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Chapter 7: Principles of Integral Evaluation Summary: The primary focus of this chapter is to introduce and explain more advanced integration techniques. With all of the techniques that are introduced, however, the focus is usually to try and find a way to use the basic integral rules (or basic list of antiderivatives) that were introduced in Chapter 4. Many of the ideas in this chapter are methods to use algebraic manipulation or special substitutions so that integrals may be simplified into an integral resembling a basic integral rule. IDEA: At this point the reader should review the integration techniques from Chapter 4. Particular emphasis should be placed upon the basic antiderivatives (§4.2) and u-substitution (§4.3). One very powerful method that is introduced in this chapter is the method of integration by parts. This method can simply be thought of as the product rule for antiderivatives. Next various trigonometric integrals are considered and then some special trigonometric substitutions that can be used to simplify a variety of integrals that involve square roots and differences and sums of squares. In the case of rational integrands, the method of partial fraction decomposition may sometimes be used to simplify the integrand. In cases where antiderivatives cannot be expressed using simple functions, the idea of numerical integration is considered for definite integrals. Finally at the end of the chapter, integrals are considered that may involve either infinite limits of integration or integrands that have an infinite discontinuity (such as a vertical asymptote). OBJECTIVES: After reading and working through this chapter you should be able to do the following: 1. be familiar with the basic integration techniques (see §4.2 and §7.1) 2. be aware of the different resources for evaluating integrals (§7.1) 3. be able to use integration by parts and recognize when it will be useful (§7.2) 4. manipulate integrands using trigonometric identities and formulas to obtain more simple integrals (§7.3) 5. be able to use trigonometric substitution and recognize when it can be used (§7.4) 141 basic antiderivatives → §4.2 142 6. use partial fraction expansion to evaluate integrals (§7.5) 7. use CAS and tables of integrals where appropriate (§7.6) 8. use numerical integration techniques where appropriate (§7.7 and also §4.4) 9. identify definite integrals as improper and determine whether they converge (to a value) or diverge (§7.8) 7.1 An Overview of Integration Methods PURPOSE: To review general integration techniques. REVIEW: 1. basic antiderivatives (§4.2) 2. u-substitution (§4.2 and §4.9) 3. F.T.C (§4.6) There are no new ideas in this section but this section does serve as a good checkpoint for the reader. There are several concepts and methods that the reader should be comfortable with before attempting to learn the material in this chapter. First, this section should serve as a reminder to the reader of all of the integration rules that have been introduced in the text so far. It is also a good idea at this point to review the integration techniques that have been introduced so far. This essentially means to review the basic integral formulas that are related to polynomials, logarithms, exponential functions, trigonometric functions, hyperbolic functions and inverse trigonometric or hyperbolic functions. Also, the reader should be comfortable with the idea of u-substitution both in indefinite integrals and with definite integrals. At this point, evaluating definite integrals using the Fundamental Theorem of Calculus should seem more natural. IDEA: The emphasis throughout this chapter will be to use tables, transformations and other techniques to evaluate integrals that do not qualify as any of the “basic” integrals seen in Chapter 4. A free CAS available online is Maxima: → maxima.sourceforge.net The emphasis in this chapter is going to be on using tables and transformation methods. For example, appropriate transformation methods can simplify an integral so that the reader only has to use the basic integral rules that were introduced in Chapter 4. Take some time now to locate the integral tables at the beginning and end of the book. Before starting the chapter the reader may also want to identify a CAS that they can use (for example, Maple, Mathematica and Derive are all good choices). This is important for two reasons. First, using a CAS can be a valuable tool for checking your answers as you progress through the material. On the other hand, there are some problems that are intended to be done using a CAS. Checklist of Key Ideas: review basic integration techniques technology; CAS tables of integrals transformation methods 143 7.2 Integration by Parts PURPOSE: To develop a versatile integration technique that allows many integrals to be rewritten involving a simpler integrand. Integration by parts can be thought of as the product rule for antiderivatives. In fact, the