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.

$\blacksquare$