Integration by Substitution/Primitive/Proof 2

Theorem

The primitive of $f$ can be evaluated by:

$\ds \int \map f x \rd x = \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $x = \map \phi u$.

Proof

We have been given that $x = \map \phi u$.

Hence:

\(\ds \int \map f x \rd x\)

\(=\)

\(\ds \int \map f x \dfrac {\d x} {\d u} \rd u\)

\(\ds \)

\(=\)

\(\ds \int \map f {\map \phi u} \dfrac {\d u} {\d x} \dfrac {\d x} {\d u} \rd x\)

Primitive of Composite Function

\(\ds \)

\(=\)

\(\ds \int \map f {\map \phi u} \dfrac {\d x} {\d u} \rd u\)

\(\ds \)

\(=\)

\(\ds \int \map f {\map \phi u} \dfrac \d {\d u} {\map \phi u} \rd u\)

$\blacksquare$