product rule for derivatives gives a good way to remember this formula. Here is the differential version of the product rule. d(uv) = u dv + v du Then the following is the integral version. Z (u dv + v du) = Solve for Z Z Z u dv + Z v du = uv u dv to find the integration by parts formula. u dv = uv − Z v du IDEA: Integration by parts is really just an antiderivative version of the product rule. This simple expression of the integration by parts formula can be deceptive, however. The variables u and v are actually supposed to represent functions of x. So the differential form of the product rule could really be written as follows. d u(x)v(x) = u(x) v′ (x)dx + v(x) u′ (x)dx Then the integral version would be as shown below. Z u(x)v′ (x) dx + Z v(x)u′ (x) dx = u(x)v(x) So the integration by parts formula is as follows. Z u(x)v′ (x) dx = u(x)v(x) − Z v(x)u′ (x) dx CAUTION: Do not forget the constant of integration with indefinite integrals. Another important thing is to not forget the constant of integration. As long as there is still an indefinite integral involved, there should be a C as a part of the antiderivative. On the other hand, it is important to remember with definite integrals that even the uv term needs to be evaluated at the limits of integration. Zb a x=b Zb u(x)v (x) dx = u(x)v(x) − v(x)u′ (x) dx ′ x=a a Note: u = u(x) dv = v′ (x) dx 144 IDEA: Integration by parts may allow a complicated integrand to be rewritten as a more basic integrand. R The basic idea behind integration by parts is that the integral u dv may not be a function that can be integrated by our basic techniques. But through an appropriate R choice of u and dv, the resulting integral of v du may be easier to handle. The toughest part about this method is picking u and dv. Finding du and v can be accomplished by taking the derivative of u and integrating dv respectively. IDEA: du is found by taking the derivative of u. v is found by integrating dv. LIATE Method: pick u towards the top ⇓ L - logarithmic I - inverse trigonometric A - algebraic, polynomial T - trigonometric E - exponential ⇓ pick dv towards the bottom The LIATE method tries to make picking u and dv easier by following a particular pattern. For example, the types of functions represented by LIATE (Logarithmic – Inverse Trigonometric – Algebraic – Trigonometric – Exponential) can be used as a guide for picking u and dv (see margin to the left). Algebraic functions really stand for polynomials or rational functions. tabular integration by parts Tabular integration by parts can simplify the process of using integration by parts. After u = u(x) and dv = v′ (x)dx are initially picked, tabular integration by parts creates two columns. In the “u” column, derivatives of u are calculated. In the “dv” column, dv terms are integrated over and over. This process should simply be continued until the derivative of u either is zero or it repeats itself as the original u. Then the appropriate terms should be matched from each column with the appropriate signs and a constant of integration added to the end. If the original R u and dv are repeated, then u dv will need to be solved for algebraically. IDEA: Integration by parts can be tried on any integral such as letting u = f (x) and dv = dx. R f (x) dx by Another difficult thing to decide is whether or not to even use integration by parts. It is a very versatile method and very often just trying it can lead to an integral with a basic formula. One choice for u and dv that can always be tried with the R integral f (x) dx is u = f (x) and dv = dx. On the other hand, there are times when integration by parts must be used. Some obvious examples are when there is a product of functions of different types that does not give rise to a basic integral form. Integration by parts also gives rise to the following useful trigonometric reduction formulas. Z 1 n−1 sinn x dx = − sinn−1 x cos x + n n Z sinn−2 x dx for n ≥ 2 n−1 1 sin x cosn−1 x + cosn−2 x dx for n ≥ 2 n n These formulas are used in the next section on trigonometric integrals (see §7.3). Z cosn x dx = Z 145 Checklist of Key Ideas: reverse of product rule guidelines for integration by parts; choosing u and dv LIATE tabular integration by parts integration by parts and definite integrals reduction formulas 7.3 Integrating Trigonometric Functions PURPOSE: To develop integration techniques for integrands that are comprised of powers or products of trigonometric functions. Some integrals are comprised solely of trigonometric functions. These are integrals where powers of trigonometric functions occur either by themselves or as products. Most often the techniques that are used here are reduction techniques. The goal is to transform the integrand into a simpler expression for which the basic methods from Chapter 4 can be applied. Sometimes this takes the form of using reduction formulas or other times, an appropriate trigonometric identity can be used with u-substitution to simplify the integrand. reduction