CHAPTERINTEGRATION 7 TECHNIQUES should recall from our brief intro-duction to differential equations...

C H A P T E R 7

7INTEGRATIONTECHNIQUESIf you have ever pondered the spin cycle of a washing machine, you have probablyrecognized that it is designed to squeeze the excess water out of your clothes before drying.The tub of the washer spins rapidly and the clothes are pressed against the wall of the tub.When the load is unusually large or a bulky item is in the washer, the distribution of weightin the tub may become unequal and cause the washing machine to visibly vibrate or evengo totally out of balance. An important concern in the engineering of a washing machine(and countless other devices, as well) is the strength and placement of springs and othershock-absorbing devices that dampen vibration. A good design must balance the compet-

Machine schematic

F

k b

ing goals of minimizing vibrations and minimizing the forcetransmitted to the floor. A simplified schematic for a machinelike the washer is shown in the figure to the right.

The figure shows a vertical force F pushing downwardagainst a spring with spring constant k (describing the stiff-ness of the spring) and a shock absorbing device with damp-ing constant b (describing how much the shock absorberresists the motion). The mathematical analysis of such a sys-tem typically begins with Newton’s second law of motion,which leads us to a differential equation describing the mo-tion of the machine. We then solve the differential equationor (if the equation is too difficult to be solved), use the equation to determine importantproperties of the solution. Engineers often use a technique based on the Laplace trans-form to analyze such differential equations (see, for example, Wan’s Mathematical Modelsand Their Analysis or an elementary differential equations text).

The mathematics that we develop in this chapter is essential for solving engineeringproblems like the analysis of a vibrating machine. You should recall from our brief intro-duction to differential equations in Chapter 6 that integration is an indispensable skill forsolving differential equations. The first five sections of this chapter are devoted to review-ing integration and developing more advanced techniques for finding antiderivatives. In thelast two sections of the chapter, we develop the ideas needed to define Laplace transformsand other important mathematical objects. In particular, we investigate how to compute def-inite integrals over an infinite domain. We define such improper integrals in section 7.7.Prior to that, we expand the use of l’Hôpital’s Rule to a wide range of indeterminate forms.

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We then use the newly expanded l’Hôpital’s Rule to calculate some of the limits arisingfrom the evaluation of improper integrals.

Although we finish the chapter in a position to define and use Laplace transforms, wemust leave an in-depth study of these to future courses. We simply offer this as one exam-ple of the many directions that mathematics classes may take after the calculus sequence.

7.1 REVIEW OF FORMULAS AND TECHNIQUES

In this brief section, we draw together all of the integration formulas and the one integra-tion technique (integration by substitution) that we have developed so far. We use these todevelop some more general formulas, as well as to solve more complicated integrationproblems. First, look over the following table of the basic integration formulas developedin Chapters 4 and 6.

Recall that each of these follows from a corresponding differentiation rule. So far, wehave expanded this list slightly by using the method of substitution, as in the followingexample.

Evaluate ∫

sin(ax) dx, for a �= 0.

Solution You may already know the answer here, but take a moment to make cer-tain that you see why that answer is correct. The obvious choice is to let u = ax, so thatdu = a dx. This gives us∫

sin(ax) dx = 1

a

∫sin(ax)︸︷︷︸ a dx︸︷︷︸ = 1

a

∫sin u du

sin u du

= −1

acos u + c = −1

acos(ax) + c.

■

A Simple SubstitutionExample 1.1

∫xn dx = xn+1

n + 1+ c, for n �= −1 (power rule)

∫1

xdx = ln |x | + c, for x �= 0∫

sin x dx = −cos x + c∫

cos x dx = sin x + c∫sec2 x dx = tan x + c

∫sec x tan x dx = sec x + c∫

csc2 x dx = −cot x + c∫

csc x cot x dx = −csc x + c∫ex dx = ex + c

∫e−x dx = −e−x + c∫

tan x dx = −ln |cos x | + c∫

1√1 − x2

dx = sin−1 x + c∫1

1 + x2dx = tan−1 x + c

∫1

|x |√x2 − 1dx = sec−1 x + c

556 Chapter 7 Integration Techniques


Observe that there is no need to memorize general rules like the one given in exam-ple 1.1, although it is often convenient to do so. Realize that you can reproduce such gen-eral rules any time you need them using substitution. Another general rule is found in thefollowing example.

