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 $I$.

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.

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:

However, also:

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:

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 $x = \map \phi u$.

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.

Also see

 * Weierstrass Substitution