formulas see §7.2 IDEA: The goal is to reduce or transform an integrand to the point where basic antiderivative techniques may be used. In many cases, trigonometric identities and definitions can be used to simplify a given integral. For example, the integral Z sin2 x cos2 x dx does not have a simple antiderivative and u-substitution does not work directly. However, the Pythagorean identity and the cosine reduction formula can be used on this integral. Using sin2 x = 1 − cos2 x leads to the following. Z (cos2 x − cos4 x) dx Then the trigonometric reduction formulas (see §7.2) can be used to evaluate these two integrals. Z cos2 x dx = 1 1 sin x cos x + 2 2 Z 1 dx = (sin x cos x + x) +C 2 Pythagorean Identity sin2 u + cos2 u = 1 146 R cos4 x dx= = 3 1 sin x cos3 x + 4 4 Z cos2 x dx 3 1 sin x cos3 x + (sin x cos x + x) +C 4 8 These are then combined to give the antiderivative below. 1 1 x sin2 x cos2 x dx = − sin x cos3 x + sin x cos x + +C 4 8 8 Z IDEA: Integrals involving trigonometric functions may often be evaluated in more than one way. There is often more than one way Z to integrate trigonometric integrals. For example, consider again the integral sin2 x cos2 x dx. By using a double-angle identity for sine we have sin2 x cos2 x = (sin x cos x)2 = 1 2 1 sin (2x) = (1 + cos (4x)) 4 8 and then it is a simple matter to integrate as follows. 1 8 Note: Many of these techniques make use of basic trig definitions, double angle-identities, and the Pythagorean identity. Z (1 + cos (4x)) dx = x sin (4x) + +C 8 32 General techniques are given for the following types of integrals in this section: Z sinm x cosn x dx Z tann x dx Z sin (mx) cos (nx) dx Z secn x dx Z tanm x secn x dx where m and n are positive integers. Checklist of Key Ideas: reduction formulas simplifying using trigonometric identities products of sines and cosines to integer powers integrating powers of tangent and secant trigonometric identities and u-substitution 147 7.4 Trigonometric Substitutions PURPOSE: To develop integration techniques to handle integrands that involve square roots. Some integrals involve square root terms where u-substitution is ineffective. In some of these cases, a special type of substitution called a trigonometric substitution may be used to simplify the square root so that the integral has a more simple form. The idea with each trigonometric substitution is to make use of the Pythagorean identity or one of its forms in order to simplify the radical. Three Forms of the Pythagorean identity: sin2 u + cos2 u = 1 tan2 u + 1 = sec2 u 1 + cot2 u = csc2 u IDEA: Trigonometric substitution involves u-substitution combined with the Pythagorean identity. √ For integrals that have a2 − x2 , a substitution of x = a sin u will often work. Notice that dx = a cos u du. This substitution can be remembered since this square root looks similar to the one which is involved in the integral for sin−1 x. (Incidentally, a substitution of x = a cos u should also work.) The substitution will get rid of the square root in the following way. p √ a2 − x2 = pa2 − a2 sin2 u = a√ 1 − sin2 u = a cos2 u = ±a cos u √ For integrals that have a2 + x2 or a2 + x2 , a substitution of x = a tan u will often work. Simplification will come from an alternative form of the Pythagorean identity: √ a2 − x2 → try x = a sin u a2 + x2 → try x = a tan u 1 + tan2 x = sec2 x This substitution x = a tan u can be remembered since the form a2 + x2 looks similar to the integral that is related to tan−1 x. Completing the square on some quadratic terms can lead to terms that look like a2 + (x − h)2 . The appropriate substitution should be x − h = a tan u and then dx = a sec2 u du. √ For integrals that have x2 − a2 , a substitution of x = a sec u will often work. Simplification will come by rearranging the version of the Pythagorean identity used above. 1 + tan2 x = sec2 x −→ sec2 x − 1 = tan2 x This substitution x = a sec u can be remembered since the form similar to the integral that relates to sec−1 x. √ x2 − a2 looks √ x2 − a2 → try x = a sec u 148 Checklist of Key Ideas: method of trigonometric substitution √ integrals involving a2 − x2 √ integrals involving a2 + x2 √ integrals involving x2 − a2 integrals involving irreducible quadratic denominators, a(x − h)2 + c2 7.5 Integrating Rational Functions by Partial Fractions PURPOSE: To develop a technique to assist in integrating rational functions. When a proper rational function is in an integral, it may sometimes be rewritten using partial fraction decomposition. This amounts to taking a fraction and writing it as a sum of fractions that do not have a common denominator. Often the resulting pieces or partial fractions are simpler and easier to handle in an integral. Partial Fraction Decomposition: 1. write as a proper rational function 2. factor the denominator 3. solve for constants in numerator To be able to use partial fraction decomposition requires three things. First, the rational function should be written as a proper rational function (degree of the polynomial on the bottom is greater than the top). Second, the denominator of the rational expressions should be factored. Last, after the partial fraction decomposition is performed, the values of the resulting constants in the numerators need to be determined. The quadratic equation is useful for factoring. If ax2 + bx + c = 0 then √ −b ± b2 − 4ac r1,2 = 2a ⇓ ax2 + bx + c = (x − r1 )(x − r2 ) There are several cases considered here. Each case depends upon the types of factors in the denominator. 