Integration by Substitution

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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 $I$.


Primitive

The primitive of $f$ can be evaluated by:

$\ds \int \map f x \rd x = \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $x = \map \phi u$.


Definite Integral

If $\map \phi a \le \map \phi b$, then the definite integral of $f$ from $a$ to $b$ can be evaluated by:

$\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $t = \map \phi u$.

If $\map \phi a > \map \phi b$, then the definite integral of $f$ from $a$ to $b$ can be evaluated by:

$\ds - \int_{\map \phi b}^{\map \phi a} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$


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


Also known as

Because the most usual substitution variable used is $u$, this method is often referred to Integration by Substitution as $u$-substitution in the source works for introductory-level calculus courses.

Some sources refer to this technique as Change of Variable, but that has a number of different meanings depending upon context.


Proof Technique

The usefulness of the technique of Integration by Substitution stems from the fact that it may be possible to choose $\phi$ such that $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$ (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.


Examples

Primitive of $\paren {2 x + 3} \sqrt {x^2 + 3 x + 2}$

$\ds \int \paren {2 x + 3} \sqrt {x^2 + 3 x + 2} \rd x = \dfrac 2 3 {\paren {\sqrt {x^2 + 3 x + 2} }^3} + C$


Primitive of $\frac {\cos x} {\paren {1 + \sin x}^2}$

$\ds \int \dfrac {\cos x} {\paren {1 + \sin x}^2} \rd x = -\dfrac 1 {1 + \sin x} + C$


Primitive of $\sqrt {1 - x}$

$\ds \int \sqrt {1 - x} \rd x = -\dfrac 2 3 \paren {1 - x}^{3 / 2} + C$


Also see


Sources

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