Integration by inverse substitution

Transcription

1 Roberto s Notes on Integral Calculus Chapter : Integration methods Section 9 Integration by inverse substitution by using the sine function What you need to know already: How to integrate through basic formulae, substitution and parts. What you can learn here: How to integrate function that involve an inside portion of the form a b. Let me begin with a reminder of two basic algebra facts that many students forget or ignore, with occasional disastrous consequences. Technical fact In general, the square root of a sum or difference CANNOT be split along such a sum or difference: a b a b a b a b In general, the square root of the square of a number is NOT equal to the original, but to its absolute value: a a ; a a Because the square root is not a distributive operation. If you are not sure of what that means, wait until you take a course in linear algebra or another abstract algebra subject or notice that those attractive equalities do not work if 4 b 1. OK, so why bring it up now? a Because we shall now look at integrals that contains a root of the type and I do not want you to make any gross errors in handling them. Can t we use a substitution? a and Sometimes yes, but not always. When we cannot, we can make a smart use of trigonometric identities. Recall that for any value of, 1sin cos. This identity allows us to change a difference of squares into a single square, thus allowing us to eliminate the square root. Hopefully this will allow us to integrate the given function. Why not? Integral Calculus Chapter : Integration methods Section 9: Integration by inverse substitution by using sine Page 1

2 Strategy for integrals that contain a b For integrands involving a factor of this form: 1. Try the inverse substitution: b asin, bd acosd. Try to compute the new integral with suitable methods. 3. If this works, change back to the original variable, by using other trigonometric identities as necessary. Why do you call it an inverse substitution? Because in a usual substitution we set the new variable to be a function of the old, but here the old variable is set to be a function of the new one,. However, besides this reversal of roles, the method of change of variable works here just as it does normally. Eample: 4 d In this case a, b 1, so we use the substitution: In this way, our integral becomes: sin, d cos d 4 4 4sin cos 4 1sin cos d d d At this point we use the basic Pythagorean identity to complete the integral: 4 1sin cos 4 cos cos 4 cos cos d d d 4 cos d 1 cos d sin c At this point we need to return to the original variable and we do so by using again trig identities. To go back from to. we observe that: 1 sin sin sin We could apply this substitution to the rest of the formula, but that would generate a messy epression. There is an easier way! Remember that: sin sin cos 1 cos 1 sin We can use this more elegant epression and conclude that: 1 sin sin 4 1 c c 1 4 sin c Other eamples are very similar to this one and you will see them in the practice questions, as well as in the net two notes, which deal with the same method in slightly different situations. However, there are a few issues that need to be carefully noted. In the eample, I assumed twice that is not true, since, in general, cos cos cos. But we know that this cos. It turns out that because the domain of the integrand is bsin a bsin, our substitution implies that cos 0, so the absolute value is not needed. So, although this is an issue of relevance and important from the mathematical point of view, it turns out not to have an effect on what we do here or in other types of inverse substitutions. So, keep in mind that, when we use these substitutions, we can safely assume that the absolute value is not needed when square rooting a square. Integral Calculus Chapter : Integration methods Section 9: Integration by inverse substitution by using sine Page

3 In order to return to the original variable, I suggested that using inverse functions is not the best approach and that using identities instead leads to simpler and more elegant conclusion. To make this step easier, it may be useful to interpret the original substitution in terms of sides and angles of a right triangle. For instance, in the last eample, the relation sin can be interpreted by seeing as the side opposite to and representing the hypotenuse, as shown here. In this way, the remaining side can be obtained through the Pythagorean Theorem and, if everything was done properly, it will consist of the same epression we had in the original integrand. Once we have all three sides, the basic trig ratios can be used to identify any trig function appearing in the solution. This same approach can and will be used for the other two inverse substitutions you will see in the net sections. This same method can be used even when the integrand involves a root of the form a b u, with u f being any function, as long as the resulting new integral can be evaluated. In that case the substitution to use is bu asin, b f ' d a cos d. In particular, if the square root is of the type a c b, the method can still be used, after completing the square and changing it to the form h b k. I think I need one more eample! 4 I will be generous and give you two! Eample: 84 3/ Integral Calculus Chapter : Integration methods Section 9: Integration by inverse substitution by using sine Page 3 d The quadratic inside a fractional power suggests the use of inverse trig substitution, but we first need to complete the square: We now let: so that: d d 3/ 3/ d / 3/ d 1 sin, d 1 cos d d 1 cosd 1 1 1sin d cos sec d 3/ 3/ 1 cosd 1 cosd 1 1sin 3/ 1 1 cos tan c To return to we use a triangle model as follows: 1 sin sin 1

4 This tells us that: tan 1 so that our integral is given by: 1 1 Eample: 1 d c Here we need to use an inverse trig substitution with a function of. We set: This can be represented as: 1 1 sin, sin, d sin cos d By implementing this inverse substitution we obtain: 1 1sin d sin cos d cos d sin 1 sin c sin cos c The substitution and the triangle model allow us to write this in terms of the original variable as: 1 Wow! This can become quite tricky! 1 d 1 sin 1 c Computationally, yes! But that is why you have been told repeatedly to become familiar with basic algebra and trigonometric formulae and operations. This is the time to put them all to work and intensely. If you feel scared, remember that it is just a matter of being uncertain: practice, practice, practice and the jitters and the errors will go away and you feel like Superman! 1 Integral Calculus Chapter : Integration methods Section 9: Integration by inverse substitution by using sine Page 4

5 Summary To deal with integrands that contain a square root of the form integral can be computed. a b, we use the inverse substitution b asin, bd a cos d and hope that the resulting To return to the original variable, standard trigonometric identities and a suitable triangle model allow us to obtain an elegant formula for the integral. Watch the algebra as you do the substitutions, both ways. Common errors to avoid Use the triangle model to return to the original variable, rather than constructing convoluted compositions of trig functions and their inverses. Learning questions for Section I -9 Review questions: 1. Describe when and how to use the method of integration by inverse substitution by using the sine function. Memory questions: 1. What is the substitution used to integrate functions containing a factor of the form a b? Integral Calculus Chapter : Integration methods Section 9: Integration by inverse substitution by using sine Page 5

6 Evaluate the integrals proposed in questions 1-1 by using a suitable inverse substitution. Computation questions: d 5. 3/ d d d 6 a 4 d d d 4 d 4 ln d ln d 3 5 3e 5 e 3 3/ 3/ d d Theory questions: 1. For what type of functions is the method of inverse substitution a good option?. What first step should have been done if the root in the denominator were 6? 3. Which general geometric theorem is behind all formulae used in inverse trig substitution? What questions do you have for your instructor? Integral Calculus Chapter : Integration methods Section 9: Integration by inverse substitution by using sine Page 6