Integration by substitution

Transcription

1 Roberto s Notes on Integral Calculus Chapter : Integration methods Section 1 Integration by substitution or by change of variable What you need to know already: What an indefinite integral is. The chain rule. Basic rules of integration allow us to obtain the general antiderivative of simple functions. We now need to devote some effort to develop methods to integrate more complicated functions. Interestingly enough, the first method, which is the simplest and most often used, is obtained by reversing the last rule of differentiation you learned: the chain rule. Technical fact If an indefinite integral can be written in the form f ' g( ) g '( ) d then it consists of the family of functions: f ' g( ) g '( ) d f g( ) c. Is this just the chain rule stated in reverse? Yes! What you can learn here: How to integrate a function that is the derivative of a composite function. Will we always be able to recognize an integral as written in that particular form? In some cases the structure we need is easily visible. d Eample: We can start by rewriting the integrand as d, thus seeing clearly that the second factor is the derivative of the inside of the first factor, which is a fifth power. Formally, we see that our integral is of the needed form f ' g( ) g '( ) d, with f d ; g. Therefore, we only need to find out what power rule to conclude that: f is. To do that, we apply the 6 d c 6 Taking the derivative of our answer confirms its correctness. Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page 1

2 But things are not always this clear, and in most other situations the chain rule structure is not so easily visible, especially when your eperience is limited. We may only be able to suspect that the integrand comes from a chain rule, but not be clear on how. To make the problem manageable and to confirm our hope, we can use the standard mathematical method of substitution. Since we are reversing the chain rule, which is applied to composite functions, we introduce a new variable, commonly denoted by u, to equal an inside portion of the integrand, a portion corresponding to what we hope is g. ( ) Once the decision is made, we need to change the given integral to a new, equivalent one whose variable of integration is u. Here is how to do that. Strategy for integration by substitution To integrate a function that we suspect to be a derivative obtained through the chain rule: 1. Re-write the integrand in a suitable calculus form, if needed.. Identify a potential candidate for the inside function g(). Usually (but not always) this will be the inside of brackets, or roots, or denominators. Denote it with a new letter: u g " inside".. Properly compute the differential of u, du g d, and from it obtain the formula to change the differential: du du gd d. g 4. Use these two formulae to change or substitute all portions of the original integral, given in terms of, by using the corresponding epressions involving the new variable u.. If the method is workable for the original integral, the new integral will contain only the new variable and will be easier to compute, so compute it. 6. Use the formula g u to epress the indefinite integral formula, found so far in terms of u, as a family of functions written in terms of the original variable. That sounds long and complicated! It may be long and complicated in some situations yes, but it is a very useful method and, for now, applicable to many nice functions in a pretty straightforward way, so do not be scared by it. 4 d Eample: This is similar to the previous one, but it is not as clear that we are dealing with a chain rule situation, since the coefficients do not seem to match. Still, apart from the coefficients, the first factor looks like the derivative of the inside of the second. So, we hope that the method works and we proceed with it. We start by rewriting the integrand in a more suitable way: 4 4 / d d Then we select the inside of the power for the substitution: u 4. Net, we compute its differential and use it to write the old differential in terms of the new: Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page

3 Net, we perform the substitution: du du d d / du 4 1 / / d u u du Indeed we now have an integral that is entirely written in terms of u and is easy to integrate with basic rules: 7/ 1 / 1 u 7/ u du c u c 7 / 1 All that is left is to switch back to the original variable and conclude that: 7/ 4 d 4 c 1 Taking the derivative of this formula will confirm that we constructed the correct indefinite integral. 7 du d u u du At this point it looks like we are stuck with both original and new variable, a situation that would not allow us to continue. However, we notice that: u u This means that we can substitute further: u 7 du u 7 u du u 8 u 7 du We can now use basic rules and switch back to for the general antiderivative we seek: u u c c Does this always work? I wish! Remember that this method is supposed to integrate a derivative obtained through the chain rule. Therefore, if the integrand comes from another method, substitution will not work. One telling sign that the method may not work is if, after the initial substitution is done, we are left with some portions that are still written in terms of the original variable, since in order to use the method, the new integral must be entirely in terms of the new variable. But don t give up too soon: it may still be possible to change the remaining portion and write it in terms of the new variable by using the original substitution. Eample: 7 d To evaluate this integral, we can try the substitution: du u, du d d This changes the original integral to: Eample: sinh cosh d We know that sinh cosh and that cosh cosh can try the substitution: u cosh, du sinh d This changes our integral to: sinh cosh d u du This is easy to integrate, thus providing: u cosh sinh cosh d c c, therefore we Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page

