# Integration by Substitution

## 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 $I$.

### Primitive

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 $x = \map \phi u$.

### Definite Integral

The definite integral of $f$ from $a$ to $b$ can be evaluated by:

- $\displaystyle \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $x = \map \phi u$.

The technique of solving an integral in this manner is called **integration by substitution**.

## 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.

## 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.