# Integration by Substitution

## Contents

## 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 $I$.

Then:

- $\displaystyle \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$

and:

- $\displaystyle \int \map f x \rd x = \int \map f {\map \phi u} \map {\phi'} u \rd u$

where $x = \map \phi u$.

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

### Corollary

Let $f : \R \to \R$ be a real function.

Let $f$ be integrable.

Let $a$, $b$, and $c$ be real numbers.

Then:

- $\displaystyle \int_{a - c}^{b - c} \map f t \rd t = \int_a^b \map f {t - c} \rd t$

## Proof for Definite Integrals

Let $\displaystyle F$ be an antiderivative of $f$.

From Derivative of Composite Function:

- $\dfrac \d {\d u} \map F {\map \phi u} = \map {F'} {\map \phi u} \map {\phi'} u = \map f {\map \phi u} \map {\phi'} u$

Hence $\map F {\map \phi u}$ is an antiderivative of $\map f {\map \phi u} \map {\phi'} u$.

Thus:

\(\displaystyle \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u\) | \(=\) | \(\displaystyle \bigintlimits {\map F {\map \phi u} } a b\) | Fundamental Theorem of Calculus: Second Part | ||||||||||

\((1):\quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle \map F {\map \phi b} - \map F {\map \phi a}\) |

However, also:

\(\displaystyle \int_{\map \phi a}^{\map \phi b} \map f t \rd t\) | \(=\) | \(\displaystyle \bigintlimits {\map F t} {\map \phi a} {\map \phi b}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map F {\map \phi b} - \map F {\map \phi a}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u\) | from $(1)$ |

which was to be proved.

$\blacksquare$

## Proof for Indefinite Integrals

Let $\displaystyle \map F u = \int \map f u \rd u$.

By definition $\map F u$ is an antiderivative of $\map f u$.

Thus by the Chain Rule for Derivatives:

\(\displaystyle \frac \d {\d u} \map F {\map \phi u}\) | \(=\) | \(\displaystyle \map {F'} {\map \phi u} \map {\phi'} u\) | |||||||||||

\((1):\quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle \map f {\map \phi u} \map {\phi'} u\) |

So $\map F {\map \phi u}$ is an antiderivative of $\map f {\map \phi u} \map {\phi'} u$.

Therefore:

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

where $x = \map \phi u$.

$\blacksquare$

## 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.

## Notes

The validity of this as a solution technique stems from the fact that it may be possible to choose $\phi$ such that $\map f {\map \phi u} \map {\phi'} u$ (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.

## Also see

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.22$ - 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards:
*Calculus*(8th ed.): $\S 4.5, \S 8.4$