Primitive of Composite Function

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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)

$\blacksquare$


Sources