Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 20/latex
\setcounter{section}{20}
We discuss now the main rules to find a primitive function, and to compute definite integrals. They rest on rules for derivation.
\zwischenueberschrift{Integration by parts}
\inputfaktbeweis
{Integration by parts/Fact}
{Theorem}
{}
{
\faktsituation {Let
\mathdisp {f,g \colon [a,b] \longrightarrow \R} { }
denote
continuously differentiable
functions.}
\faktfolgerung {Then
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f(t)g'(t) \, d t
}
{ =} { fg | _{ a } ^{ b } - \int_{ a }^{ b } f'(t)g(t) \, d t
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}}
\faktzusatz {}
}
{
Due to
the product rule,
the function $fg$ is a
primitive function
for \mathl{fg'+f'g}{.} Therefore,
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f(t) g'(t) \, d t + \int_{ a }^{ b } f'(t) g(t) \, d t
}
{ =} { \int_{ a }^{ b } { \left( fg'+f'g \right) } (t) \, d t
}
{ =} { fg | _{ a } ^{ b }
}
{ } {
}
{ } {
}
}
{}{}{.}
In using integration by parts, two things are to be considered. Firstly, the function to be integrated is usually not in the form \mathl{fg'}{,} but just as a product $uv$
\zusatzklammer {if there is no product, then this rule will probably not help, however, sometimes the trivial product $1 u$ might help} {} {.}
Then for one factor, we have to find a primitive function, and we have to differentiate the other factor. If $V$ is a primitive function of $v$, then the formula reads
\mavergleichskettedisp
{\vergleichskette
{ \int uv
}
{ =} { uV- \int u' V
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
Secondly, integration by parts only helps when the integral on the right, i.e. \mathl{\int_{ a }^{ b } f'(t)g(t) \, d t}{,} can be integrated.
\inputbeispiel{}
{
We determine a
primitive function
for the
natural logarithm
\mathl{\ln x}{,} with
integration by parts.
We write
\mavergleichskette
{\vergleichskette
{ \ln x
}
{ = }{ 1 \cdot \ln x
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
and we integrate the constant function $1$, and we differentiate the logarithm. Then
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } \ln x \, d x
}
{ =} { (x \cdot \ln x) | _{ a } ^{ b } - \int_{ a }^{ b } x \cdot { \frac{ 1 }{ x } } \, d x
}
{ =} { (x \cdot \ln x) | _{ a } ^{ b } - \int_{ a }^{ b } 1 \, d x
}
{ =} { (x \cdot \ln x) | _{ a } ^{ b } - x | _{ a } ^{ b }
}
{ } {
}
}
{}{}{.}
So a primitive function is \mathl{x \cdot \ln x - x}{.}
}
\inputbeispiel{}
{
A
primitive function
for the
sine function
\mathl{\sin x}{} is \mathl{- \cos x}{.} In order to find a primitive function for \mathl{\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 $0$ and have the value $0$ there. For
\mavergleichskette
{\vergleichskette
{n
}
{ \geq }{2
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
with
integration by parts,
we get
\mavergleichskettealign
{\vergleichskettealign
{ \int_{ 0 }^{ x } \sin^{ n } t \, d t
}
{ =} { \int_{ 0 }^{ x } \sin^{ n-2 } t \cdot \sin^{ 2 } t \, d t
}
{ =} { \int_{ 0 }^{ x } \sin^{ n-2 } t \cdot { \left( 1- \cos^{ 2 } t \right) } \, d t
}
{ =} { \int_{ 0 }^{ x } \sin^{ n-2 } t \, d t - \int_{ 0 }^{ x } { \left( \sin^{ n-2 } t \cos t \right) } \cos t \, d t
}
{ =} { \int_{ 0 }^{ x } \sin^{ n-2 } t \, d t - \frac{ \sin^{ n-1 } t }{ n-1} \cos t | _{ 0 } ^{ x } - \frac{1}{n-1} { \left(\int_{ 0 }^{ x } \sin^{ n } t \, d t\right) }
}
}
{}
{}{.}
Multiplication with \mathl{n-1}{} and rearranging yields
\mavergleichskettedisp
{\vergleichskette
{ n \int_{ 0 }^{ x } \sin^{ n } t \, d t
}
{ =} {(n-1) \int_{ 0 }^{ x } \sin^{ n-2 } t \, d t - \sin^{ n-1 } x \cos x
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
In particular, for
\mavergleichskette
{\vergleichskette
{n
}
{ = }{2
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
we have
\mavergleichskettedisp
{\vergleichskette
{ \int_{ 0 }^{ x } \sin^{ 2 } t \, d t
}
{ =} { \frac{1}{2} { \left(x- \sin x \cos x\right) }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
}
\zwischenueberschrift{Integration of inverse function}
\inputfaktbeweis
