# Primitive function/Substitution/Section

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## Theorem

Suppose that ${\displaystyle {}I}$ denotes a real interval, and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

denote a continuous function. Let

${\displaystyle g\colon [a,b]\longrightarrow I}$

be a continuously differentiable function. Then

${\displaystyle {}\int _{a}^{b}f(g(t))g'(t)\,dt=\int _{g(a)}^{g(b)}f(s)\,ds\,}$

holds.

### Proof

Since ${\displaystyle {}f}$ is continuous and ${\displaystyle {}g}$ is continuously differentiable, both integrals exist. Let ${\displaystyle {}F}$ denote a primitive function for ${\displaystyle {}f}$, which exists, due to fact. Because of the chain rule, the composite function

${\displaystyle {}t\mapsto F(g(t))=(F\circ g)(t)\,}$

has the derivative ${\displaystyle {}F'(g(t))g'(t)=f(g(t))g'(t)}$. Therefore,

${\displaystyle {}\int _{a}^{b}f(g(t))g'(t)\,dt=(F\circ g)|_{a}^{b}=F(g(b))-F(g(a))=F|_{g(a)}^{g(b)}=\int _{g(a)}^{g(b)}f(s)\,ds\,.}$
${\displaystyle \Box }$

## Example

Typical examples, where one sees immediately that one can apply substitution, are

${\displaystyle \int g^{n}g',}$

with the primitive function

${\displaystyle {\frac {1}{n+1}}g^{n+1}}$

or

${\displaystyle \int {\frac {g'}{g}},}$

with the primitive function

${\displaystyle \ln g.}$

Often, the indefinite integral is not in a form where one can apply the preceding rule directly. Then the following variant is more appropriate.

## Corollary

Suppose that

${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$

is a continuous function, and let

${\displaystyle \varphi \colon [c,d]\longrightarrow [a,b],s\longmapsto \varphi (s),}$

denote a bijective continuously differentiable function. Then

${\displaystyle {}\int _{a}^{b}f(t)\,dt=\int _{\varphi ^{-1}(a)}^{\varphi ^{-1}(b)}f(\varphi (s))\cdot \varphi '(s)\,ds\,}$

holds.

### Proof

Because of fact, we have

${\displaystyle {}\int _{\varphi ^{-1}(a)}^{\varphi ^{-1}(b)}f(\varphi (s))\varphi '(s)\,ds=\int _{\varphi {\left(\varphi ^{-1}(a)\right)}}^{\varphi {\left(\varphi ^{-1}(b)\right)}}f(t)\,dt=\int _{a}^{b}f(t)\,dt\,.}$
${\displaystyle \Box }$

## Remark

The substitution is applied in the following way: suppose that the integral

${\displaystyle \int _{a}^{b}f(t)\,dt}$

has to be computed. Then one needs an idea that the integral gets simpler by the substitution

${\displaystyle {}t=\varphi (s)\,}$

(taking into account the derivative ${\displaystyle {}\varphi '(s)}$ and that the inverse function ${\displaystyle {}\varphi ^{-1}}$ has to be determined). Setting ${\displaystyle {}c=\varphi ^{-1}(a)}$ and ${\displaystyle {}d=\varphi ^{-1}(b)}$, we have the situation

${\displaystyle [c,d]{\stackrel {\varphi }{\longrightarrow }}[a,b]{\stackrel {f}{\longrightarrow }}\mathbb {R} .}$

In certain cases, some standard substitutions help.

In order to make a substitution, three operations have to be done.

1. Replace ${\displaystyle {}f(t)}$ by ${\displaystyle {}f(\varphi (s))}$.
2. Replace ${\displaystyle {}dt}$ by ${\displaystyle {}\varphi '(s)ds}$.
3. Replace the integration bounds ${\displaystyle {}a}$ and ${\displaystyle {}b}$ by ${\displaystyle {}\varphi ^{-1}(a)}$ and ${\displaystyle {}\varphi ^{-1}(b)}$.

To remember the second step, think of

${\displaystyle {}dt=d\varphi (s)=\varphi '(s)ds\,,}$

which in the framework "differential forms“, has a meaning.

## Example

The upper curve of the unit circle is the set

${\displaystyle {\left\{(x,y)\mid x^{2}+y^{2}=1,\,-1\leq x\leq 1,\,y\geq 0\right\}}.}$

For a given ${\displaystyle {}x}$, ${\displaystyle {}-1\leq x\leq 1}$, there exists exactly one ${\displaystyle {}y}$ fulfilling this condition, namely ${\displaystyle {}y={\sqrt {1-x^{2}}}}$. Hence, the area of the upper half of the unit circle is the area beneath the graph of the function ${\displaystyle {}x\mapsto {\sqrt {1-x^{2}}}}$, above the interval ${\displaystyle {}[-1,1]}$, that is

${\displaystyle \int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx.}$

Applying substitution with

${\displaystyle x=\cos t{\text{ and }}t=\arccos x}$

(where ${\displaystyle {}\cos :[0,\pi ]\rightarrow [-1,1]}$ is bijective, due to fact), we obtain, using example, the identities

{\displaystyle {}{\begin{aligned}\int _{a}^{b}{\sqrt {1-x^{2}}}\,dx&=\int _{\arccos a}^{\arccos b}{\sqrt {1-\cos ^{2}t}}(-\sin t)\,dt\\&=-\int _{\arccos a}^{\arccos b}\sin ^{2}t\,dt\\&={\frac {1}{2}}(\sin t\cos t-t)|_{\arccos a}^{\arccos b}.\end{aligned}}}

In particular, we get that

${\displaystyle {}{\frac {1}{2}}{\left(x\cdot \sin {\left(\arccos x\right)}-\arccos x\right)}={\frac {1}{2}}{\left(x\cdot {\sqrt {1-x^{2}}}-\arccos x\right)}\,}$

is a primitive function for ${\displaystyle {}{\sqrt {1-x^{2}}}}$. Therefore,

{\displaystyle {}{\begin{aligned}\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx&={\frac {1}{2}}{\left(x\cdot {\sqrt {1-x^{2}}}-\arccos x\right)}|_{-1}^{1}\\&={\frac {1}{2}}(-\arccos 1+\arccos(-1))\\&=\pi /2.\end{aligned}}}