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:

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


where $u = \map f x$.


Corollary

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

\(\ds \map F x\) \(=\) \(\ds \int \map {\paren {g \circ f} } x \rd x\)
\(\ds \) \(=\) \(\ds \int \map g {\map f x} \rd x\) 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: $\dfrac {\d x} {\d u} = \dfrac 1 {\d u / \d x}$

$\blacksquare$


Sources

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