{Primitive function/Inverse function/Fact}
{Theorem}
{}
{
\faktsituation {Let
$f \colon [a,b] \rightarrow [c,d]$
denote a
bijective
differentiable function,
and let $F$ denote a
primitive function
for $f$.}
\faktfolgerung {Then
\mavergleichskettedisp
{\vergleichskette
{ G(y)
}
{ \defeq} { y f^{-1} (y) - F { \left( f^{-1}(y) \right) }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
is a primitive function for the
inverse function
\mathl{f^{-1}}{.}}
\faktzusatz {}
}
{
Differentiating,
using
Lemma 14.7
and
Theorem 14.8
,
yields
\mavergleichskettealign
{\vergleichskettealign
{ { \left( y f^{-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)
}
{ } {
}
{ } {
}
}
{}
{}{.}
For this statement, there exists also an easy geometric explanation. When
$f \colon [a,b] \rightarrow \R_+$
is a strictly increasing continuous function
\zusatzklammer {and therefore induces a bijection between
\mathkor {} {[a,b]} {and} {[f(a),f(b)]} {}} {} {,}
then the following relation between the areas holds:
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f(s) \, d s + \int_{ f(a) }^{ f(b) } f^{-1}(t) \, d t
}
{ =} { bf(b)-a f(a)
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{,}
or, equivalently,
\mavergleichskettedisp
{\vergleichskette
{ \int_{ f(a) }^{ f(b) } f^{-1}(t) \, d t
}
{ =} { bf(b)-a f(a) - \int_{ a }^{ b } f(s) \, d s
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
For the primitive function $G$ of \mathl{f^{-1}}{} with starting point \mathl{f(a)}{,} we have, if $F$ denotes a primitive function for $f$, the relation
\mavergleichskettealign
{\vergleichskettealign
{ G(y)
}
{ =} { \int_{ f(a) }^{ y } f^{-1}(t) \, d t
}
{ =} { \int_{ f(a) }^{ f(f^{-1}(y)) } f^{-1}(t) \, d t
}
{ =} { f^{-1}(y) f( f^{-1}(y))-a f(a) - \int_{ a }^{ f^{-1}(y) } f(s) \, d s
}
{ =} { y f^{-1}(y) - a f(a) - F(f^{-1}(y)) + F(a)
}
}
{
\vergleichskettefortsetzungalign
{ =} { y f^{-1}(y) - F(f^{-1}(y)) - a f(a) + F(a)
}
{ } {
}
{ } {}
{ } {}
}
{}{,}
where \mathl{- a f(a) + F(a)}{} is a constant of integration.
\inputbeispiel{}
{
We compute a
primitive function
for \mathl{\arctan x}{,} using
Theorem 20.4
.
A primitive function of tangent is
\mavergleichskettedisp
{\vergleichskette
{ \int_{ }^{ } \tan t \, d t
}
{ =} { - \ln (\cos x)
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
Hence,
\mathdisp {x \cdot \arctan x + \ln (\cos (\arctan x))} { }
is a primitive function for \mathl{\arctan x}{.}
}
\zwischenueberschrift{Substitution}
\inputfaktbeweis
{Integration/Substitution/Fact}
{Theorem}
{}
{
\faktsituation {Suppose that $I$ denotes a
real interval,
and let
\mathdisp {f \colon I \longrightarrow \R} { }
denote a
continuous function.
Let
\mathdisp {g \colon [a,b] \longrightarrow I} { }
be a
continuously differentiable
function.}
\faktfolgerung {Then
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f(g(t)) g'(t) \, d t
}
{ =} { \int_{ g(a) }^{ g(b) } f ( s) \, d s
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
holds.}
\faktzusatz {}
}
{
Since $f$ is continuous and $g$ is continuously differentiable, both integrals exist. Let $F$ denote a
primitive function
for $f$, which exists, due to
Theorem 18.17
.
Because of
the chain rule,
the composite function
\mavergleichskettedisp
{\vergleichskette
{ t \mapsto F(g(t))
}
{ =} { (F \circ g)(t)
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
has the derivative
\mavergleichskette
{\vergleichskette
{ F'(g(t)) g'(t)
}
{ = }{ f(g(t))g'(t)
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Therefore,
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f(g(t)) g'(t) \, d t
}
{ =} { (F \circ g) | _{ a } ^{ b }
}
{ =} { F(g(b)) - F(g(a))
}
{ =} { F | _{ g(a) } ^{ g(b) }
}
{ =} { \int_{ g(a) }^{ g(b) } f(s) \, d s }
}
{}{}{.}
\inputbeispiel{}
{
Typical examples, where one sees immediately that one can apply
substitution,
are
\mathdisp {\int g^n g'} { , }
with the primitive function
\mathdisp {{ \frac{ 1 }{ n+1 } } g^{n+1}} { }
or
\mathdisp {\int { \frac{ g' }{ g } }} { , }
with the primitive function
\mathdisp {\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.
