Section 6.2 Integration by Parts

Transcription

1 Section 6 Integration by Parts AP E XC II Version 0 Authors Gregory Hartman, PhD Department of Applied Mathema cs Virginia Military Ins tute Brian Heinold, PhD Department of Mathema cs and Computer Science Mount Saint Mary s University Troy Siemers, PhD Department of Applied Mathema cs Virginia Military Ins tute Dimplekumar Chalishajar, PhD Department of Applied Mathema cs Virginia Military Ins tute Editor Jennifer Bowen, PhD Department of Mathema cs and Computer Science The College of Wooster

2 Copyright 04 Gregory Hartman Licensed to the public under Crea ve Commons A ribu on-noncommercial 30 United States License

3 Contents Preface Table of Contents iii v 5 Integra on 85 5 An deriva ves and Indefinite Integra on 85 5 The Definite Integral Riemann Sums The Fundamental Theorem of Calculus 55 Numerical Integra on 33 6 Techniques of An differen a on 47 6 Subs tu on 47 6 Integra on by Parts Trigonometric Integrals Trigonometric Subs tu on Par al Frac on Decomposi on Hyperbolic Func ons L Hôpital s Rule Improper Integra on 3 7 Applica ons of Integra on Area Between Curves Volume by Cross-Sec onal Area; Disk and Washer Methods The Shell Method Arc Length and Surface Area Work Fluid Forces Sequences and Series Sequences Infinite Series Integral and Comparison Tests Ra o and Root Tests 49

4 85 Alterna ng Series and Absolute Convergence Power Series Taylor Polynomials Taylor Series 457 A Solu ons To Selected Problems A Index A

5 Chapter 6 Techniques of An differen a on 6 Integra on by Parts Here s a simple integral that we can t yet do: x cos x dx It s a simple ma er to take the deriva ve of the integrand using the Product Rule, but there is no Product Rule for integrals However, this sec on introduces Integra on by Parts, a method of integra on that is based on the Product Rule for deriva ves It will enable us to evaluate this integral The Product Rule says that if u and v are func ons of x, then (uv) = u v+uv For simplicity, we ve wri en u for u(x) and v for v(x) Suppose we integrate both sides with respect to x This gives (uv) dx = (u v + uv ) dx By the Fundamental Theorem of Calculus, the le side integrates to uv The right side can be broken up into two integrals, and we have uv = u v dx + uv dx Solving for the second integral we have uv dx = uv u v dx Using differen al nota on, we can write du = u (x)dx and dv = v (x)dx and the expression above can be wri en as follows: u dv = uv v du This is the Integra on by Parts formula For reference purposes, we state this in a theorem Theorem 48 Integra on by Parts Let u and v be differen able func ons of x on an interval I containing a and b Then u dv = uv v du, and x=b x=a u dv = uv b a x=b x=a v du 66

6 6 Integra on by Parts Let s try an example to understand our new technique Example 57 Integra ng using Integra on by Parts Evaluate x cos x dx S The key to Integra on by Parts is to iden fy part of the integrand as u and part as dv Regular prac ce will help one make good iden- fica ons, and later we will introduce some principles that help For now, let u = x and dv = cos x dx It is generally useful to make a small table of these values as done below Right now we only know u and dv as shown on the le of Figure 63; on the right we fill in the rest of what we need If u = x, then du = dx Since dv = cos x dx, v is an an deriva ve of cos x We choose v = sin x u = x v =? dv = cos x dx u = x du = dx v = sin x dv = cos x dx Figure 63: Se ng up Integra on by Parts Now subs tute all of this into the Integra on by Parts formula, giving x cos x = x sin x sin x dx We can then integrate sin x to get cos x + C and overall our answer is x cos x dx = x sin x + cos x + C Note how the an deriva ve contains a product, x sin x This product is what makes Integra on by Parts necessary The example above demonstrates how Integra on by Parts works in general We try to iden fy u and dv in the integral we are given, and the key is that we usually want to choose u and dv so that du is simpler than u and v is hopefully not too much more complicated than dv This will mean that the integral on the right side of the Integra on by Parts formula, v du will be simpler to integrate than the original integral u dv In the example above, we chose u = x and dv = cos x dx Then du = dx was simpler than u and v = sin x is no more complicated than dv Therefore, instead of integra ng x cos x dx, we could integrate sin x dx, which we know how to do A useful mnemonic for helping to determine u is LIATE, where L = Logarithmic, I = Inverse Trig, A = Algebraic (polynomials), T = Trigonometric, and E = Exponen al 67