1. distinct linear factors A B ax + b = + (x − r1 )(x − r2 ) x − r1 x − r2 2. repeated linear factors ax + b A B = + (x − r)2 x − r (x − r)2 ax2 + bx + c A B C = + + (x − r)3 x − r (x − r)2 (x − r)3 149 3. distinct quadratic factors b3 x3 + b2 x2 + b1 x + b0 A1 x + A0 B1 x + B0 = + (a2 x2 + a1 x + a0 )(c2 x2 + c1 x + c0 ) a2 x2 + a1 x + a0 (c2 x2 + c1 x + c0 ) 4. repeated quadratic factors A1 x + A0 B1 x + B0 b3 x3 + b2 x2 + b1 x + b0 = + (a2 x2 + a1 x + a0 )2 a2 x2 + a1 x + a0 (a2 x2 + a1 x + a0 )2 There are two ways to evaluate the constants in the numerators after partial fractions have been determined based upon the factors of the denominator. A new numerator involving the new, unknown coefficients can be found by recombining the partial fraction terms into one rational term. Then several equations for the coefficients can be obtained by equating the numerators of the terms with the full denominators. Then the coefficients can either be determined by trying to match the coefficients of the terms with the corresponding powers of x or by substituting different values for x into the equations. Checklist of Key Ideas: partial fraction decomposition proper rational function determining coefficients by matching terms determining coefficients by substitution linear factors quadratic factors quadratic formula using long division (or synthetic division) when possible 7.6 Using Computer Algebra Systems and Tables of Integrals PURPOSE: To introduce alternative methods for evaluating integrals. There are an infinite variety of integrals that can arise. It will not always be a simple process to find a substitution or transformation that will allow a simple antiderivative to be found. Using a table of integrals can shorten the effort required to evaluate a given integral. Likewise, a Computer Algebra System (CAS) can greatly shorten the time required to evaluate an integral. It is still useful to evaluating constants: 1. substitution 2. matching terms 150 understand how integration works, however, as sometimes the results that are returned by a table of integrals or CAS may seem more mysterious then the original integral. CAUTION: Sometimes no perfect match or no possible match for an integrand can be found → u-substitution may help. When using a table of integrals, sometimes a perfect match cannot be found for the integral that is being considered. In some cases, an appropriate u-substitution can be used to find a form in the tables that does match. One example of this is integrals that involve fractional powers of x. In these cases, a good substitution to try is u = x1/r where r is the LCD of all of the fractional powers of x. This can have the effect of removing all of the fractional powers of x so that other methods may then be employed. Checklist of Key Ideas: be familiar with what is in the table of integrals perfect matches and imperfect matches using substitution to create a match integrals involving fractional powers of x rational trigonometric functions understanding and using integration with CAS 7.7 Numerical Integration; Simpson’s Rule PURPOSE: To develop techniques for approximating the value of a definite integral. see also §4.4 midpoint method Definite integrals will often arise that cannot be evaluated using the Fundamental Theorem of Calculus since a simple antiderivative is not known for the integrand. In these cases, numerical integration may be the only way to find a value for the integral. Some forms of numerical integration have already been encountered: left endpoint approximations, right endpoint approximations and midpoint approximations. More sophisticated methods are introduced here. The midpoint method has already been introduced (i.e., the rectangle method with heights of rectangles determined at the midpoints of the intervals). This is sometimes called the tangent line approximation because it is equivalent to drawing a tangent line to a curve at the midpoint of each subinterval. The trapezoidal area underneath this tangent line is the same as the area of the rectangle determined by the midpoint method. This is due to the symmetry of the tangent line at the midpoint of the subinterval. 