Evaluate ∫

1

a2 + x2dx, for a �= 0.

Solution Notice that this is nearly the same as ∫

1

1 + x2dx and we can write

∫1

a2 + x2dx = 1

a2

∫1

1 + (xa

)2 dx .

Now, letting u = xa , we have du = 1

a dx and so, ∫1

a2 + x2dx = 1

a2

∫1

1 + (xa

)2 dx = 1

a

∫1

1 + (xa

)2︸︷︷︸(

1

a

)dx︸︷︷︸

1 + u2 du

= 1

a

∫1

1 + u2du = 1

atan−1 u + c = 1

atan−1

(x

a

)+ c.

■

The basic integral formulas given above together with substitution and some ingenu-ity will allow you to evaluate a large number of integrals. However, substitution will notresolve all of your integration difficulties, as we see in the following example.

Evaluate ∫(x2 − 5)2 dx .

Solution Note that you can’t evaluate this as it stands. Your first impulse might beto substitute u = x2 − 5. However, this fails, as we don’t have du = 2x dx in the integral.(We can force the constant 2 into the integral, but we can’t get the x in there.) On theother hand, you can always multiply out the binomial to obtain ∫

(x2 − 5)2 dx =∫

(x4 − 10x2 + 25) dx = x5

5− 10

x3

3+ 25x + c.

■

The moral of example 1.3 is to make certain you don’t overlook simpler methods. Themost general rule in integration is to keep trying. Sometimes, you will need to do somealgebra before you can recognize the form of the integrand.

Evaluate ∫

1√−5 + 6x − x2dx .

An Integral Where We Must Complete the SquareExample 1.4

An Integrand That Must be ExpandedExample 1.3

Generalizing a Basic Integration RuleExample 1.2

Section 7.1 Review of Formulas and Techniques 557


Solution Probably very little comes to mind here. Substitution for either the entiredenominator or the quantity under the square root does not work. (Why not?) So, what’sleft? Recall that there are essentially only two things you can do to a quadratic polyno-mial: either factor it or complete the square. Here, doing the latter sheds some light onthe integral. We have ∫

1√−5 + 6x − x2dx =

∫1√

−5 − (x2 − 6x + 9) + 9dx =

∫1√

4 − (x − 3)2dx .

Notice how much this looks like ∫

1√1 − x2

dx = sin−1 x + c. If we factor out the 4 in the square root, we get ∫

1√−5 + 6x − x2dx =

∫1√

4 − (x − 3)2dx =

∫1√

1 − (x − 3

2

)2

1

2dx .

Now, let u = x − 32

, so that du = 12 dx . This gives us∫

1√−5 + 6x − x2dx =

∫1√

1 − (x − 3

2

)2︸︷︷︸1

2dx︸︷︷︸ =

∫1√

1 − u2du

√1 − u2

du

= sin−1 u + c = sin−1

(x − 3

2

)+ c.

■

The following example illustrates the value of perseverance.

Evaluate ∫

4x + 1

2x2 + 4x + 10dx .

Solution As with most integrals, you cannot evaluate this as it stands. Notice thatthe numerator is very nearly the derivative of the denominator (but not quite). So askyourself what else you can do with an algebraic expression such as this. In the denomi-nator, there’s a quadratic polynomial, but it doesn’t factor. The only remaining algebraictechnique, then, is to complete the square. Notice that ∫

4x + 1

2x2 + 4x + 10dx =

∫4x + 1

2(x2 + 2x + 1) − 2 + 10dx =

∫4x + 1

2(x + 1)2 + 8dx .

So, you’ve completed the square. Now what can you do? When the next step is notobvious, we recommend that you look for a pattern. Here, you should focus on the

denominator. It nearly looks like the denominator in ∫

1

1 + x2dx = tan−1 x + c. If we

factor out an 8, it will look even more like this, as follows. ∫4x + 1

2x2 + 4x + 10dx =

∫4x + 1

2(x + 1)2 + 8dx

= 1

8

∫4x + 1

14 (x + 1)2 + 1

dx

= 1

8

∫4x + 1(

x + 12

)2 + 1dx .

An Integral Requiring Some ImaginationExample 1.5



Now, make the substitution u = x + 12

, so that du = 12 dx and x = 2u − 1. We then

have 4(2u − 1) + 1∫

4x + 1

2x2 + 4x + 10dx = 1

8

∫4x + 1(

x + 12

)2 + 1dx = 1

4

∫ ︷︸︸︷4x + 1(

x + 12

)2 + 1︸︷︷︸1

2dx︸︷︷︸

u2 + 1du

= 1

4

∫4(2u − 1) + 1

u2 + 1du = 1

4

∫8u − 3

u2 + 1du

= 4

4

∫2u

u2 + 1du − 3

4

∫1

u2 + 1du

= ln(u2 + 1) − 3

4tan−1 u + c

= ln

[(x + 1

2

)2

+ 1

]− 3

4tan−1

(x + 1

2

)+ c.

■

This last example was tedious, but reasonably straightforward. The real issue in inte-gration is to recognize what pieces are present in a given integral and to see how you mightrewrite the integral in a more familiar form. In the next section, we develop a powerful toolfor evaluating integrals.

Section 7.1 Review of Formulas and Techniques 559

EXERCISES 7.1

1. In example 1.2, explain how you should know to write

the denominator as a2[1 + (

xa

)2]. Would this still be a

good first step if the numerator was x instead of 1? What wouldyou do if the denominator was

√a2 − x2 ?

2. In both examples 1.4 and 1.5, we completed the squareand found antiderivatives involving sin−1 x , tan−1 x and

ln(x2 + 1). Briefly describe how the presence of an x in thenumerator or a square root in the denominator affects which ofthese functions will be in the antiderivative.

In exercises 3–40, evaluate the integral.

3.∫

sin 6x dx 4.∫

3 cos 4x dx

5.∫

sec 2x tan 2x dx 6.∫

x sec x2 tan x2 dx

7.∫

e3−2x dx 8.∫

3

e6xdx

9.∫

sin√

x√x

dx 10.∫

cos 1/x

x2dx

11.∫ π

0cos xesin x dx 12.

∫ π/4

0sec2 xetan x dx

13.∫ 0

−π/4

sin x

cos2 xdx 14.

∫ π/2

π/4

1

sin2 xdx

15.∫

3

16 + x2dx 16.

∫2

4 + 4x2dx

17.∫

x2

1 + x6dx 18.

∫x5

1 + x6dx

19.∫

1√4 − x2

dx 20.∫

ex

√1 − e2x

dx

21.∫

x√1 − x4

dx 22.∫

2x3

√1 − x4

dx

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23.∫

4

5 + 2x + x2dx 24.

∫4x + 4

5 + 2x + x2dx

25.∫

4x

5 + 2x + x2dx 26.

∫x + 1

x2 + 2x + 4dx

27.∫

(x2 + 4)2 dx 28.∫

x(x2 + 4)2 dx

29.∫

1√3 − 2x − x2

dx 30.∫

x + 1√3 − 2x − x2

dx

31.∫

2√4x − x2

dx 32.∫

3√5 − 4x − x2

dx

33.∫ −1

−2eln(x 2+1) dx 34.