7 Chapter 6 Techniques of An differen a on If the integrand contains both a logarithmic and an algebraic term, in general le ng u be the logarithmic term works best, as indicated by L coming before A in LIATE We now consider another example Example 58 Integra ng using Integra on by Parts Evaluate xe x dx S The integrand contains an algebraic term (x) and an exponen al term (e x ) Our mnemonic suggests le ng u be the algebraic term, so we choose u = x and dv = e x dx Then du = dx and v = e x as indicated by the tables below u = x v =? dv = e x dx u = x du = dx v = e x dv = e x dx Figure 64: Se ng up Integra on by Parts We see du is simpler than u, while there is no change in going from dv to v This is good The Integra on by Parts formula gives xe x dx = xe x e x dx The integral on the right is simple; our final answer is xe x dx = xe x e x + C Note again how the an deriva ves contain a product term Example 59 Integra ng using Integra on by Parts Evaluate x cos x dx S The mnemonic suggests le ng u = x instead of the trigonometric func on, hence dv = cos x dx Then du = x dx and v = sin x as shown below u = x v =? dv = cos x dx u = x du = x dx v = sin x dv = cos x dx Figure 65: Se ng up Integra on by Parts 68

8 6 Integra on by Parts The Integra on by Parts formula gives x cos x dx = x sin x x sin x dx At this point, the integral on the right is indeed simpler than the one we started with, but to evaluate it, we need to do Integra on by Parts again Here we choose u = x and dv = sin x and fill in the rest below u = x v =? dv = sin x dx u = x du = dx v = cos x dv = sin x dx Figure 66: Se ng up Integra on by Parts (again) ( x cos x dx = x sin x x cos x ) cos x dx The integral all the way on the right is now something we can evaluate It evaluates to sin x Then going through and simplifying, being careful to keep all the signs straight, our answer is x cos x dx = x sin x + x cos x sin x + C Example 60 Integra ng using Integra on by Parts Evaluate e x cos x dx S This is a classic problem Our mnemonic suggests le ng u be the trigonometric func on instead of the exponen al In this par cular example, one can let u be either cos x or e x ; to demonstrate that we do not have to follow LIATE, we choose u = e x and hence dv = cos x dx Then du = e x dx and v = sin x as shown below u = e x v =? dv = cos x dx u = e x du = e x dx v = sin x dv = cos x dx Figure 67: Se ng up Integra on by Parts No ce that du is no simpler than u, going against our general rule (but bear with us) The Integra on by Parts formula yields e x cos x dx = e x sin x e x sin x dx 69

9 Chapter 6 Techniques of An differen a on The integral on the right is not much different than the one we started with, so it seems like we have go en nowhere Let s s ck keep working and apply Integra on by Parts to the new integral, using u = e x and dv = sin x dx This leads us to the following: u = e x v =? dv = sin x dx u = e x du = e x dx v = cos x dv = sin x dx Figure 68: Se ng up Integra on by Parts (again) The Integra on by Parts formula then gives: ( ) e x cos x dx = e x sin x e x cos x e x cos x dx = e x sin x + e x cos x e x cos x dx It seems we are back right where we started, as the right hand side contains e x cos x dx But this actually a good thing Add e x cos x dx to both sides This gives Now divide both sides by : e x cos x dx = e x sin x + e x cos x e x cos x dx = ( e x sin x + e x cos x ) Simplifying a li le and adding the constant of integra on, our answer is thus e x cos x dx = ex (sin x + cos x) + C Example 6 Integra ng using Integra on by Parts: an deriva ve of ln x Evaluate ln x dx S One may have no ced that we have rules for integra ng the familiar trigonometric func ons and e x, but we have not yet given a rule for integra ng ln x That is because ln x can t easily be done with any of the rules we have learned up to this point But it can be done by a clever applica on of Integra on by Parts Set u = ln x and dv = dx This is a good, sneaky trick to learn 70