151 The trapezoidal approximation is the average of the left and right endpoint approximations. In essence, each subinterval is approximated by the area of a trapezoid. The trapezoid is formed by connecting the function values at the left and right endpoints of the subinterval with a straight line. Then the area of the resulting trapezoid is used to approximate the area under the curve on the subinterval. trapezoidal method Simpson’s rule is the result of drawing a quadratic function over the interval and then finding the area under the quadratic function on the interval. The quadratic function that is used is required to go through the function values at the left endpoint, right endpoint and the midpoint of the subinterval. As it turns out, the following is true. Simpson’s rule 1 1 Sn = (Ln/2 + 4Mn/2 + Rn/2 ) = (Tn/2 + 2Mn/2 ) 6 3 Usually Simpson’s Rule is applied on an even number of intervals so that the midpoints occur at the nodes of the partition that is being used and extra function values do not need to be calculated. Thus, in the formula given above, n and n/2 should be integers. Checklist of Key Ideas: trapezoidal approximations midpoint approximations; tangent line approximation Simpson’s Rule absolute errors; |EM |, |ET | and |ES | overestimates and underestimates error bounds 7.8 Improper Integrals PURPOSE: To evaluate integrals with discontinuous integrands or infinite limits of integration. When an infinite discontinuity (such as a vertical asymptote) or an infinite limit of integration appears within a definite integral then we have an improper integral. In some of these cases, a value for the integral can be found. Then the definite integral is said to converge. In cases where an improper integral does not evaluate to some finite number, then the integral is said to diverge. To evaluate improper integrals, the definite integral is rewritten using a dummy variable such as b. There are two situations that may arise: if the limits of integration are infinite or if the integrand is discontinuous in some way (particularly at the endpoints of the interval). In the case of infinite limits of integration, the converge vs. diverge Types of Improper Integrals: 1. infinite limit of integration 2. discontinuous integrand 152 limits are rewritten using a dummy variable (such as b). The integral is evaluated with the dummy variable and then the limit is taken as the dummy variable (i.e., b) approaches the appropriate limiting value. For example, the following integral has an infinite upper limit of integration. Z b 0 e−x dx = lim Z b b→∞ 0 e−x dx IDEA: First, evaluate the integral with a limit of integration of b. Then take the limit involving b. The definite integral is first evaluated using a limit of integration of b and then the limit involving b is evaluated. If the limit exists, then the improper integral is said to converge. Otherwise, the integral is said to diverge. In the case where the integrand is discontinuous, the point of discontinuity may occur either at an endpoint or in the middle of the interval. If the discontinuity occurs at an endpoint then a dummy variable is again used in place of the appropriate limit of integration. The following integral has an integrand that is discontinuous as it approaches the lower limit of integration. x=1 Z 1 Z 1 √ 1/2 −1/2 −1/2 = lim 2 − 2 b = 2 x dx = lim x dx = lim 2x b→0+ b 0 b→0+ x=b b→0+ Again, the integral is evaluated first with the dummy variable, then the limit is evaluated. If the limit exists and is finite then the improper integral converges, otherwise it diverges. IDEA: More than one dummy variable and more than one integral may be required if 1. there are two infinite limits of integration. 2. there is a discontinuity in the middle of the interval. discontinuity in middle of interval If a discontinuity occurs in the middle of the interval then the integral needs to be written as two integrals and two different limits taken. If either of the resulting integrals diverges then the whole improper integral will diverge. For example, Z 1 1 −1 x dx = lim Z a 1 a→0− −1 x dx + lim Z 1 1 b→0+ b x dx which diverges because both integrals diverge. Notice that in writing two different integrals, two different dummy variables are used. A similar situation can also arise with infinite limits of integration. For example, Z ∞ −∞ 2 e−x dx = lim Z 0 a→−∞ a 2 e−x dx + lim Z b b→∞ 0 2 e−x dx Again, notice that two dummy variables are used, one for each infinite limit of integration. In this example, the fact that x = 0 was chosen as the “middle” limit of integration was arbitrary, although the same number had to be used in both integrals. 153 Checklist of Key Ideas: improper integrals; infinite discontinuities; infinite limits of integration rewriting as a proper integral and taking a limit convergence and divergence applications to arc length and surface area 154 Chapter 7 Sample Tests 1 cos9 x +C 9 1 (b) − cos9 x +C 9 Section 7.1 1. Evaluate Z (a) (8 − 2x)5 dx. (8 − 2x)6 +C 6 −(8 − 2x)6 (b) +C 6 −(8 − 2x)6 (c) +C 12 −(8 − 2x)6 +C (d) 3 Z √ 2. Evaluate 4x + 9 dx. (c) 9 cos9 x +C (a) 1 √ +C 8 4x + 9 2 +C (b) √ 4x + 9 1 (c) (4x + 9)3/2 +C 6 1 (d) (4x + 9)3/2 +C 2 (d) −9 cos9 x +C 7. Z Z (c) (d) e 8. Z ex dx 5 + ex x +C +C = 5ex +C 5 + ex (b) 5 ln (5 + ex ) +C (a) 9. Z (c) 2 ln |4 − x2 | +C 1 x2 e +C 2 cos x sin x e dx = (a) cos x ecos x +C (b) −ecos x +C (c) − sin x ecos x +C (d) −xecos x +C 8 cos x sin x dx = 4x dx = 4 − x2 (b) −2 ln |4 − x2 | +C 2 2 ex +C 5 + ex (a) 4 ln |4 − x4 | +C x ex dx = (c) ex +C 2 1 (d) x2 ex +C 2 6. 