Integration by Substitution/Primitive

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 $I$.

The primitive of $f$ can be evaluated by:


 * $\displaystyle \int \map f x \rd x = \int \map f {\map \phi u} \map {\phi'} u \rd u$

where $x = \map \phi u$.

Proof
Let $\map F u = \int \map f u \rd u$.

By definition $\map F u$ is a primitive of $\map f u$.

Thus by the Chain Rule for Derivatives:

So $\map F {\map \phi u}$ is an antiderivative of $\map f {\map \phi u} \map {\phi'} u$.

Therefore:


 * $\displaystyle \int \map f {\map \phi u} \map {\phi'} u \rd u = \map F {\map \phi u} = \int \map f x \rd x$

where $x = \map \phi u$.