4 Eample: d 1 At first, this integral does not seem a good candidate for the method of substitution, since the integrand includes only one factor, making it unclear what the g would be. But since we have no other method available yet, and there is an inside piece, we try it anyway. We let: 1 u 1, du d d du thus changing the integral to: d 1 du 1/ u We still have the old variable, but we can eliminate it by going back to the substitution: Therefore: u 1 u 1 du u 1 1/ 1/ u u / 1/ du u u du We now integrate by using the power and addition rules: Finally we switch back to : 6 / / u u c / / c Eample: tan d Given the definition of the tangent function we can write: tan d sin d cos Since the cosine function is inside the denominator, we can use the substitution u cos, du sin d, thus getting: sin sin 1 d du du cos u sin u ln u c ln cos c Notice that, since the tangent function is discontinuous, a different constant should be used on each interval of continuity. Notice also that an alternative epression of this antiderivative can be obtained as follows: 1 ln cos c ln cos c ln sec c Some people prefer this formula as it avoids the negative in front and keeps the relation between tan and sec that eists for the derivative formula. Knot on your finger The general antiderivative of the tangent function is: tan d ln cos c ln sec c As last eample of this section, we shall use the method of substitution to obtain the integral of a basic and familiar function. There are literally infinitely many other possible eamples and cases, and you will discover many of them through practice and in conjunction with later methods. For now I will conclude this section with some important aspects of the method of substitution that you need to keep in mind in order to avoid problems. Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page 4

5 Warning bells The method of substitution is a method because it consists of several steps. Skipping or mishandling any one of these steps can create errors and lead to the wrong conclusion or to a dead end. This method only works if the integrand comes from the chain rule. If it does not, you will either get a new integral that is not easier than the original, or an integrand containing both old and new variables with no possibility of further changes. In that case, you cannot proceed further. There may be more than one candidate for the substitution. Do not assume that the first choice you make will work. If it does not, try a different one before giving up. If it does, great, but still try the others, since this may show you why the others do not work, or may allow you to discover more efficient choices. Finally, let me state that this method looks reasonable especially since it works but it relies on several technicalities that can be tricky to check. In fact, mathematicians still argue about the proper way to interpret this method and its steps. Summary By changing variable, we may be able to change a given integral into a much easier one, thus allowing us to determine the required antiderivative. When applying the method of substitution, it is essential to remember that the differential is an important part of the integral and must be changed as well. Common errors to avoid Do not leave behind any pieces of the integral. Do not even try to integrate the new version of the integral until all instances of the original variable are changed to the new one. Remember to switch back to the original variable when done. Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page

6 Learning questions for Section I -1 Review questions: 1. Eplain how the method of integration by substitution works.. Describe how we select the portion of the integrand to be used for the substitution. Memory questions: 1. Which formula describes the substitution method? 4. What is the general antiderivative of y tan?. Which rule of differentiation is inverted by the method of substitution?. Which part of the integrand is a good candidate for the u substitution? In questions 1-6 use the method of substitution to compute the given integral ( ) d 1 d. d Computation questions: 8 d 8. d z z 1dz 7. t t 1dt Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page d z 4z dz. e d 1 d e

7 1. e d e 4 t e 1. dt t 14. d ln 1. tanh d 16. coth d sinh d sinh cosh d 1 sinh d cosh d 1. cos e d. cos d 1 tan sec d e sec e tan e d cos 9 sin 1 cos d. 1 d In each of questions 7-0, determine the function whose derivative is given and whose graph contains the given point. sin 7. y, P,1 9. y 4 9, P, e e 8. y, P1, y 1, P 1,1 Theory questions: 1. What is the first method to consider when trying to integrate a function that is not a recognizable derivative?. For what type of functions does the method of substitution work? 4. For what functions is the method of substitution not effective?. Provide an eample of an integral for which the method of substitution does not work.. Does the method of integration by substitution work on every function? 6. If you are left with the original variable as a factor after making a substitution, can you move it outside the integral? Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page 7

8 7. Is it true that the method of substitution can be used whenever the integrand is a composite function? 9. Why is integration by substitution called a method? 8. Identify two reasons, besides computational errors, why the method of substitution may not work. Proof questions: 1. If F is an antiderivative for f, determine the indefinite integral f 1 sin d sec Application questions: 1. An object starts moving from the origin along the ais so that after t seconds t e its velocity is given by the function v.what is its position second t 1 e after starting?. A supply canister is dropped from a height of 000m and falls under the effect of gravity and friction so that its velocity is given by the function /0 v00 e t 1 in m/sec. Which formula describes the height of the canister t seconds after its release? Templated questions: 1. Identify an alternative substitution to compute any integral for which you used the method of substitution and determine whether this alternative is better, worse or equally valid and for what reason. What questions do you have for your instructor? Integral Calculus Chapter : Integration methods Section 1: Integration by substitution Page 8