Primitive of Reciprocal of p plus q by Tangent of a x/Proof 1
Jump to navigation
Jump to search
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
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. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |