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

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

denote continuously differentiable functions. Then

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

Proof

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

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

$\Box$

In using integration by parts, two things are to be considered. Firstly, the function to be integrated is usually not in the form ${}fg'$, but just as a product ${}uv$ (if there is no product, then this rule will probably not help, however, sometimes the trivial product ${}1u$ 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

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

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

Example

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

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

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

Example

A primitive function for the sine function ${}\sin x$ is ${}-\cos x$. In order to find a primitive function for ${}\sin ^{n}x$, 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 ${}n\geq 2$, with integration by parts, we get

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

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

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

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

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

Integration of inverse function

Theorem

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

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

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

Proof

Differentiating, using Lemma 14.7 and Theorem 14.8 , yields

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

$\Box$

The graph with its inverse function, and the areas relevant for the computation of the integral of the inverse function.

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

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

or, equivalently,

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

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

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

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

Example

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

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

Hence,

$x\cdot \arctan x+\ln(\cos(\arctan x))$

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

Substitution

Theorem

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

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

denote a continuous function. Let

$g\colon [a,b]\longrightarrow I$

be a continuously differentiable function. Then

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

holds.

Proof

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

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

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

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

$\Box$

Example

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

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

with the primitive function

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

or

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

with the primitive function

$\ln g.$

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

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

is a continuous function, and let

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

denote a bijective continuously differentiable function. Then

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

holds.

Proof

Because of Theorem 20.6 , we have

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

$\Box$

Remark

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

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

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

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

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

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

In certain cases, some standard substitutions help.

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

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

To remember the second step, think of

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

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

Example

The upper curve of the unit circle is the set

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

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

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

Applying substitution with

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

(where ${}\cos :[0,\pi ]\rightarrow [-1,1]$ is bijective, due to Corollary 16.14 ), we obtain, using Example 20.3 , the identities

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

In particular, we get that

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

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

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

Example

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

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

yields

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

A primitive function of the hyperbolic sine squared follows from

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

Therefore,

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

and hence

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

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

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

Example

We want to find a primitive function for the function

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

We first determine the derivative of

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

This is

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

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

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

Hence, we have

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

Footnotes

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