# 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 $I$.

### 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 as **$u$-substitution** in the source works for introductory-level calculus courses.

Some sources refer to this technique as **change of variable**.

## 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 $\dfrac {\cos x} {\paren {1 + \sin x}^2}$

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

## Also see

## Sources

- 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**change of variable** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**change of variable**