2.2 Integration durch Substitution

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Zeile 29: Zeile 29:
The chain rule <math>\ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \, u'(x)\ </math> can be written in integral form as
The chain rule <math>\ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \, u'(x)\ </math> can be written in integral form as
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{{Displayed math||<math>\int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C</math>}}
+
{{Abgesetzte Formel||<math>\int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C</math>}}
or,
or,
<div class="regel">
<div class="regel">
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{{Displayed math||<math>\int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,}</math>}}
+
{{Abgesetzte Formel||<math>\int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,}</math>}}
</div>
</div>
where ''F'' is a primitive function of ''f''. We compare this with the formula
where ''F'' is a primitive function of ''f''. We compare this with the formula
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{{Displayed math||<math>\int f(u) \, du = F(u) + C\,\mbox{.}</math>}}
+
{{Abgesetzte Formel||<math>\int f(u) \, du = F(u) + C\,\mbox{.}</math>}}
We can see that we have replaced the term <math>u(x)</math> with variable <math>u</math> and the <math>u'(x)\, dx</math> with <math>du</math>. One thus can transform the more complicated integrand <math>f(u(x)) \, u'(x)</math> (with <math>x</math> as the variable) to the simpler (and possibly more tractable) <math>f(u)</math> (with the <math>u</math> as the variable). The method is called substitution, or change of variable, and can be used when the integrand can be written in the form <math>f(u(x)) \, u'(x)</math>.
We can see that we have replaced the term <math>u(x)</math> with variable <math>u</math> and the <math>u'(x)\, dx</math> with <math>du</math>. One thus can transform the more complicated integrand <math>f(u(x)) \, u'(x)</math> (with <math>x</math> as the variable) to the simpler (and possibly more tractable) <math>f(u)</math> (with the <math>u</math> as the variable). The method is called substitution, or change of variable, and can be used when the integrand can be written in the form <math>f(u(x)) \, u'(x)</math>.
Zeile 48: Zeile 48:
'' Note 2'' Replacing <math>u'(x) \, dx</math> with <math>du</math> also may be justified by studying the transition from the increment ratio to the derivative:
'' Note 2'' Replacing <math>u'(x) \, dx</math> with <math>du</math> also may be justified by studying the transition from the increment ratio to the derivative:
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{{Displayed math||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}}
+
{{Abgesetzte Formel||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}}
which, as <math>\Delta x</math> goes towards zero can be considered as a formal transition between variables
which, as <math>\Delta x</math> goes towards zero can be considered as a formal transition between variables
-
{{Displayed math||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}}
+
{{Abgesetzte Formel||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}}
ie., a small change, <math>dx</math>, in the variable <math>x</math> gives rise to an approximate change <math>u'(x)\,dx</math> in the variable <math>u</math>.
ie., a small change, <math>dx</math>, in the variable <math>x</math> gives rise to an approximate change <math>u'(x)\,dx</math> in the variable <math>u</math>.
Zeile 65: Zeile 65:
If one puts <math>u(x)= x^2</math>, one gets <math>u'(x)= 2x</math>. The variable substitution replaces <math>e^{x^2}</math> with <math>e^u</math> and <math>u'(x)\,dx</math>, i.e. <math>2x\,dx</math>, with <math>du</math>
If one puts <math>u(x)= x^2</math>, one gets <math>u'(x)= 2x</math>. The variable substitution replaces <math>e^{x^2}</math> with <math>e^u</math> and <math>u'(x)\,dx</math>, i.e. <math>2x\,dx</math>, with <math>du</math>
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{{Displayed math||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \times 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</math>}}
+
{{Abgesetzte Formel||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \times 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</math>}}
</div>
</div>
Zeile 77: Zeile 77:
Put <math>u=x^3 + 1</math>. This means <math>u'=3x^2</math>, or <math>du= 3x^2\, dx</math>, and
Put <math>u=x^3 + 1</math>. This means <math>u'=3x^2</math>, or <math>du= 3x^2\, dx</math>, and
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{{Displayed math||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \times 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \times 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Zeile 89: Zeile 89:
After rewriting <math>\tan x</math> as <math>\sin x/\cos x</math> we substitute <math>u=\cos x</math>,
After rewriting <math>\tan x</math> as <math>\sin x/\cos x</math> we substitute <math>u=\cos x</math>,
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{{Displayed math||<math>\begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Zeile 108: Zeile 108:
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\,dx</math>
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\,dx</math>
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{{Displayed math||<math>\begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}</math>}}
+
{{Abgesetzte Formel||<math>\begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}</math>}}
Note that the limits of integration must be written in the form <math>x = 0</math> and <math>x = 2</math> when the variable of integration is not <math>x</math>. it is wrong to write
Note that the limits of integration must be written in the form <math>x = 0</math> and <math>x = 2</math> when the variable of integration is not <math>x</math>. it is wrong to write
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{{Displayed math||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.}</math>}}
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{{Abgesetzte Formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.}</math>}}
Zeile 119: Zeile 119:
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\, dx</math>. The limit of integration <math>x=0</math> is equivalent to <math>u=e^0 = 1</math> and <math>x=2</math> is equivalent to <math>u=e^2</math>
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\, dx</math>. The limit of integration <math>x=0</math> is equivalent to <math>u=e^0 = 1</math> and <math>x=2</math> is equivalent to <math>u=e^2</math>
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{{Displayed math||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}</math>}}
+
{{Abgesetzte Formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}</math>}}
</div>
</div>
Zeile 131: Zeile 131:
The substitution <math>u=\sin x</math> gives <math>du=\cos x\,dx</math> and the limits of integration become <math>u=\sin 0=0</math> and <math>u=\sin(\pi/2)=1</math>. The integral is
The substitution <math>u=\sin x</math> gives <math>du=\cos x\,dx</math> and the limits of integration become <math>u=\sin 0=0</math> and <math>u=\sin(\pi/2)=1</math>. The integral is
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{{Displayed math||<math>\int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}</math>}}
+
{{Abgesetzte Formel||<math>\int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}</math>}}
Zeile 146: Zeile 146:
Examine the following calculation
Examine the following calculation
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{{Displayed math||<math>\int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}</math>}}
+
{{Abgesetzte Formel||<math>\int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}</math>}}
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Version vom 12:49, 10. Mär. 2009

