# Primitive of Cosine of a x + b

## 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

$\blacksquare$

## Proof 2

Let $u = a x + b$.

Then:

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

Then:

 $\ds \int \map \cos {a x + b} \rd x$ $=$ $\ds \int \dfrac {\cos u} a \rd u$ 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 $u$

$\blacksquare$

## 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.

$\blacksquare$