# Integration by Substitution/Primitive

## Theorem

Let $\phi$ be a real function which has a derivative on the closed interval $\closedint a b$.

Let $I$ be an open interval which contains the image of $\closedint a b$ under $\phi$.

Let $f$ be a real function which is continuous on $I$.

The primitive of $f$ can be evaluated by:

- $\ds \int \map f x \rd x = \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $x = \map \phi u$.

## Proof 1

Let $\map F x = \ds \int \map f x \rd x$.

Thus by definition $\map F x$ is a primitive of $\map f x$.

\(\ds \map {\frac \d {\d u} } {\map F x}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\map F {\map \phi u} }\) | Definition of $\map \phi u$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac \d {\d x} \map F {\map \phi u} \dfrac \d {\d u} \map \phi u\) | Chain Rule for Derivatives | |||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac \d {\d x} \map F x \dfrac \d {\d u} \map \phi u\) | Definition of $\map \phi u$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f x \dfrac \d {\d u} \map \phi u\) | as $\map F x = \ds \int \map f x \rd x$ |

So $\map F x$ is an antiderivative of $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$.

Therefore:

\(\ds \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u\) | \(=\) | \(\ds \map F x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int \map f x \rd x\) |

where $x = \map \phi u$.

$\blacksquare$

## Proof 2

We have been given that $x = \map \phi u$.

Hence:

\(\ds \int \map f x \rd x\) | \(=\) | \(\ds \int \map f x \dfrac {\d x} {\d u} \rd u\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int \map f {\map \phi u} \dfrac {\d u} {\d x} \dfrac {\d x} {\d u} \rd x\) | Primitive of Composite Function | |||||||||||

\(\ds \) | \(=\) | \(\ds \int \map f {\map \phi u} \dfrac {\d x} {\d u} \rd u\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int \map f {\map \phi u} \dfrac \d {\d u} {\map \phi u} \rd u\) |

$\blacksquare$

## Also presented as

This can also be seen in the form:

- $\ds \int \map f x = \int \map \phi u \dfrac {\d x} {\d u} \rd u$

and:

- $\ds \int \map F {\map \phi x} \rd x = \int \dfrac {\map F u} {\map {f'} x} \rd u$

where $x = \map \phi u$.

## Also known as

Because the most usual substitution variable used is $u$, this method is often referred to as **$u$-substitution** in the source works for introductory-level calculus courses.

Some sources refer to this technique as **change of variable**.

## Proof Technique

The usefulness of the technique of Integration by Substitution stems from the fact that it may be possible to choose $\phi$ such that $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$ (despite its seeming complexity in this context) may be easier to integrate.

If $\phi$ is a trigonometric function, the use of trigonometric identities to simplify the integrand is called **integration by trigonometric substitution** (or simply **trig substitution**).

Care must be taken to address the domain and image of $\phi$.

This consideration frequently arises when inverse trigonometric functions are involved.

## Sources

- 1944: R.P. Gillespie:
*Integration*(2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Integration of Elementary Functions: $\S 8$. Change of Variable - 1960: Margaret M. Gow:
*A Course in Pure Mathematics*... (previous) ... (next): Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text {V}$. - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**change of variable** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**change of variable**

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- 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards:
*Calculus*(8th ed.): $\S 4.5, \S 8.4$