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:

$\displaystyle \int g \circ f \left({x}\right) \ \mathrm d x = \int \frac {g \left({u}\right)}{f' \left({x}\right)} \ \mathrm d u$

where $u = f \left({x}\right)$.


Proof

\(\displaystyle F \left({x}\right)\) \(=\) \(\displaystyle \int g \circ f \left({x}\right) \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int g \left({f \left({x}\right)}\right) \ \mathrm d x\) Definition of Composite Function
\(\displaystyle \) \(=\) \(\displaystyle \int g \left({u}\right) \ \mathrm d x\) where $u = f \left({x}\right)$
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d F} {\mathrm d x}\) \(=\) \(\displaystyle g \left({u}\right)\) Definition of Primitive
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d F} {\mathrm d u} \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle g \left({u}\right)\) Chain Rule for Derivatives
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d F} {\mathrm d u} \frac {\mathrm d} {\mathrm d x} f \left({x}\right)\) \(=\) \(\displaystyle g \left({u}\right)\) Definition of $u$
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d F} {\mathrm d u} f' \left({x}\right)\) \(=\) \(\displaystyle g \left({u}\right)\) Definition of Derivative
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d F} {\mathrm d u}\) \(=\) \(\displaystyle \frac {g \left({u}\right)} {f' \left({x}\right)}\)
\(\displaystyle \implies \ \ \) \(\displaystyle F\) \(=\) \(\displaystyle \int \frac {g \left({u}\right)} {f' \left({x}\right)} \mathrm d u\) Definition of Primitive

$\blacksquare$


Sources