∫ 3

1e2 ln x dx

35.∫ 4

3x√

x − 3 dx 36.∫ 1

0x(x − 3)2 dx

37.∫ 2

0

ex

1 + e2xdx 38.

∫ 0

−1ex cot(ex ) csc(ex ) dx

39.∫ 4

1

x2 + 1√x

dx 40.∫ 0

−2xe−x 2

dx

In exercises 41–48, evaluate the integral (using formulas not re-viewed in this section, but covered elsewhere in the text).

41.∫

sinh 2x dx 42.∫

cosh 4x dx

43.∫

tanh 3x dx 44.∫

ex cosh ex dx

45.∫

x cosh x2 dx 46.∫

sec2 x cosh(tan x) dx

47.∫

3√x2 − 1

dx 48.∫

3√x2 + 1

dx

49.∫

3x

x2√

x4 − 1dx 50.

∫cos x√

sin2 x + 1dx

In exercises 51–56, you are given a pair of integrals. Evaluate theintegral that can be worked using the techniques covered so far(the other cannot).

51.∫

5

3 + x2dx and

∫5

3 + x3dx

52.∫

sin 2x dx and∫

sin2 x dx

53.∫

ln x dx and∫

ln x

2xdx

54.∫

x3

1 + x8dx and

∫x4

1 + x8dx

55.∫

e−x 2dx and

∫xe−x 2

dx

56.∫

sec x dx and∫

sec2 x dx

57. Find ∫ 2

0f (x) dx , where f (x) =

{x/(x2 + 1) if x ≤ 1

x2/(x2 + 1) if x > 1

58. Find ∫ 2

−2f (x) dx , where f (x) =

{xex 2

if x < 0

x2ex 3if x ≥ 0

59. Evaluate both integrals in exercise 51 with your CAS.

Also try ∫

5

3 + x4dx and

∫5

3 + x5dx . Describe any

patterns that you see. Try to decide whether or not an antideriv-

ative can be found for ∫

5

3 + xndx , for any power n.

60. Evaluate both integrals in exercise 52 with your CAS.Also try

∫sin3 x dx and

∫sin4 x dx . Describe any pat-

terns that you see. Try to decide whether or not antiderivativescan be found for any power of sin x .

7.2 INTEGRATION BY PARTS

At this point, you will have recognized that although differentiation and integration are in-timately connected, computing derivatives is vastly different from computing integrals.You have at your disposal rules for computing the derivative of nearly any function you canimagine. By contrast, you can compute antiderivatives for a relatively small number offunctions. Thus far, we have developed a collection of basic integration formulas, most of


Section 7.2 Integration by Parts 561

which followed directly from the corresponding differentiation formulas. Beyond the basicformulas, you have one technique, integration by substitution, that you have used to rewritecertain integrals in a simpler form. Unfortunately, this leaves us with many integrals thatcannot be evaluated given your current knowledge. For instance,∫

x sin x dx

cannot be evaluated with what you already know. We improve this situation in the currentsection by introducing a powerful tool called integration by parts.

We have said any number of times that every differentiation rule gives rise to a corre-sponding integration rule. So, recall the product rule:

d

dx[ f (x)g(x)] = f ′(x)g(x) + f (x)g′(x).

If we integrate both sides of this equation, we get∫d

dx[ f (x)g(x)] dx =

∫f ′(x)g(x) dx +

∫f (x)g′(x) dx .

Of course, the integral on the left-hand side is simply f (x)g(x). Solving for the secondintegral on the right-hand side then yields∫

f (x)g′(x) dx = f (x)g(x) −∫

f ′(x)g(x) dx .

This rule is called integration by parts. (It may be beneficial for you to differentiate bothsides to verify the equality directly.) You’re probably wondering what this is all about.Admittedly, the significance of this new rule is hard to see at first. We’ll let the examplesconvince you of the power of this technique. First, it’s usually convenient to write thisusing the notation u = f(x) and v = g(x). Notice that this gives us

du = f ′(x) dx and dv = g′(x) dx,

so that the integration by parts algorithm becomes

(2.1)

To apply integration by parts, you need to make a judicious choice of u and dv so that theintegral on the right-hand side of (2.1) is one that you know how to evaluate.

