Primitive of Cosine of a x over Sine of a x plus phi

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int \frac {\cos a x \rd x} {\map \sin {a x + \phi} } = \frac {\ln \size {\map \sin {a x + \phi} } } {a \cos \phi} + \tan \phi \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} } + C$


Proof

First note that:

\(\text {(1)}: \quad\) \(\ds \map {\frac \d {\d x} } {\map \sin {a x + \phi} }\) \(=\) \(\ds a \map \cos {a x + \phi}\) Derivative of $\sin a x$ etc.


Then:

\(\ds \int \frac {\cos a x \rd x} {\map \sin {a x + \phi} }\) \(=\) \(\ds \int \frac {\cos a x \cos \phi \rd x} {\cos \phi \map \sin {a x + \phi} }\) multiplying top and bottom by $\cos \phi$
\(\ds \) \(=\) \(\ds \int \frac {\paren {\cos a x \cos \phi - \sin a x \sin \phi + \sin a x \sin \phi} \rd x} {\cos \phi \map \sin {a x + \phi} }\)
\(\ds \) \(=\) \(\ds \int \frac {\paren {\map \cos {a x + \phi} + \sin a x \sin \phi} \rd x} {\cos \phi \map \sin {a x + \phi} }\) Cosine of Sum
\(\ds \) \(=\) \(\ds \frac 1 {\cos \phi} \int \frac {\map \cos {a x + \phi} \rd x} {\map \sin {a x + \phi} } + \frac {\sin \phi} {\cos \phi} \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} }\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {\cos \phi} \ln \size {\map \sin {a x + \phi} } + \frac {\sin \phi} {\cos \phi} \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} }\) Primitive of Function under its Derivative and $(1)$
\(\ds \) \(=\) \(\ds \frac {\ln \size {\map \sin {a x + \phi} } } {a \cos \phi} + \tan \phi \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} } + C\) Tangent is Sine divided by Cosine

$\blacksquare$