2.2 Substitution

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{{Mall:Vald flik|[[2.2 Variabelsubstitution|Teori]]}}
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{{Selected tab|[[2.2 Substitution|Theory]]}}
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{{Mall:Ej vald flik|[[2.2 Övningar|Övningar]]}}
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{{Not selected tab|[[2.2 Exercises|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Contents:'''
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* Variabelsubstitution
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* Integration by substitution
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned to:
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* Förstå härledningen av formeln för variabelsubstitution.
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* Understand the derivation of the formula for substitution .
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* Lösa enklare integrationsproblem som kräver omskrivning och/eller substitution i ett steg.
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* Solve simple integration problems that require rewriting and / or substitution in one of the steps.
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* Veta hur integrationsgränserna ändras under variabelsubstitution.
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* Know how the limits of integration are to be changed after a variable substitution.
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* Veta när en variabelsubstitution är tillåten.
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* Know when substitution is allowed.
}}
}}
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== Variabelsubstitution ==
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== Substitution, or change of variable ==
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När man inte direkt kan bestämma en primitiv funktion genom att utnyttja de vanliga deriveringsreglerna ”baklänges”, behöver man andra metoder eller tekniker. En sådan är ''variabelsubstitution'', vilken kan sägas baseras på regeln för derivering av sammansatta funktioner — den s.k. ''kedjeregeln''.
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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''.
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Kedjeregeln <math>\ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \cdot u'(x)\ </math> kan i integralform skrivas
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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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{{Fristående formel||<math>\int f^{\,\prime}(u(x)) \cdot u'(x) \, dx = f(u(x)) + C</math>}}
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{{Displayed math||<math>\int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C</math>}}
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eller,
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or,
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>\int f(u(x)) \cdot u'(x) \, dx = F (u(x)) + C\,\mbox{,}</math>}}
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{{Displayed math||<math>\int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,}</math>}}
</div>
</div>
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där ''F'' är en primitiv funktion till ''f''. Jämför vi denna formel med
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where ''F'' is a primitive function of ''f''. We compare this with the formula
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{{Fristående formel||<math>\int f(u) \, du = F(u) + C\,\mbox{,}</math>}}
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{{Displayed math||<math>\int f(u) \, du = F(u) + C\,\mbox{.}</math>}}
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så kan vi se det som att vi ersätter uttrycket <math>u(x)</math> med variabeln <math>u</math> och <math>u'(x)\, dx</math> med <math>du</math>. Man kan alltså omvandla den krångligare integranden <math>f(u(x)) \cdot u'(x)</math> (med <math>x</math> som variabel) med den förhoppningsvis enklare <math>f(u)</math> (med <math>u</math> som variabel). Metoden kallas variabelsubstitution och kan användas när integranden kan skrivas på formen <math>f(u(x)) \cdot u'(x)</math>.
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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>.
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''Anm. 1'' Metoden bygger naturligtvis på att alla förutsättningar för integrering är uppfyllda; att <math>u(x)</math> är deriverbar i det aktuella intervallet, samt att <math>f</math> är kontinuerlig i värdemängden till <math>u</math>, dvs. för alla värden som <math>u</math> kan anta i intervallet.
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''Note 1'' The method is based on the assumption that all the conditions for integration are satisfied; that is, <math>u(x)</math> is differentiable in the interval in question, and that <math>f</math> is continuous for all values of <math>u</math> in the range, that is, for all the values that <math>u</math> can take on in the interval.
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''Anm. 2'' Att ersätta <math>u'(x) \, dx</math> med <math>du</math> kan också motiveras genom att studera övergången från differenskvot till derivata:
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'' 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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{{Fristående formel||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}}
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{{Displayed math||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}}
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vilket när <math>\Delta x</math> går mot noll kan betraktas som en formell gränsövergång
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which, as <math>\Delta x</math> goes towards zero can be considered as a formal transition between variables
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{{Fristående formel||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}}
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{{Displayed math||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}}
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dvs., en liten ändring, <math>dx</math>, i variabeln <math>x</math> ger upphov till en ungefärlig ändring <math>u'(x)\,dx</math> i variabeln <math>u</math>.
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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>.
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
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Bestäm integralen <math>\ \int 2 x\, e^{x^2} \, dx</math>.
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Determine the integral<math>\ \int 2 x\, e^{x^2} \, dx</math>.
<br>
<br>
<br>
<br>
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Om man sätter <math>u(x)= x^2</math>, så blir <math>u'(x)= 2x</math>. Vid variabelbytet ersätts då <math>e^{x^2}</math> med <math>e^u</math> och <math>u'(x)\,dx</math>, dvs. <math>2x\,dx</math>, med <math>du</math>
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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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{{Fristående formel||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \cdot 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</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>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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Bestäm integralen <math>\ \int (x^3 + 1)^3 \cdot x^2 \, dx</math>.
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Determine the integral <math>\ \int (x^3 + 1)^3 \, x^2 \, dx</math>.
<br>
<br>
<br>
<br>
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Sätt <math>u=x^3 + 1</math>. Då blir <math>u'=3x^2</math>, eller <math>du= 3x^2\, dx</math>, och
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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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{{Fristående formel||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \cdot 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>}}
