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2.2 Integration durch Substitution

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Version vom 13:56, 17. Sep. 2008

       Theory          Exercises      

Contents:

  • Integration by substitution

Learning outcomes:

After this section, you will have learned to:

  • Understand the derivation of the formula for variable 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 variable substitution is allowed.

Variable substitution

When you cannot directly determine a primitive function by using the usual rules of differentiation ”in the opposite direction” method, other methods or techniques are needed. One such is variable substitution, which can be said to be based on the rule for the differentiation of composite functions — the so-called chain rule.

The chain rule  ddxf(u(x))=f(u(x))u(x)  can be written in integral form as

f(u(x))u(x)dx=f(u(x))+C 

or,

f(u(x))u(x)dx=F(u(x))+C, 

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

f(u)du=F(u)+C. 

We can see that we have replaced the term u(x) with variable u and the u(x)dx with du. One thus can transform the more complicated integrand f(u(x))u(x) (with x as the variable) to the, let us hope, easier f(u) (with the u as the variable). The method is called variable substitution and can be used when the integrand can be written in the form f(u(x))u(x).


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


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

limx0xu=dxdu=u(x),

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

uu(x)xdu=u(x)dx,

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


Example 1

Determine the integral 2xex2dx .

If one puts u(x)=x2, one gets u(x)=2x. The variable substitution replaces ex2 with eu and u(x)dx, i.e. \displaystyle 2x\,dx, with \displaystyle du

\displaystyle \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{.}

Example 2

Determine the integral \displaystyle \ \int (x^3 + 1)^3 \cdot 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} \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*}

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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