       Theory          Exercises      

Contents:

  • Integration by substitution

Learning outcomes:

After this section, you will have learned to:

  • Understand the derivation of the formula for substitution .
  • Solve simple integration problems that require rewriting and / or substitution in one of the steps.
  • Know how the limits of integration are to be changed after a variable substitution.
  • Know when substitution is allowed.

Substitution, or change of variable

When you cannot directly determine an indefinite integral by inspection (that is, by simple "differentiation in reverse"), other methods or techniques are needed. One such is substitution (sometimes called change of variable), which can be said to be based on the rule for the differentiation of composite functions — the so-called chain rule.

The chain rule \displaystyle \ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \, u'(x)\ can be written in integral form as

\displaystyle \int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C

or,

\displaystyle \int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,}

where F is a primitive function of f. We compare this with the formula

\displaystyle \int f(u) \, du = F(u) + C\,\mbox{.}

We can see that we have replaced the term \displaystyle u(x) with variable \displaystyle u and the \displaystyle u'(x)\, dx with \displaystyle du. One thus can transform the more complicated integrand \displaystyle f(u(x)) \, u'(x) (with \displaystyle x as the variable) to the simpler (and possibly more tractable) \displaystyle f(u) (with the \displaystyle u as the variable). The method is called substitution, or change of variable, and can be used when the integrand can be written in the form \displaystyle f(u(x)) \, u'(x).


