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:

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

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.

$\blacksquare$

Proof for Indefinite Integrals

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

As by definition $\displaystyle F \left({u}\right)$ is an antiderivative of $\displaystyle f \left({u}\right)$, the Chain Rule gives:

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

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

Therefore:

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

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.

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 Edition (2005): $\S 4.5, \S 8.4$