# 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}
\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}
\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}
\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.
\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 } { } }{.}