10 6 Integra on by Parts as it can help in other situa ons This determines du = (/x) dx and v = x as shown below u = ln x v =? dv = dx u = ln x du = /x dx v = x dv = dx Figure 69: Se ng up Integra on by Parts Pu ng this all together in the Integra on by Parts formula, things work out very nicely: ln x dx = x ln x x x dx The new integral simplifies to dx, which is about as simple as things get Its integral is x + C and our answer is ln x dx = x ln x x + C Example 6 Integra ng using Int by Parts: an deriva ve of arctan x Evaluate arctan x dx S The same sneaky trick we used above works here Let u = arctan x and dv = dx Then du = /( + x ) dx and v = x The Integra on by Parts formula gives arctan x dx = x arctan x x + x dx The integral on the right can be done by subs tu on Taking u = + x, we get du = x dx The integral then becomes arctan x dx = x arctan x u du The integral on the right evaluates to ln u + C, which becomes ln( + x ) + C Therefore, the answer is arctan x dx = x arctan x ln( + x ) + C 7

11 Chapter 6 Techniques of An differen a on Subs tu on Before Integra on When taking deriva ves, it was common to employ mul ple rules (such as, using both the Quo ent and the Chain Rules) It should then come as no surprise that some integrals are best evaluated by combining integra on techniques In par cular, here we illustrate making an unusual subs tu on first before using Integra on by Parts Example 63 Integra on by Parts a er subs tu on Evaluate cos(ln x) dx S The integrand contains a composi on of func ons, leading us to think Subs tu on would be beneficial Le ng u = ln x, we have du = /x dx This seems problema c, as we do not have a /x in the integrand But consider: du = dx x du = dx x Since u = ln x, we can use inverse func ons and conclude that x = e u Therefore we have that dx = x du = e u du We can thus replace ln x with u and dx with e u du Thus we rewrite our integral as cos(ln x) dx = e u cos u du We evaluated this integral in Example 60 Using the result there, we have: cos(ln x) dx = e u cos u du = eu( sin u + cos u ) + C = eln x( sin(ln x) + cos(ln x) ) + C = x( sin(ln x) + cos(ln x) ) + C Definite Integrals and Integra on By Parts So far we have focused only on evalua ng indefinite integrals Of course, we can use Integra on by Parts to evaluate definite integrals as well, as Theorem 7

12 6 Integra on by Parts 48 states We do so in the next example Example 64 Definite integra on using Integra on by Parts Evaluate x ln x dx S Once again, our mnemonic suggests we let u = ln x (We could let u = x and dv = ln x dx, as we now know the an deriva ves of ln x However, le ng u = ln x makes our next integral much simpler as it removes the logarithm from the integral en rely) So we have u = ln x and dv = x dx We then get du = (/x) dx and v = x 3 /3 as shown below u = ln x v =? dv = x dx u = ln x v = x 3 /3 du = /x dx dv = x dx Figure 60: Se ng up Integra on by Parts The Integra on by Parts formula then gives x ln x dx = x3 3 ln x x 3 3 = x3 3 ln x = x3 3 ln x x3 9 ( ) x 3 x3 = ln x 3 9 ( 8 = 3 ln 8 ) 9 = 8 3 ln 7 9 x dx x 3 dx ( 3 ln 9 ) 07 In general, Integra on by Parts is useful for integra ng certain products of func ons, like xe x dx or x 3 sin x dx It is also useful for integrals involving logarithms and inverse trigonometric func ons As stated before, integra on is generally more difficult than deriva on We are developing tools for handling a large array of integrals, and experience will 73

13 Chapter 6 Techniques of An differen a on tell us when one tool is preferable/necessary over another For instance, consider the three similar looking integrals xe x dx, xe x dx and xe x3 dx While the first is calculated easily with Integra on by Parts, the second is best approached with Subs tu on Taking things one step further, the third integral has no answer in terms of elementary func ons, so none of the methods we learn in calculus will get us the exact answer Regardless of these issues, Integra on by Parts is a very useful method, second only to subs tu on 74