\inputfaktbeweis
{Integration/Substitution/dx version/Fact}
{Corollary}
{}
{
\faktsituation {Suppose that
\mathdisp {f \colon [a,b] \longrightarrow \R} { }
is a
continuous function,
and let
\mathdisp {\varphi \colon [c,d] \longrightarrow [a,b]
, s \longmapsto \varphi(s)} { , }
denote a
bijective
continuously differentiable
function.}
\faktfolgerung {Then
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } f ( t) \, d t
}
{ =} { \int_{ \varphi^{-1}(a) }^{ \varphi^{-1}(b) } f( \varphi(s)) \cdot \varphi'(s) \, d s
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
holds.}
\faktzusatz {}
}
{
Because of
Theorem 20.6
,
we have
\mavergleichskettedisp
{\vergleichskette
{ \int_{ \varphi^{-1}(a) }^{ \varphi^{-1}(b) } f( \varphi(s)) \varphi'(s) \, d s
}
{ =} { \int_{ \varphi { \left( \varphi^{-1}(a) \right) } }^{ \varphi { \left( \varphi^{-1}(b) \right) } } f( t) \, d t
}
{ =} { \int_{ a }^{ b } f(t) \, d t
}
{ } {
}
{ } {
}
}
{}{}{.}
\inputremark {}
{
The
substitution
is applied in the following way: suppose that the integral
\mathdisp {\int_{ a }^{ b } f ( t) \, d t} { }
has to be computed. Then one needs an idea that the integral gets simpler by the substitution
\mavergleichskettedisp
{\vergleichskette
{ t
}
{ =} { \varphi(s)
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
\zusatzklammer {taking into account the derivative \mathl{\varphi'(s)}{} and that the inverse function \mathl{\varphi^{-1}}{} has to be determined} {} {.}
Setting
\mathkor {} {c= \varphi^{-1}(a)} {and} {d= \varphi^{-1}(b)} {,}
we have the situation
\mathdisp {[c,d] \stackrel{\varphi}{\longrightarrow} [a,b] \stackrel{f}{\longrightarrow} \R} { . }
In certain cases, some standard substitutions help.
In order to make a substitution, three operations have to be done. \aufzaehlungdrei {Replace \mathl{f(t)}{} by \mathl{f(\varphi(s))}{.} } {Replace \mathl{dt}{} by \mathl{\varphi'(s)ds}{.} } {Replace the integration bounds \mathkor {} {a} {and} {b} {} by \mathkor {} {\varphi^{-1}(a)} {and} {\varphi^{-1}(b)} {.} }
To remember the second step, think of
\mavergleichskettedisp
{\vergleichskette
{ dt
}
{ =} { d \varphi(s)
}
{ =} { \varphi'(s)ds
}
{ } {
}
{ } {
}
}
{}{}{,}
which in the framework \anfuehrung{differential forms}{,} has a meaning.
}
\inputbeispiel{}
{
The upper curve of the unit circle is the set
\mathdisp {{ \left\{ (x,y) \mid x^2+y^2 = 1 , \, -1 \leq x \leq 1 , \, y \geq 0 \right\} }} { . }
For a given
\mathbed {x} {}
{-1 \leq x \leq 1} {}
{} {} {} {,}
there exists exactly one $y$ fulfilling this condition, namely
\mavergleichskette
{\vergleichskette
{ 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 \mathl{x \mapsto \sqrt{1-x^2}}{,} above the interval \mathl{[-1,1]}{,} that is
\mathdisp {\int_{ -1 }^{ 1 } \sqrt{1-x^2} \, d x} { . }
Applying
substitution
with
\mathdisp {x = \cos t \text{ and } t = \arccos x} { }
\zusatzklammer {where \mathl{\cos :[0, \pi] \rightarrow [-1,1]}{} is bijective, due to
Corollary 16.14
} {} {,}
we obtain, using
Example 20.3
,
the identities
\mavergleichskettealign
{\vergleichskettealign
{ \int_{ a }^{ b } \sqrt{1-x^2} \, d x
}
{ =} { \int_{ \arccos a }^{ \arccos b } \sqrt{1- \cos^{ 2 } t } (- \sin t ) \, d t
}
{ =} { - \int_{ \arccos a }^{ \arccos b } \sin^{ 2 } t \, d t
}
{ =} { \frac{1}{2} ( \sin t \cos t -t ) | _{ \arccos a } ^{ \arccos b }
}
{ } {
}
}
{}
{}{.}
In particular, we get that
\mavergleichskettedisp
{\vergleichskette
{ { \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 \mathl{\sqrt{1-x^2}}{.} Therefore,
