Primitive of Reciprocal of p plus q by Tangent of a x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\displaystyle \int \frac {\mathrm d x}{p + q \tan a x} = \frac {p x} {p^2 + q^2} + \frac q {a \left({p^2 + q^2}\right)} \ln \left\vert{q \sin a x + p \cos a x}\right\vert + C$


Proof

First, let $\arctan \dfrac p q = \phi$.

Let $z = a x + \phi$.

\(\displaystyle z\) \(=\) \(\displaystyle \sin \left({a x + \phi}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d z} {\mathrm d x}\) \(=\) \(\displaystyle a \cos \left({a x + \phi}\right)\) Derivative of $\sin a x$ etc.
\(\displaystyle \) \(=\) \(\displaystyle a \cos z\)


Then:

\(\displaystyle \int \frac {\mathrm d x} {p + q \tan a x}\) \(=\) \(\displaystyle \int \frac {\mathrm d x} {p + q \dfrac {\sin a x} {\cos a x} }\) Tangent is Sine divided by Cosine
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {\cos a x \ \mathrm d x}{p \cos a x + q \sin a x}\) multiplying top and bottom by $\cos a x$
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {\cos a x \ \mathrm d x} {\sqrt {p^2 + q^2} \sin \left({a x + \phi}\right)}\) Multiple of Sine plus Multiple of Cosine: Sine Form
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt {p^2 + q^2} } \int \frac {\cos a x \ \mathrm d x} {\sin \left({a x + \phi}\right)}\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt {p^2 + q^2} } \left({\frac {\ln \left\vert {\sin \left({a x + \phi}\right)}\right\vert} {a \cos \phi} + \tan \phi \int \frac {\sin a x \ \mathrm d x} {\sin \left({a x + \phi}\right)} + C}\right)\) Primitive of $\dfrac {\cos a x} {\sin \left({a x + \phi}\right)}$



Also see


Sources