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 $u = \map f x$.

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

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: General Rules of Integration: $14.6$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 16$: Indefinite Integrals: General Rules of Integration: $16.6.$