# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 20

We discuss now the main rules to find a primitive function, and to compute definite integrals. They rest on rules for derivation.

Integration by parts

## Theorem

Let

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

denote continuously differentiable functions. Then

${\displaystyle {}\int _{a}^{b}f(t)g'(t)\,dt=fg|_{a}^{b}-\int _{a}^{b}f'(t)g(t)\,dt\,.}$

### Proof

Due to the product rule the function ${\displaystyle {}fg}$ is a primitive function for ${\displaystyle {}fg'+f'g}$. Therefore,

${\displaystyle {}\int _{a}^{b}f(t)g'(t)\,dt+\int _{a}^{b}f'(t)g(t)\,dt=\int _{a}^{b}{\left(fg'+f'g\right)}(t)\,dt=fg|_{a}^{b}\,.}$
${\displaystyle \Box }$

In using integration by parts, two things are to be considered. Firstly, the function to be integrated is usually not in the form ${\displaystyle {}fg'}$, but just as a product ${\displaystyle {}uv}$ (if there is no product, then this rule will probably not help, however, sometimes the trivial product ${\displaystyle {}1u}$ might help). Then for one factor, we have to find a primitive function, and we have to differentiate the other factor. If ${\displaystyle {}V}$ is a primitive function of ${\displaystyle {}v}$, then the formula reads

${\displaystyle {}\int uv=uV-\int u'V\,.}$

Secondly, integration by parts only helps when the integral on the right, i.e. ${\displaystyle {}\int _{a}^{b}f'(t)g(t)\,dt}$, can be integrated.

## Example

We determine a primitive function for the natural logarithm ${\displaystyle {}\ln x}$, with integration by parts We write ${\displaystyle {}\ln x=1\cdot \ln x}$, and we integrate the constant function ${\displaystyle {}1}$, and we differentiate the logarithm. Then

${\displaystyle {}\int _{a}^{b}\ln x\,dx=(x\cdot \ln x)|_{a}^{b}-\int _{a}^{b}x\cdot {\frac {1}{x}}\,dx=(x\cdot \ln x)|_{a}^{b}-\int _{a}^{b}1\,dx=(x\cdot \ln x)|_{a}^{b}-x|_{a}^{b}\,.}$

So a primitive function is ${\displaystyle {}x\cdot \ln x-x}$.

## Example

A primitive function for the sine function ${\displaystyle {}\sin x}$ is ${\displaystyle {}-\cos x}$. In order to find a primitive function for ${\displaystyle {}\sin ^{n}x}$, we use integration by parts to get a recursive relation to a power with a smaller exponent. To make this more precise, we work over an interval, the primitive function shall start at ${\displaystyle {}0}$ and have the value ${\displaystyle {}0}$ there. For ${\displaystyle {}n\geq 2}$, with integration by parts we get

{\displaystyle {}{\begin{aligned}\int _{0}^{x}\sin ^{n}t\,dt&=\int _{0}^{x}\sin ^{n-2}t\cdot \sin ^{2}t\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\cdot {\left(1-\cos ^{2}t\right)}\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\,dt-\int _{0}^{x}{\left(\sin ^{n-2}t\cos t\right)}\cos t\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\,dt-{\frac {\sin ^{n-1}t}{n-1}}\cos t|_{0}^{x}-{\frac {1}{n-1}}{\left(\int _{0}^{x}\sin ^{n}t\,dt\right)}.\end{aligned}}}

Multiplication with ${\displaystyle {}n-1}$ and rearranging yields

${\displaystyle {}n\int _{0}^{x}\sin ^{n}t\,dt=(n-1)\int _{0}^{x}\sin ^{n-2}t\,dt-\sin ^{n-1}x\cos x\,.}$

In particular, for ${\displaystyle {}n=2}$, we have

${\displaystyle {}\int _{0}^{x}\sin ^{2}t\,dt={\frac {1}{2}}{\left(x-\sin x\cos x\right)}\,.}$

Integration of inverse function

## Theorem

Let ${\displaystyle {}f\colon [a,b]\rightarrow [c,d]}$ denote a bijective differentiable function, and let ${\displaystyle {}F}$ denote a primitive function for ${\displaystyle {}f}$. Then

${\displaystyle {}G(y):=yf^{-1}(y)-F{\left(f^{-1}(y)\right)}\,}$

is a primitive function for the inverse function ${\displaystyle {}f^{-1}}$.