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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>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
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Bestäm integralen <math>\ \int \tan x \, dx\ \ </math> där <math>-\pi/2 < x < \pi/2</math>.
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Determine the integral <math>\ \int \tan x \, dx\,\mbox{,}\ \ </math> where <math>-\pi/2 < x < \pi/2</math>.
<br>
<br>
<br>
<br>
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Efter en omskrivning av <math>\tan x</math> som <math>\sin x/\cos x</math> substituerar vi <math>u=\cos x</math>
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After rewriting <math>\tan x</math> as <math>\sin x/\cos x</math> we substitute <math>u=\cos x</math>,
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{{Fristående 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>}}
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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>}}
</div>
</div>
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== Integrationsgränser vid variabelbyte ==
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== The limits of integration during variable substitution. ==
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Vid beräkning av bestämda integraler, t.ex. en area, där man använder variabelsubstitution kan man gå till väga på två sätt. Antingen beräknar man integralen som vanligt, byter tillbaka till den ursprungliga variabeln och sätter in de ursprungliga integrationsgränserna. Alternativt ändrar man integrationsgränser samtidigt som man gör variabelbytet. De båda metoderna illustreras i följande exempel.
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 +
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.
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
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Beräkna integralen <math>\ \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx</math>.
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Determine the integral <math>\ \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx</math>.
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''Metod 1''
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'' Method 1''
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Sätt <math>u=e^x</math> vilket ger att <math>u'= e^x</math> och <math>du= e^x\,dx</math>
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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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{{Fristående 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>}}
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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>}}
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Observera att integrationsgränserna måste skrivas <math>x = 0</math> och <math>x = 2</math> när integrationsvariabeln inte är <math>x</math>. Det vore fel att skriva
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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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{{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ osv.}</math>}}
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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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''Metod 2''
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'' Method 2''
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Sätt <math>u=e^x</math> vilket ger att <math>u'= e^x</math> och <math>du= e^x\, dx</math>. Integrationsgränsen <math>x=0</math> motsvaras då av <math>u=e^0 = 1</math> och <math>x=2</math> motsvaras av <math>u=e^2</math>
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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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{{Fristående 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>}}
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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>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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Beräkna integralen <math> \ \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx</math>.
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Determine the integral <math> \ \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx</math>.
<br>
<br>
<br>
<br>
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Substitutionen <math>u=\sin x</math> ger att <math>du=\cos x\,dx</math> och integrationsgränserna förändras till <math>u=\sin 0=0</math> och <math>u=\sin(\pi/2)=1</math>. Integralen blir
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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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{{Fristående 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>}}
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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>}}
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<center>{{:2.2 - Figur - Area under y = sin³x cos x resp. y = u³}}</center>
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<center>{{:2.2 - Figure - The area under y = sin³x cos x and y = u³, respectively}}</center>
{| width="80%" align="center"
{| width="80%" align="center"
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||<small>Figuren till vänster visar grafen till integranden sin³''x'' cos ''x'' och figuren till höger grafen till integranden ''u''³ som fås efter variabelsubstitutionen. Vid variabelbytet ändras integranden och integrationsintervallet. Integralens värde, storleken på arean, ändras dock inte.</small>
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||<small> The figure on the left shows the graph of the integrand sin³''x'' cos ''x'' and the figure on the right the graph of integrand ''u''³ 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. </small>
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<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
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Betrakta beräkningen
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Examine the following calculation
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{{Fristående 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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{{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>}}
{| width="100%"
{| width="100%"
| width="95%" |
| width="95%" |
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Denna uträkning är dock felaktig, vilket beror på att <math>f(u)=1/u^2</math> inte är kontinuerlig i '''hela''' intervallet <math>[-1,1]</math>.
+
This calculation, however, is wrong, which is due to the fact that <math>f(u)=1/u^2</math> is not continuous '''throughout''' the interval <math>[-1,1]</math>.
-
Villkoret att <math>f(u(x))</math> ska vara definierad och kontinuerlig för alla värden som <math>u(x)</math> kan anta i det aktuella intervallet behövs om man vill vara säker på att substitutionen <math>u=u(x)</math> ska fungera.
+
A necessary condition in the theory is that <math>f(u(x))</math> be defined and continuous for all values which <math>u(x)</math> can take in the interval in question. Otherwise one cannot be certain that the substitution <math>u=u(x)</math> will work.
| width="5%" |
| width="5%" |
||
||
{|
{|
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||{{:2.2 - Figur - Grafen till f(u) = 1/u²}}
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||{{:2.2 - Figure - The graph of f(u) = 1/u²}}
|-
|-
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||<small>Grafen till ''f''(''u'') = 1/''u''²</small>
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||<small>Graph of ''f''(''u'') = 1/''u''²</small>
|}
|}
|}
|}
</div>
</div>

Current revision

       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{.}


[Image]

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.

[Image]

Graph of f(u) = 1/u²
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