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:
 * $\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \ \mathrm d t = \int_a^b f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) \ \mathrm d u$

and:


 * $\displaystyle \int f \left({x}\right) \ \mathrm d x = \int f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) \ \mathrm d u$

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 $\displaystyle F \left({x}\right) = \int_{\phi \left({a}\right)}^x f \left({t}\right) \ \mathrm d t$.

From Derivative of a Composite Function and the first part of the Fundamental Theorem of Calculus:


 * $\displaystyle \frac {\mathrm d} {\mathrm d u} F \left({\phi \left({u}\right)}\right) = F^{\prime} \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) = f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right)$.

Thus from the second part of the Fundamental Theorem of Calculus:


 * $\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) \ \mathrm d u = \left[{F \left({\phi \left({u}\right)}\right)}\right]_a^b$

which is what we wanted to prove.