\) \(\displaystyle \map \tan {\frac {a x} 2 + \frac 1 2 \arctan \frac {-p} q}\) substituting for $z$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\tan \dfrac {a x} 2 + \map \tan {\frac 1 2 \arctan \frac {-p} q} } {1 - \tan \dfrac {a x} 2 \map \tan {\frac 1 2 \arctan \frac {-p} q} }\) Tangent of Sum


Also see


Sources