Primitive of Composite Function

Theorem

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

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

Then:

 $\ds \int \map {\paren {g \circ f} } x \rd x$ $=$ $\ds \int \map g u \frac {\d x} {\d u} \rd u$ $\ds$ $=$ $\ds \int \frac {\map g u} {\map {f'} x} \rd u$

where $u = \map f x$.

Proof

 $\ds \map F x$ $=$ $\ds \int \map {\paren {g \circ f} } x \rd x$ $\ds$ $=$ $\ds \int \map g {\map f x} \rd x$ Definition of Composition of Mappings $\ds$ $=$ $\ds \int \map g u \rd x$ where $u = \map f x$ $\ds \leadsto \ \$ $\ds \frac {\d F} {\d x}$ $=$ $\ds \map g u$ Definition of Primitive (Calculus) $\ds \leadsto \ \$ $\ds \frac {\d F} {\d u} \frac {\d u} {\d x}$ $=$ $\ds \map g u$ Chain Rule for Derivatives $\ds \leadsto \ \$ $\ds \frac {\d F} {\d u} \frac \d {\d x} \map f x$ $=$ $\ds \map g u$ Definition of $u$ $\ds \leadsto \ \$ $\ds \frac {\d F} {\d u} \map {f'} x$ $=$ $\ds \map g u$ Definition of Derivative $\ds \leadsto \ \$ $\ds \frac {\d F} {\d u}$ $=$ $\ds \frac {\map g u} {\map {f'} x}$ $\ds \leadsto \ \$ $\ds F$ $=$ $\ds \int \frac {\map g u} {\map {f'} x} \rd u$ Definition of Primitive (Calculus) $\ds$ $=$ $\ds \int \map g u \frac {\d x} {\d u} \rd u$ Derivative of Inverse Function: $\dfrac {\d x} {\d u} = \dfrac 1 {\d u / \d x}$

$\blacksquare$