Integration by Substitution

From ProofWiki
Jump to navigation Jump to search


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 $I$.


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


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

where $x = \map \phi u$.

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


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

Let $f$ be integrable.

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


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

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 $\map f {\map \phi u} \map {\phi'} u$.


\(\displaystyle \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u\) \(=\) \(\displaystyle \bigintlimits {\map F {\map \phi u} } a b\) 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.


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 $\map f {\map \phi u} \map {\phi'} u$.


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

where $x = \map \phi u$.


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.


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