### Proof

Differentiating, using Lemma 14.7 and Theorem 14.8 , yields

{\displaystyle {}{\begin{aligned}{\left(yf^{-1}(y)-F{\left(f^{-1}(y)\right)}\right)}'&=f^{-1}(y)+y{\frac {1}{f'(f^{-1}(y))}}-f{\left(f^{-1}(y)\right)}{\frac {1}{f'{\left(f^{-1}(y)\right)}}}\\&=f^{-1}(y).\end{aligned}}}
${\displaystyle \Box }$

For this statement, there exists also an easy geometric explanation. When ${\displaystyle {}f\colon [a,b]\rightarrow \mathbb {R} _{+}}$ is a strictly increasing continuous function (and therefore induces a bijection between ${\displaystyle {}[a,b]}$ and ${\displaystyle {}[f(a),f(b)]}$), then the following relation between the areas holds:

${\displaystyle {}\int _{a}^{b}f(s)\,ds+\int _{f(a)}^{f(b)}f^{-1}(t)\,dt=bf(b)-af(a)\,,}$

or, equivalently,

${\displaystyle {}\int _{f(a)}^{f(b)}f^{-1}(t)\,dt=bf(b)-af(a)-\int _{a}^{b}f(s)\,ds\,.}$

For the primitive function ${\displaystyle {}G}$ of ${\displaystyle {}f^{-1}}$ with starting point ${\displaystyle {}f(a)}$, we have, if ${\displaystyle {}F}$ denotes a primitive function for ${\displaystyle {}f}$, the relation

{\displaystyle {}{\begin{aligned}G(y)&=\int _{f(a)}^{y}f^{-1}(t)\,dt\\&=\int _{f(a)}^{f(f^{-1}(y))}f^{-1}(t)\,dt\\&=f^{-1}(y)f(f^{-1}(y))-af(a)-\int _{a}^{f^{-1}(y)}f(s)\,ds\\&=yf^{-1}(y)-af(a)-F(f^{-1}(y))+F(a)\\&=yf^{-1}(y)-F(f^{-1}(y))-af(a)+F(a),\end{aligned}}}

where ${\displaystyle {}-af(a)+F(a)}$ is a constant of integration.

## Example

We compute a primitive function for ${\displaystyle {}\arctan x}$, using Theorem 20.4 . A primitive function of tangent is

${\displaystyle {}\int _{}^{}\tan t\,dt=-\ln(\cos x)\,.}$

Hence,

${\displaystyle x\cdot \arctan x+\ln(\cos(\arctan x))}$

is a primitive function for ${\displaystyle {}\arctan x}$.

Substitution

## 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 Theorem 18.17 . 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 Theorem 20.6 , 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

$\displaystyle { \left\{ (x,y) \mid x^2+y^2 [[Category:Wikiversity soft redirects|Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 20]] __NOINDEX__ 1 , \, -1 \leq x \leq 1 , \, y \geq 0 \right\} } .$

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 Corollary 16.14 ), we obtain, using Example 20.3 , 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}}}

## Example

We determine a primitive function for ${\displaystyle {}{\sqrt {x^{2}-1}}}$, using the hyperbolic functions ${\displaystyle {}\sinh t}$ and ${\displaystyle {}\cosh t}$, for which the relation ${\displaystyle {}\cosh ^{2}t-\sinh ^{2}t=1}$ holds. The substitution

${\displaystyle x=\cosh t{\text{ with }}dx=\sinh tdt}$

yields[1]