\mavergleichskettealign
{\vergleichskettealign
{ \int_{ -1 }^{ 1 } \sqrt{1-x^2} \, d x
}
{ =} { { \frac{ 1 }{ 2 } } { \left( x \cdot \sqrt{1-x^2} - \arccos x \right) } | _{ -1 } ^{ 1 }
}
{ =} { { \frac{ 1 }{ 2 } } ( - \arccos 1 + \arccos (-1) )
}
{ =} { \pi/2
}
{ } {}
}
{}
{}{.}
}
\inputbeispiel{}
{
We determine a
primitive function
for \mathl{\sqrt{x^2-1}}{,} using the hyperbolic functions
\mathkor {} {\sinh t} {and} {\cosh t} {,}
for which the relation
\mavergleichskette
{\vergleichskette
{ \cosh^{ 2 } t - \sinh^{ 2 } t
}
{ = }{ 1
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
holds. The
substitution
\mathdisp {x= \cosh t \text{ with } dx = \sinh t dt} { }
yields\zusatzfussnote {The inverse function of the hyperbolic cosine is called \stichwort {area hyperbolic cosine} {,} and is denoted by $\, \operatorname{arcosh} \, x \,$} {.} {}
\mavergleichskettedisp
{\vergleichskette
{ \int_{ a }^{ b } \sqrt{x^2-1} \, d x
}
{ =} { \int_{ \, \operatorname{arcosh} \, a \, }^{ \, \operatorname{arcosh} \, b \, } \sqrt{ \cosh^{ 2 } t-1 } \cdot \sinh t \, d t
}
{ =} { \int_{ \, \operatorname{arcosh} \, a \, }^{ \, \operatorname{arcosh} \, b \, } \sinh^{ 2 } t \, d t
}
{ } {
}
{ } {
}
}
{}{}{.}
A primitive function of the hyperbolic sine squared follows from
\mavergleichskettedisp
{\vergleichskette
{ \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,
\mavergleichskettedisp
{\vergleichskette
{ \int_{ }^{ } \sinh^{ 2 } u \, d t
}
{ =} { { \frac{ 1 }{ 4 } } { \left( { \frac{ 1 }{ 2 } } e^{2u} - { \frac{ 1 }{ 2 } } e^{-2u} - 2u\right) }
}
{ =} { { \frac{ 1 }{ 4 } } \sinh 2 u - { \frac{ 1 }{ 2 } } u
}
{ } {
}
{ } {
}
}
{}{}{,}
and hence
\mavergleichskettedisp
{\vergleichskette
{ \int_{ }^{ } \sqrt{x^2-1} \, d x
}
{ =} { { \frac{ 1 }{ 4 } } \sinh (2 \, \operatorname{arcosh} \, x \,) - { \frac{ 1 }{ 2 } } \, \operatorname{arcosh} \, x \,
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
Due to
the addition theorem
for hyperbolic sine, we have
\mavergleichskette
{\vergleichskette
{ \sinh 2u
}
{ = }{ 2 \sinh u \cosh u
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
and therefore this primitive function can also be written as
\mavergleichskettealign
{\vergleichskettealign
{ { \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) }
}
{ } {
}
{ } {
}
}
{}
{}{.}
}
\inputbeispiel{}
{
We want to find a
primitive function
for the function
\mavergleichskettedisp
{\vergleichskette
{f(x)
}
{ =} { { \frac{ x^2 }{ (x \cos x - \sin x )^2 } }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
We first determine the derivative of
\mathdisp {{ \frac{ 1 }{ x \cos x - \sin x } }} { . }
This is
\mavergleichskettedisp
{\vergleichskette
{ - { \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 $f$ as a product
\mavergleichskette
{\vergleichskette
{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
\mavergleichskettedisp
{\vergleichskette
{ { \left({ \frac{ x }{ \sin x } }\right) }'
}
{ =} { { \frac{ \sin x - x \cos x }{ \sin^{ 2 } x } }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
Hence, we have
\mavergleichskettealign
{\vergleichskettealign
{ \int_{ }^{ } f ( x) \, d x
}
{ =} { { \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 } } \, d x
}
{ =} { { \frac{ 1 }{ x \cos x - \sin x } } \cdot { \frac{ x }{ \sin x } } + \int_{ }^{ } { \frac{ 1 }{ \sin^{ 2 } x } } \, d x
}
{ =} { { \frac{ 1 }{ x \cos x - \sin x } } \cdot { \frac{ x }{ \sin x } } - \cot x
}
{ } {}
}
{}
{}{.}
}