Integration by Substitution

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:

However, also:

which was to be proved.

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:

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

Also see

 * Weierstrass Substitution