Note 1 The method is based on the assumption that all the conditions for integration are satisfied; that is, \displaystyle u(x) is differentiable in the interval in question, and that \displaystyle f is continuous for all values of \displaystyle u in the range, that is, for all the values that \displaystyle u can take on in the interval.


Note 2 Replacing \displaystyle u'(x) \, dx with \displaystyle du also may be justified by studying the transition from the increment ratio to the derivative:

\displaystyle \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}

which, as \displaystyle \Delta x goes towards zero can be considered as a formal transition between variables

\displaystyle \Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}

ie., a small change, \displaystyle dx, in the variable \displaystyle x gives rise to an approximate change \displaystyle u'(x)\,dx in the variable \displaystyle u.


Example 1

Determine the integral\displaystyle \ \int 2 x\, e^{x^2} \, dx.

If one puts \displaystyle u(x)= x^2, one gets \displaystyle u'(x)= 2x. The variable substitution replaces \displaystyle e^{x^2} with \displaystyle e^u and \displaystyle u'(x)\,dx, i.e. \displaystyle 2x\,dx, with \displaystyle du

\displaystyle \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \times 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}

Example 2

Determine the integral \displaystyle \ \int (x^3 + 1)^3 \, x^2 \, dx.

Put \displaystyle u=x^3 + 1. This means \displaystyle u'=3x^2, or \displaystyle du= 3x^2\, dx, and

\displaystyle \begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \times 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}

Example 3

Determine the integral \displaystyle \ \int \tan x \, dx\,\mbox{,}\ \ where \displaystyle -\pi/2 < x < \pi/2.

After rewriting \displaystyle \tan x as \displaystyle \sin x/\cos x we substitute \displaystyle u=\cos x,

\displaystyle \begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}


The limits of integration during variable substitution.

When calculating definite integrals, such as an area, one can go about using variable substitution in two ways. Either one can calculate the integral as usual and then switch back to the original variable and insert the original limits of integration. Alternatively one can change the limits of integration simultaneously with the variable substitution. The two methods are illustrated in the following example.

Example 4

Determine the integral \displaystyle \ \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx.


Method 1

Put \displaystyle u=e^x which gives that \displaystyle u'= e^x and \displaystyle du= e^x\,dx

\displaystyle \begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}

Note that the limits of integration must be written in the form \displaystyle x = 0 and \displaystyle x = 2 when the variable of integration is not \displaystyle x. it is wrong to write

\displaystyle \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.}


Method 2

Put \displaystyle u=e^x which gives that \displaystyle u'= e^x and \displaystyle du= e^x\, dx. The limit of integration \displaystyle x=0 is equivalent to \displaystyle u=e^0 = 1 and \displaystyle x=2 is equivalent to \displaystyle u=e^2

\displaystyle \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}

Example 5

Determine the integral \displaystyle \ \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx.

The substitution \displaystyle u=\sin x gives \displaystyle du=\cos x\,dx and the limits of integration become \displaystyle u=\sin 0=0 and \displaystyle u=\sin(\pi/2)=1. The integral is

\displaystyle \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}


2.2 - Figure - The area under y = sin³x cos x and y = u³, respectively
The figure on the left shows the graph of the integrand sin³x cos x and the figure on the right the graph of integrand u³ which is obtained after the variable substitution. The change of variable modifies the integrand and the interval of the integration. The integrals value, the size of the area, is not changed however.

Example 6

Examine the following calculation

\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}

This calculation, however, is wrong, which is due to the fact that \displaystyle f(u)=1/u^2 is not continuous throughout the interval \displaystyle [-1,1].

A necessary condition in the theory is that \displaystyle f(u(x)) be defined and continuous for all values which \displaystyle u(x) can take in the interval in question. Otherwise one cannot be certain that the substitution \displaystyle u=u(x) will work.

2.2 - Figure - The graph of f(u) = 1/u²
Graph of f(u) = 1/u²
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