${\displaystyle {}\int _{a}^{b}{\sqrt {x^{2}-1}}\,dx=\int _{\,\operatorname {arcosh} \,a\,}^{\,\operatorname {arcosh} \,b\,}{\sqrt {\cosh ^{2}t-1}}\cdot \sinh t\,dt=\int _{\,\operatorname {arcosh} \,a\,}^{\,\operatorname {arcosh} \,b\,}\sinh ^{2}t\,dt\,.}$

A primitive function of the hyperbolic sine squared follows from

${\displaystyle {}\sinh ^{2}t={\left({\frac {1}{2}}{\left(e^{t}-e^{-t}\right)}\right)}^{2}={\frac {1}{4}}{\left(e^{2t}+e^{-2t}-2\right)}\,.}$

Therefore,

${\displaystyle {}\int _{}^{}\sinh ^{2}u\,dt={\frac {1}{4}}{\left({\frac {1}{2}}e^{2u}-{\frac {1}{2}}e^{-2u}-2u\right)}={\frac {1}{4}}\sinh 2u-{\frac {1}{2}}u\,,}$

and hence

${\displaystyle {}\int _{}^{}{\sqrt {x^{2}-1}}\,dx={\frac {1}{4}}\sinh(2\,\operatorname {arcosh} \,x\,)-{\frac {1}{2}}\,\operatorname {arcosh} \,x\,\,.}$

Due to the addition theorem for hyperbolic sine, we have ${\displaystyle {}\sinh 2u=2\sinh u\cosh u}$, and therefore this primitive function can also be written as

{\displaystyle {}{\begin{aligned}{\frac {1}{2}}{\left(\sinh {\left(\,\operatorname {arcosh} \,x\,\right)}\cosh {\left(\,\operatorname {arcosh} \,x\,\right)}-\,\operatorname {arcosh} \,x\,\right)}&={\frac {1}{2}}{\left({\sqrt {\cosh {\left(\,\operatorname {arcosh} \,x\,\right)}^{2}-1}}\cdot x-\,\operatorname {arcosh} \,x\,\right)}\\&={\frac {1}{2}}{\left({\sqrt {x^{2}-1}}\cdot x-\,\operatorname {arcosh} \,x\,\right)}.\end{aligned}}}

## Example

We want to find a primitive function for the function

${\displaystyle {}f(x)={\frac {x^{2}}{(x\cos x-\sin x)^{2}}}\,.}$

We first determine the derivative of

${\displaystyle {\frac {1}{x\cos x-\sin x}}.}$

This is

${\displaystyle {}-{\frac {\cos x-x\sin x-\cos x}{(x\cos x-\sin x)^{2}}}={\frac {x\sin x}{(x\cos x-\sin x)^{2}}}\,.}$

Therefore, we write ${\displaystyle {}f}$ as a product ${\displaystyle {}f(x)={\frac {x\sin x}{(x\cos x-\sin x)^{2}}}\cdot {\frac {x}{\sin x}}}$. We apply integration by parts where we integrate the first factor and differentiate the second factor. The derivative of the second factor is

${\displaystyle {}{\left({\frac {x}{\sin x}}\right)}'={\frac {\sin x-x\cos x}{\sin ^{2}x}}\,.}$

Hence, we have

{\displaystyle {}{\begin{aligned}\int _{}^{}f(x)\,dx&={\frac {1}{x\cos x-\sin x}}\cdot {\frac {x}{\sin x}}-\int _{}^{}{\frac {1}{x\cos x-\sin x}}\cdot {\frac {\sin x-x\cos x}{\sin ^{2}x}}\,dx\\&={\frac {1}{x\cos x-\sin x}}\cdot {\frac {x}{\sin x}}+\int _{}^{}{\frac {1}{\sin ^{2}x}}\,dx\\&={\frac {1}{x\cos x-\sin x}}\cdot {\frac {x}{\sin x}}-\cot x.\end{aligned}}}

Footnotes
1. The inverse function of the hyperbolic cosine is called area hyperbolic cosine, and is denoted by ${\displaystyle {}\,\operatorname {arcosh} \,x\,}$.

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