# Integration by Substitution

## Contents

## Theorem

Let $\phi$ be a real function which has a derivative on the closed interval $\left[{a \,.\,.\, b}\right]$.

Let $I$ be an open interval which contains the image of $\left[{a \,.\,.\, b}\right]$ under $\phi$.

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

Then:

- $\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \ \mathrm d t = \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u$

and:

- $\displaystyle \int f \left({x}\right) \ \mathrm d x = \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u$

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

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

## Proof for Definite Integrals

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

From Derivative of Composite Function:

- $\dfrac {\mathrm d} {\mathrm d u} F \left({\phi \left({u}\right)}\right) = F' \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) = f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$

Hence $F \left({\phi \left({u}\right)}\right)$ is an antiderivative of $f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$.

Thus:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left[{F \left({\phi \left({u}\right)}\right)}\right]_a^b\) | \(\displaystyle \) | \(\displaystyle \) | Fundamental Theorem of Calculus: Second Part | ||

\((1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle F \left({\phi \left({b}\right)}\right) - F \left({\phi \left({a}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) |

However, also:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \ \mathrm d t\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left[{F \left({t}\right)}\right]_{\phi \left({a}\right)}^{\phi \left({b}\right)}\) | \(\displaystyle \) | \(\displaystyle \) | |||

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u\) | \(\displaystyle \) | \(\displaystyle \) | from $(1)$ |

which was to be proved.

$\blacksquare$

## Proof for Indefinite Integrals

Let $\displaystyle F \left({u}\right) = \int f \left({u}\right) \ \mathrm d u$.

By definition $F \left({u}\right)$ is an antiderivative of $f \left({u}\right)$.

Thus by the Chain Rule:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {\mathrm d} {\mathrm d u} F \left({\phi \left({u}\right)}\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle F' \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\((1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)\) | \(\displaystyle \) | \(\displaystyle \) |

So $F \left({\phi \left({u}\right)}\right)$ is an antiderivative of $f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$.

Therefore:

- $\displaystyle \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u = F \left({\phi \left({u}\right)}\right) = \int f \left({x}\right) \ \mathrm d x$

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

$\blacksquare$

## Notes

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

Its validity as a solution technique stems from the fact that it may be possible to choose $\phi$ such that $f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right)$ (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

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