14 Exercises 6 Terms and Concepts T/F: Integra on by Parts is useful in evalua ng integrands that contain products of func ons T/F: Integra on by Parts can be thought of as the opposite of the Chain Rule 3 For what is LIATE useful? Problems In Exercises 4 33, evaluate the given indefinite integral 4 x sin x dx 5 xe x dx 6 x sin x dx 7 x 3 sin x dx 8 xe x dx 9 x 3 e x dx 0 xe x dx e x sin x dx e x cos x dx 3 e x sin(3x) dx 4 e 5x cos(5x) dx 5 sin x cos x dx 6 sin x dx 7 tan (x) dx 8 x tan x dx 9 sin x dx 0 x ln x dx (x ) ln x dx x ln(x ) dx 3 x ln(x ) dx 4 x ln x dx 5 (ln x) dx 6 (ln(x + )) dx 7 x sec x dx 8 x csc x dx 9 x x dx 30 x x dx 3 sec x tan x dx 3 x sec x tan x dx 33 x csc x cot x dx In Exercises 34 38, evaluate the indefinite integral a er first making a subs tu on 34 sin(ln x) dx 35 sin( x) dx 36 ln( x) dx 37 e x dx 38 e ln x dx In Exercises 39 47, evaluate the definite integral Note: the corresponding indefinite integrals appear in Exercises π 0 π/4 π/4 π/ π/ ln 0 0 π 0 π/ π/ x sin x dx xe x dx x sin x dx x 3 sin x dx xe x dx x 3 e x dx xe x dx e x sin x dx e x cos x dx

15 Solutions to Odd Exercises x + 3x + ln x ) + C x 3 3 x + x ln x + + C 3 37 x 8x + 5 ln x + + C ( ) 39 7 tan x + C sin ( x 5 ) + C sec ( x /4) + C tan ( ) x 7 + C sin ( ) 4 x 5 + C 49 3(x 3 +3) + C 5 x + C 53 3 cos 3 (x) + C 55 7 ln 3x + + C 3 57 ln x + 7x C 59 x + ln x 7x + + 7x + C 6 tan (x) + C 63 ) 3 sin 3x 4 + C 65 9 ) 5 tan x+6 5 ln x + x C 67 x 9 ln x C 69 tan (cos(x)) + C 7 ln sec x + tan x + C (integrand simplifies to sec x) 73 x 6x C 75 35/5 77 /5 79 π/ 8 π/6 Sec on 6 T 3 Determining which func ons in the integrand to set equal to u and which to set equal to dv 5 e x xe x + C 7 x 3 cos x + 3x sin x + 6x cos x 6 sin x + C 9 x 3 e x 3x e x + 6xe x 6e x + C /e x (sin x cos x) + C 3 /3e x ( sin(3x) 3 cos(3x)) + C 5 / cos x + C 7 x tan (x) 4 ln 4x + + C 9 x + x sin x + C x 4 + x ln x + x x ln x + C 3 x ln ( x ) x + C 5 x + x (ln x ) x ln x + C 7 x tan(x) + ln cos(x) + C 9 5 (x )5/ (x )3/ + C 3 sec x + C 33 x csc x ln csc x + cot x + C 35 sin ( x ) x cos ( x ) + C 37 xe x e x + C 39 π / e 5 4e 4 47 /5 ( e π + e π) Sec on 63 F 3 F sin4 (x) + C 6 cos6 x 4 cos4 x + C 9 9 sin9 (x) + 3 sin7 (x) 3 sin5 (x) + sin3 (x) + C ( 8 cos(8x) cos(x)) + C ( 3 4 sin(4x) 0 sin(0x)) + C ( 5 sin(x) + 3 sin(3x)) + C tan 5 (x) 5 + C tan 6 (x) 6 + tan4 (x) 4 + C sec 5 (x) 5 sec3 (x) 3 + C 3 tan3 x tan x + x + C (sec x tan x ln sec x + tan x ) + C /35 3 /3 33 6/5 Sec on 64 backwards 3 (a) tan θ + = sec θ (b) 9 sec θ ( x x + + ln ) x + + x + C (sin x + x ) x + C x x ln x + x + C x x + /4 + 4 ln x + /4 + x + C = x 4x ln 4x + + x + C ( ) 3 4 x x /6 3 ln 4x + 4 x /6 + C = x 6x 8 ln 4x + 6x + C ( ) 5 3 sin x + C (Trig Subst is not needed) 7 7 x sec (x/ ) + C 9 x 3 + C (Trig Subst is not needed)