# 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:

$\displaystyle \int \map {g \circ f} x \rd x = \int \frac {\map g u} {\map {f'} x} \rd u$

where $u = \map f x$.

## Proof

 $\displaystyle \map F x$ $=$ $\displaystyle \int \map {g \circ f} x \rd x$ $\displaystyle$ $=$ $\displaystyle \int \map g {\map f x} \rd x$ Definition of Composition of Mappings $\displaystyle$ $=$ $\displaystyle \int \map g u \rd x$ where $u = \map f x$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d F} {\d x}$ $=$ $\displaystyle \map g u$ Definition of Primitive (Calculus) $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d F} {\d u} \frac {\d u} {\d x}$ $=$ $\displaystyle \map g u$ Chain Rule for Derivatives $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d F} {\d u} \frac \d {\d x} \map f x$ $=$ $\displaystyle \map g u$ Definition of $u$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d F} {\d u} \map {f'} x$ $=$ $\displaystyle \map g u$ Definition of Derivative $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d F} {\d u}$ $=$ $\displaystyle \frac {\map g u} {\map {f'} x}$ $\displaystyle \leadsto \ \$ $\displaystyle F$ $=$ $\displaystyle \int \frac {\map g u} {\map {f'} x} \rd u$ Definition of Primitive (Calculus)

$\blacksquare$