4 √ (d) 2 Z e (a) 3 cos (x2 ) +C (b) 2ex +C 5. 3e x +C 4x3/2 (c) ln (5 + ex ) +C (a) Z (b) 3x sin (x2 ) dx = (c) −6 cos (x2 ) +C 3 (d) − cos (x2 ) +C 2 4. e x +C 2x3/2 √ x (b) 6 cos (x2 ) +C Z √ (a) √ (a) 3. √ e x √ dx = 2 x (d) 4 ln |4 − x2 | +C 10. Z 4 sinh2 x cosh x dx = (a) 4 sinh3 x +C 3 (b) 12 sinh3 x +C (c) 4 sinh2 x +C (d) 2 sinh2 x +C 11. Answer true or false. In evaluating for u is u = x2 . Z 12. Answer true or false. In evaluating choice for u would be sin x. 2 x 7x dx a good choice Z cos6 x sin x dx a good 155 13. Answer true or false. In evaluating choice for u would be ex + 8. 14. Answer true or false. In evaluating for u would be sin x. Z 15. Answer true or false. In evaluating choice for u would be x4 . Z 1 1 cos (2x) − x2 cos (2x) +C 2 4 1 1 (b) x sin (2x) + cos (2x) − x2 cos (2x) +C 4 2 1 (c) x sin (2x) + cos (2x) − x2 cos (2x) +C 2 1 (d) 2x sin (2x) + cos (2x) − x2 cos (2x) +C 2 ex (ex + 8) dx a good (a) x sin (2x) + 4 sin x dx a good choice cos x Z x3 cos (x4 ) dx a good 6. Z Section 7.2 1. Z (a) 2. Z 7. (d) 8. (d) 3x3 e3x +C x cos (9x) dx = x2 sin (9x) +C 18 sin (9x) (b) +C 9 cos (9x) x sin (9x) + +C (c) 81 9 cos (9x) x cos (9x) (d) + +C 81 9 (a) 4. Z 2x sin x dx = 9. 5e4x sin (3x) dx = (a) 5e4x 3 sin (3x) + cos (3x) +C (b) 5e4x 3 sin (3x) − cos (3x) +C 5 (c) e4x 3 sin (3x) − 3 cos (3x) +C 7 1 (d) e4x 4 sin (3x) − 3 cos (3x) +C 5 Z 1 4xe6x dx = (b) 224.2 (c) 224.7 (d) 225.1 10. Z 3 1 x2 ln x dx = (a) 7.00 (b) 2 sin x − 2 cos x +C (b) 7.03 (d) 2 cos x + 2x cos x +C (d) 6.92 2x2 sin (2x) dx = √ cos−1 (6x) +C 6 (a) 223.9 (c) 6.96 (c) 2 cos x − 2x cos x +C 5. Z 0 (a) 2 sin x − 2x cos x +C Z sin−1 (6x) dx = 1 − 36x2 +C 6 √ 1 − 36x2 +C (b) x cos−1 (6x) + 6 (c) 6 cos−1 (6x) +C (c) 27e3x +C 3. (d) x2 ln (2x) − x2 +C (a) x sin−1 (6x) + e3x (9x2 − 6x + 2) +C 3 (b) xe3x +C Z Z 9x2 e3x dx = (a) 1 2 x2 x ln (2x) − +C 4 4 5 2 5x2 (b) x ln (2x) − +C 2 4 (c) 5x2 ln (2x) − 5x2 +C (a) x e6x dx = e6x (6x − 1) +C 6 e6x (6x − 1) +C (b) 36 1 (c) x2 e6x +C 2 e6x (d) +C 6 5x ln (2x) dx = 11. Z 3 1 cos (ln x) dx = 156 1 3x + sin (2x) + sin 4x +C 2 8 3x 3 1 − sin (2x) + cos3 x sin x +C (b) 2 4 8 3 3 (c) sin (2x) + sin x cos x +C 4 3x 3 + cos (2x) + sin3 x cos x +C (d) 2 4 (a) 1.57 (a) (b) 1.52 (c) 1.48 (d) 1.42 12. Answer true or false. Z π /4 0 x sin (2x) dx = 1/4. 13. Answer true or false. Z 3π /4 14. Answer true or false. Z 1 π /4 0 15. Answer true or false. Z 1 0 x tan x dx = 1. 5. Z 4xe−3x dx = 0. 1. 13 cos x sin x dx = cos14 x +C 14 cos14 x +C (b) − 14 cos14 x (c) +C 13 cos14 x +C (d) − 13 (a) 2. Z 2 sin2 (5x) dx = sin (10x) (a) x − +C 10 (b) 2x − 2 sin (5x) +C 6. Z ln |cos (9x)| +C 9 ln |cos (9x)| (b) − +C 9 tan2 (9x) +C (c) 18 tan2 (9x) +C (d) − 18 7. Z Answer true or false. sin (6x) cos (4x) dx = − 8. Z − sin3 (2x) dx = (a) (b) (c) (d) 4. Z 1 1 cos (2x) + sin3 (2x) +C 2 6 1 1 cos (2x) − cos3 (2x) +C 2 6 1 1 − cos (2x) + cos3 (2x) +C 2 6 1 1 − cos (2x) − sin3 (2x) +C 2 6 4 cos4 x dx = sec2 (5x + 8) dx = tan (5x + 8) +C 5 tan (5x + 8) +C (b) 5x + 8 tan (5x + 8) (c) − +C 5 tan (5x + 8) (d) − +C 5x + 8 (a) sin (5x) +C 5 sin (5x) (d) x − +C 10 Z tan (9x) dx = (a) (c) x − 3. x sin (16x) − +C 8 128 x cos (16x) (b) − +C 8 128 x sin (8x) (c) − +C 8 128 x cos (8x) (d) − +C 8 128 (a) ln (x2 + 20) dx = 1. Section 7.3 Z sin2 (4x) cos2 (4x) dx = 9. Z csc (6x) dx = ln |tan (3x)| +C 6 ln |tan (3x)| (b) − +C 6 ln |tan (6x)| +C (c) 6 ln |tan (6x)| (d) − +C 12 (a) cos (2x) cos (10x) − +C 4 20 157 10. Answer true or false. 11. Z Z tan11 x sec2 x dx = tan12 x +C 12 (b) 1.388 tan x sec5 x dx = (c) 3.271 (a) sec5 x +C (d) 1.398 sec6 x (b) +C 6 sec5 x +C (c) 5 (d) sec6 x +C 12. Answer − (a) 1.381 true 4. Z 3 2 (−x)2 dx p = (−x)2 − 1 (a) 14.941 (b) 0.077 (c) 0.093 or Z false. cot6 (4x) +C 24 13. Answer true or false. Z π /4 Z π /4 0 15. Answer true or false. 5. Z 2 4dx √ = 4 x x2 + 5 (a) 2.003 1 0 14. Answer true or false. (d) 17.01 cot5 (4x) csc2 (4x) dx = Z π /3 tan2 (6x) dx = 1.00. (b) 1.684 (c) 1.692 sec2 (x) dx = 0. −π /3 (d) 1.698 tan (2x) dx = 0. 6. Z 1 0 x3 dx = (9 + x2 )5/2 (a) 0.00086 Section 7.4 1. Z p (c) 0.00041 9 − x2 dx = √ x x 9 − x2 + sin−1 +C 6 9 √ x 9 − x2 9 −1 x (b) + sin +C 2 2 3 √ x x 9 − x2 + 2 sin−1 +C (c) 2 3 √ x x 9 − x2 (d) + 3 sin−1 +C 2 3 Z dx √ = 2. 5 − x2 √ sin−1 5 x (a) +C 5 −1 sin (5x) (b) +C 5 √ ! 5x +C (c) sin−1 5 √ √ ! 5 −1 5x (d) sin +C 5 5 (a) 3. (b) 0.00031 Z 0 −1 3ex p 4 − 2e2x dx = (d) 0.00082 7. Z 4√ 2 3x − 2 dx x (a) 4.72 1 = (b) 4.79 (c) 3.12 (d) 3.17 8. Z 2π π cos θ d θ p = 4 − sin2 θ (a) 1 (b) 0 (c) −1 (d) 4 9. Z dx = x2 + 3x + 9 √ ! √ √ 2 3 −1 2 3 x + 3 3 tan +C (a) 9 9 2x + 3 2 +C (b) tan−1 3 3 2 (c) tan−1 (2x + 3) +C 3 158 (d) 2 tan−1 11 2x + 3 11 10. Z Answer true or false. Z dx dx √ = p 2 x − 4x + 2 (x − 2)2 − 2 11. Z Answer true or false. Z Z Z dx dx dx dx = −7 + 2 2 x 3 x − 7x + 3 x Z 1 dx √ = 12. 