Primitive of Composite Function/Corollary

Theorem

Let $f$ and $g$ be a real functions which are integrable.

Let the composite function $g \circ f$ also be integrable.

Then:

\(\ds \int \map {\paren {g \circ f} } x \map {f'} x \rd x\)

\(=\)

\(\ds \int \map g u \rd u\)

where $u = \map f x$.

Proof

\(\ds \map F x\)

\(=\)

\(\ds \int \map {\paren {g \circ f} } x \map {f'} x \rd x\)

\(\ds \)

\(=\)

\(\ds \int \map g {\map f x} \map {f'} x \rd x\)

Definition of Composition of Mappings

\(\ds \)

\(=\)

\(\ds \int \map g u \map {f'} x \rd x\)

where $u = \map f x$

\(\ds \leadsto \ \ \)

\(\ds \frac {\d F} {\d x}\)

\(=\)

\(\ds \map g u \map {f'} x\)

Definition of Primitive (Calculus)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d F} {\d u} \frac {\d u} {\d x}\)

\(=\)

\(\ds \map g u \map {f'} x\)

Chain Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds \frac {\d F} {\d u} \frac \d {\d x} \map f x\)

\(=\)

\(\ds \map g u \map {f'} x\)

Definition of $u$

\(\ds \leadsto \ \ \)

\(\ds \frac {\d F} {\d u} \map {f'} x\)

\(=\)

\(\ds \map g u \map {f'} x\)

Definition of Derivative

\(\ds \leadsto \ \ \)

\(\ds \frac {\d F} {\d u}\)

\(=\)

\(\ds \map g u\)

\(\ds \leadsto \ \ \)

\(\ds F\)

\(=\)

\(\ds \int \map g u \rd u\)

Definition of Primitive (Calculus)

$\blacksquare$