Primitive of Cosine of a x + b

From ProofWiki
Jump to navigation Jump to search

Corollary to Primitive of Cosine Function

$\ds \int \map \cos {a x + b} \rd x = \frac {\map \sin {a x + b} } a + C$

where $a$ is a non-zero constant.

Proof 1

\(\ds \int \cos x \rd x\) \(=\) \(\ds \sin x + C\) Primitive of $\cos x$
\(\ds \leadsto \ \ \) \(\ds \int \map \cos {a x + b} \rd x\) \(=\) \(\ds \frac 1 a \paren {\map \sin {a x + b} } + C\) Primitive of Function of $a x + b$
\(\ds \) \(=\) \(\ds \frac {\map \sin {a x + b} } a + C\) simplifying


Proof 2

Let $u = a x + b$.


$\dfrac {\d u} {\d x} = a$


\(\ds \int \map \cos {a x + b} \rd x\) \(=\) \(\ds \int \dfrac {\cos u} a \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \dfrac {\sin u} a\) Primitive of $\cos u$
\(\ds \) \(=\) \(\ds \frac {\map \sin {a x + b} } a + C\) substituting back for $u$


Proof 3

\(\ds \map {\dfrac \d {\d x} } {\frac {\map \sin {a x + b} } a}\) \(=\) \(\ds \dfrac 1 a \map \cos {a x + b} \map {\dfrac \d {\d x} } {a x + b}\) Derivative of Sine Function, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac 1 a \cdot a \map \cos {a x + b}\) Power Rule for Derivatives
\(\ds \) \(=\) \(\ds \cos {a x + b}\) simplifying

The result follows by definition of primitive.