0 2 5x − x2 (a) 0.46 (b) 0.47 (c) 1 (d) 0 13. Z 2 1 3 2 + x−1 x 3. Which of the following is the partial fraction decomposition 3x2 − 2x + 2 of 2 ? (x + 2)(x − 1) (d) +C dx = 3x2 + 9x + 4 1 2 + x2 + 2 x − 1 2 1 + (b) 2 x +2 x−1 2x 1 (c) 2 + x +2 x−1 x 2 (d) 2 + x +2 x−1 Z 4x + 10 dx = 4. x2 + 5x − 6 (a) (a) 2 ln |x2 + 5x − 5| +C (b) 2 ln |x2 + 5x − 6| +C (a) 0.021 (c) 2 ln |x + 6| + ln |x − 1| +C (b) 0.034 (d) 2 ln |x + 2| + ln |x + 3| +C (c) 0.043 (d) 0.053 14. Answer true or false. 5. Z 2 0 15. Answer true or false. Z π 0 √ 4x 16x − 1 dx = 90.9. cos x sin x 1. Which of the following is the the partial fraction decomposi5x + 10 tion of ? (x − 2)(x + 3) (b) 2 ln |x2 + 1| + ln |x + 2| +C (c) 2 tan−1 x + ln |x + 2| +C (d) tan−1 x + ln |x + 2| +C 6. Answer true or false. Z ln |x + 2| + ln |x + 4| +C 7. Answer true or false. ln |x − 2| +C (a) 2. Which of the following is the partial fraction decomposition 5x − 2 ? of 2 x −x 1 2 (a) + x−1 x 1 2 + (b) x−1 x 3 2 (c) + x−1 x x2 + 2x + 5 dx = x3 + 2x2 + x + 2 (a) ln |4x2 + 1| + ln |2x + 4| +C p 1 − sin2 x dx = 0. Section 7.5 4 1 + x+3 x−2 2 3 (b) + x+3 x−2 4 1 (c) + x+3 x−2 3 2 (d) + x+3 x−2 Z 8. Answer true or false. ln |x2 + 2| +C 9. Z x2 + 4 dx = (x − 1)3 (a) ln |x − 1| − Z x2 + 2x + 1 x3 dx = + x2 + x + (x + 2)(x + 4) 3 Z dx = ln |x + 3| + (x + 3)(x − 2) 2x3 + 4x2 + 2x + 2 dx = tan−1 x + (x2 + 4)(x2 + 2) 5 2 +C − x − 1 2(x − 1)2 (b) ln |x − 1| +C x3 + 2x + ln3 |x − 1| +C 3 x3 1 (d) +C + 2x − 3 2(x − 1)2 (c) 10. Z Answer true or false. x3 + x + 3 dx = ln |x| + ln |x + 5| +C x(x + 5) 159 11. Answer true or false. Z 12. Z Answer true or false. 2x + 1 dx = ln |x2 + 2| + ln |x + 2| +C (x2 + 2)(x + 2) 13. Z Answer true or false. x2 − 3x − 17 3 dx = ln |x + 7| − tan−1 (x/2) +C 2 (x + 7)(x2 + 4) 14. Z Answer true or false. dx 1 = tan−1 x + tan−1 (x/2) +C 2 (x2 + 1)(x2 + 4) 15. Answer true or false. Z x6 1 + +C 6 x 1 ln (3x) +C − (b) x6 6 36 dx = ln3 |x − 6| +C (x − 6)3 (a) x6 ln (3x) 1 − +C 6 36 x6 1 (d) − +C 6 x Z √ 5. 4 x ln x dx = (c) 16 8x3/2 ln x − x3/2 +C 3 9 8 3/2 3/2 (b) 4x ln x − x +C 3 16 8x3/2 ln x − +C (c) 3 9 8x3/2 (d) ln x +C 3 x dx = ln |x + 3| +C (x + 3)2 (a) Section 7.6 1. Z 6x dx = 4x + 3 (a) (b) (c) (d) 2. Z 3 3 + ln |4x + 3| +C 2 8 3 3 − ln |4x + 3| +C 2 8 x2 6 ln |4x + 3| + +C 2 3x 9 − ln |4x + 3| +C 2 8 (c) (d) Z 1 4 − ln |4 − 5x| +C 25(4 − 5x) 25 2 ln |4 − 5x| +C 25 sin (3x) sin (7x) dx = 1 1 sin (4x) − sin (10x) +C 8 20 1 1 (b) sin (7x) − sin (3x) +C 8 20 1 1 (c) cos (7x) − sin (3x) +C 7 3 1 1 (d) cos (7x) + sin (3x) +C 7 3 (a) 4. Z Z 7. Z 16x dx = (4 − 5x)2 2 (a) ln |4 − 5x| +C 5 64 + 16(4 − 5x) ln |4 − 5x| +C (b) 25(4 − 5x) 3. 6. x5 ln (3x) dx = e4x cos (2x) dx = sin (2x) 4x cos (2x) +C + (a) e 3 2 e4x (b) cos (2x) − sin (2x) +C 20 e4x 2 cos (2x) + sin (2x) +C (c) 10 e4x 3 cos (2x) + 2 sin (2x) +C (d) 6 e−3x sin (4x) dx = e−3x − 3 sin (4x) − 4 sin (4x) +C 5 e−3x − 3 cos (4x) + 4 sin (4x) +C (b) 5 e−3x 3 cos (4x) + 4 sin (4x) +C (c) 25 −e−3x 3 sin (4x) + 4 cos (4x) +C (d) 25 Z dx √ 8. = 2 x 2x2 + 6 −6x (a) √ +C 2x2 + 6 6x (b) √ +C 2x2 + 6 √ 2x2 + 6 +C (c) 6x √ − 2x2 + 6 +C (d) 6x (a) 160 9. Z 3. Use n = 4 subintervals with the midpoint rule to approximate ln (4x + 2) dx = the integral 2 (a) 2 ln (4x + 2) +C 0 (b) 2.5430 (c) 2.5621 ln2 (4x + 2) (d) +C 2 (d) 2.5745 10. Answer true or false. For 11. Answer true or false. 12. Answer true or false. 13. √ Z Z Z Z x ln (6 − 2x2 ) dx a good choice √ √ cos x dx = 2 sin x +C. √ e x √ dx = e x √ ( x + 1) +C Z 1 0 |x − 1/3| dx. (a) 0.2733 (b) 0.2708 (d) 0.2917 5. Use 2n = 6 subintervals with Simpson’s rule to approximate the given integral: Z 5 1 dx. 1 x (a) 1.6131 2 (3x − 14)(x + 7)3/2 +C 15 (b) (x − 7)3/2 +C 2 (c) (x − 7)3/2 + x +C 3 5 2 (d) (x + 7)5/2 − (x − 7)3/2 +C 3 2 14. Answer true or false. The area enclosed by y = π y = 0, x = 0 and x = 4 is 128 3 . 15. Answer true or false. 4. Use n = 5 subintervals to approximate the following integral using the trapezoidal rule: (c) 0.2867 x x + 7 dx = (a) ln (x + 1) dx. (a) 2.5257 (4x + 2) ln (4x + 2) − x +C 4 (c) 4x ln (4x + 2) − 4x +C (b) for u is 6 − 2x2 . Z 3 Z x 3 (b) 1.6094 (c) 1.7351 √ 16 − x2 , 1 dt √ = + x. 3 t 3t − 5 (d) 1.6436 6. Use Simpson’s rule with 2n = 8 subintervals to approximate the given integral: Z 4 1 |3x − 4| dx. (a) 10.875 (b) 10.833 Section 7.7 (c) 10.828 (d) 10.750 1. Use n = 10 to approximate the integral by the midpoint rule: Z 1 2x3/2 dx 0 (a) 0.804 (b) 0.799 (c) 4.98 (d) 8.04 2. Use n = 10 to approximate the integral by the midpoint rule: Z 1 0 (x3 + 1) dx (a) 3.49 7. Use the midpoint rule with n = 5 subintervals to approximate the given integral: Z 1 0 (x3 + 4) dx. (a) 4.24500 (b) 4.24875 (c) 4.24219 (d) 4.26000 8. Use n = 6 subintervals to approximate the integral by the Z 3 dx trapezoidal rule: 3 1 x +1 (a) 0.31546 (b) 1.249 (b) 0.32546 (c) 1.257 (c) 0.31398 (d) 3.51 (d) 0.31888 161 9. Use the midpoint rule with n = 5 subintervals to approximate Z 1 dx the integral: . −1 1 + x2 (a) 1.57746 (b) 1.55747 (c) 1.58118 Section 7.8 1. Answer true or false. Z 8 dx is an improper integral. x−4 Z 4 dx is an improper integral. 2. Answer true or false. 0 x−1 0 3. Answer true or false. (d) 1.57542 4. Z ∞ dx x4 1 10. Use Simpson’s rule with 2n = 6 subintervals to approximate Z 2p x3 + 1 dx. the integral: Z 2 −∞ e5x dx is an improper integral. = (a) 1/3 0 (b) 1/6 (a) 3.23200 (c) 1/2 (b) 3.24594 (d) Diverges. (c) 3.25988 5. Z ∞ dx √ = x 6 (d) 3.24109 (a) 1/2 11. Use the trapezoidal rule with n = 6 subintervals to approximate the integral: Z 5 2 (b) 1/6 x sin (1/x) dx. (c) 2 (d) Diverges. (a) 2.95032 (b) 2.95071 6. (c) 2.94954 (c) 6 (d) Diverges. x sin (1/x) dx. 7. (a) 2.877695 dx 1 − x2 = (b) −3/2 (c) 0 (d) 2.880575 (d) −π /2 13. Answer true or false. If the trapezoidal rule is used with n = 38 subintervals to approximate the integral | ≤ 1 × 10−3 . Z 2 0 2 e−x dx, 8. Z π /2 0 cot x dx = (a) 0 If the midpoint rule is used with n = 19 subintervals to approximate the integral then |EM | ≤ 1 × 10−5 . Z 2 0 2 e−x dx, subintervals to approximate the integral (b) 1 (c) −30.08 (d) Diverges. 15. Answer true or false. If Simpson’s rule is used with 2n = 8 |ES | ≤ 1 × 10−3 . √ (a) 3/2 (c) 2.871765 14. Answer true or false. Z 0 1 (b) 2.877027 then |ET e6x dx = (b) 1/6 12. Use Simpson’s Rule with 2n = 6 to approximate the integral: 1 −∞ (a) −1/6 (d) 2.94730 Z 4 Z 0 Z 2 0 2 e−x dx, then 9. Answer true or false. Z 3 0 10. Answer true or false. Z ∞ 1 x−2 dx diverges. x−2 dx diverges. 162 11. Answer true or false. Z 1 0 12. Answer true or false. ln (2x) dx diverges. Z ∞ (d) cos x dx diverges. 6. Answer true or false. 0 13. Answer true or false. Z ∞ 0 14. Answer true or false. 15. Answer true or false. e−6x dx diverges. Z 0 −∞ Z 4 −∞ 7. e−5x dx diverges. Z x−3 dx = 1 − lim b→−∞ −2 = 1. b2 sinh12 x cosh x dx = (a) Z 1 sinh13 x +C 13 choice for u is 4ex + 5. 4. Z x 1 p x 9 − x2 + 9 sin−1 +C 2 9 x 9 1 p +C (b) x 9 − x2 + sin−1 2 2 3 p 1 x +C (c) x 9 − x2 − 9 sin−1 2 9 p 9 1 x (d) x 9 − x2 − sin−1 +C 2 2 3 Z 1 2 dx √ = 10. 0 6x − x2 (a) 1.68 (b) 1.66 (c) 1.72 (d) 1.84 Z 3x cos x dx = x x e (4e + 5)dx, a good 2 1 + is the partial fraction dex+4 x−8 3x − 12 . composition of (x + 4)(x − 8) 11. Answer true or false. 12. Z (b) 3 sin x + 3x sin x +C (a) e3x 3 sin (6x) + 6 cos (6x) +C (b) e3x 3 sin (6x) − 6 cos (6x) +C (c) −e3x sin (6x) − 2 cos (6x) +C dx = (a) ln |9x2 + 1| + ln |2x + 4| +C (c) 9 tan−1 x + 2 ln |x + 2| +C (d) 3 cos x + 3x cos x +C e3x sin (6x) dx = 2x2 + 9x + 20 x3 + 2x2 + x + 2 (b) 9 ln |x2 + 1| + 2 ln |x + 2| +C (c) 3 cos x − 3x sin x +C 5. tan (6x) dx = 1 ln |cos (6x)| +C 6 1 (b) − ln |cos (6x)| +C 6 1 (c) tan2 (6x) +C 12 1 (d) − tan2 (6x) +C 12 8. Z Answer true or false. (a) 3 cos x + 3x sin x +C Z x cot x dx = 0. (a) 2x dx √ = 9 − x4 x +C (a) sin−1 3 2 x (b) sin−1 +C 3 x +C (c) cos−1 3 2 x (d) cos−1 +C 3 3. Answer true or false. In evaluating 5π /4 1 cos (14x) sin (8x) cos (6x) dx = − cos (2x) − +C 4 28 Z p 9 − x2 dx = 9. (b) 13 sinh13 x +C 1 sinh11 x +C (c) 11 (d) 11 sinh11 x +C 2. Z Z 7π /4 (a) Chapter 7 Test 1. e3x sin (6x) − 2 cos (6x) +C 15 13. Z (d) 3 tan−1 x + 2 ln |x + 2| +C x8 ln2 (x) dx = x9 81 ln2 (x) − 18 ln (x) + 2 +C 729 1 x9 ln2 (x) − +C (b) 9 81 (a) 163 (c) x9 ln2 (x) +C 9 17. Use 2n = 10 subdivisions to approximate the integral by Simpson’s Rule: x9 ln2 (x) 1 (d) − +C 9 9 0 midpoint rule: 0 (cos x + 1) dx (a) 4.1667 (c) 4.1995 (d) 4.2001 18. Z ∞ 0 e−5x dx = (a) 1.8424 (a) 1/5 (b) 1.8422 (b) −1/5 (c) 5 (c) 1.8420 (d) Diverges. (d) 1.8418 16. Use n = 10 subdivisions to approximate the integral by the trapezoid rule: Z 2 1 (a) 31.24 (b) 31.32 (c) 31.29 (d) 31.20 (x5 + 4) dx. (b) 4.1892 14. Z Answer true or false. x2 x sin (4x) cos (4x) x sin (4x) dx = − − . 4 4 16 15. Use n = 10 subdivisions to approximate the integral by the Z 1 Z 1 7 (x − 1) dx 19. Z 3 0 √ dx 9 − x2 = (a) π /2 (b) −π /2 (c) 0 (d) Diverges. 20. Answer true or false. Z ∞ 4 e2x dx diverges. 164 Chapter 7: Answers to Sample Tests Section 7.1 1. c 9. b 2. c 10. a 3. d 11. true 4. a 12. false 5. b 13. true 6. b 14. false 7. d 15. true 8. c 2. a 10. a 3. c 11. b 4. a 12. true 5. c 13. false 6. b 14. false 7. a 15. false 8. d 2. a 10. true 3. b 11. c 4. a 12. true 5. a 13. false 6. b 14. false 7. true 15. true 8. a 2. c 10. true 3. c 11. false 4. b 12. a 5. d 13. c 6. a 14. true 7. a 15. true 8. b 2. c 10. false 3. c 11. false 4. b 12. false 5. c 13. true 6. false 14. false 7. false 15. false 8. false 2. b 10. true 3. a 11. false 4. b 12. false 5. a 13. a 6. c 14. false 7. d 15. false 8. d 2. b 10. d 3. c 11. c 4. c 12. b 5. a 13. true 6. d 14. false 7. a 15. true 8. b 2. true 10. false 3. true 11. false 4. a 12. true 5. d 13. false 6. b 14. true 7. d 15. false 8. d 2. b 10. a 18. a 3. true 11. true 19. a 4. a 12. c 20. true 5. d 13. a 6. false 14. false 7. b 15. d 8. true 16. c Section 7.2 1. b 9. b Section 7.3 1. b 9. a Section 7.4 1. b 9. a Section 7.5 1. a 9. a Section 7.6 1. d 9. b Section 7.7 1. b 9. a Section 7.8 1. true 9. true Chapter 7 Test 1. a 9. b 17. a