Evaluate ∫

x sin x dx .

Solution First, observe that you cannot evaluate this as it stands. That is, it is notone of our basic integrals and no substitution will help. To use integration by parts, youwill need to choose u (something to differentiate) and dv (something to integrate).Notice that if we let

u = x and dv = sin x dx,

then du = dx and integrating dv, we have

v =∫

sin x dx = −cos x + k.

Integration by PartsExample 2.1

∫u dv = uv −

∫v du.

H I S T O R I C A L N O T E S

Brook Taylor (1685–1731) AnEnglish mathematician who iscredited with devising integrationby parts. Taylor made importantcontributions to probability, thetheory of magnetism and the use ofvanishing lines in linearperspective. However, he is bestknown for Taylor’s Theorem (seesection 8.7), in which hegeneralized results of Newton,Halley, the Bernoullis and others.Personal tragedy (both his wivesdied during childbirth) and poorhealth limited the mathematicaloutput of this brilliantmathematician.

Integration by Parts


In performing integration by parts, we drop this constant of integration (think about whyit makes sense to do this). Also, we usually write this information as the block:

u = x dv = sin x dx

du = dx v = −cos x

This gives us ∫x︸︷︷︸ sin x dx︸︷︷︸ =

∫u dv = uv −

∫v du

u dv

= −x cos x −∫

(−cos x) dx

= −x cos x + sin x + c. (2.2)

It’s a simple matter to differentiate the expression on the right-hand side of (2.2) andverify directly that you have indeed found an antiderivative of x sin x .

■

You should quickly realize that the choice of u and dv is critical. Observe what hap-pens if we switch the choice of u and dv made in example 2.1.

Consider ∫

x sin x dx as in example 2.1, but this time, reverse the choice of u and dv.

Solution Here, we let

u = sin x dv = x dx

du = cos x dx v = 12 x2

This gives us∫sin x︸︷︷︸ x dx︸︷︷︸ = uv −

∫v du = 1

2 x2 cos x − 12

∫x2 cos x dx .

u dv

Notice that the last integral is one that we do not know how to calculate any better thanthe original one. In fact, we have made the situation worse in that the power of x in thenew integral is higher than in the original integral.

■

When using integration by parts, keep in mind that you are splitting up the integrand into two pieces.One of these pieces, corresponding to u, will be differentiated and the other, corresponding to dv, willbe integrated. Since you can differentiate virtually every function you run across, you should think interms of choosing a dv for which you know an antiderivative, as well as a choice of both that willresult in an easier integral. Unfortunately, it’s not always so easy to see the problem through frombeginning to end. You will learn what works best by working through lots of problems. Even if youdon’t see how the problem is going to end up, try something! (At least you’ll learn what doesn’twork.)

Remark 2.1

A Poor Choice of u and dvExample 2.2



Evaluate ∫

ln x dx .

Solution This may look like it should be simple, but it’s not one of our basic inte-grals and there’s no substitution that will simplify it. That leaves us with integration byparts. Remember that you must pick u (to be differentiated) and dv (to be integrated).You obviously can’t pick dv = ln x dx, since the problem here is to find a way to inte-grate this very term. So, try

u = ln x dv = dx

du = 1x dx v = x

From the integration by parts algorithm, we have∫ln x︸︷︷︸ dx︸︷︷︸ = uv −

∫v du = x ln x −

∫x

(1

x

)dx

u dv

= x ln x −∫

1 dx = x ln x − x + c.

■

Frequently, an integration by parts results in an integral that we cannot evaluate directly,but instead, one that we can evaluate only by repeating integration by parts one or more times.

Evaluate ∫

x2 sin x dx .

Solution Certainly, you cannot evaluate this as it stands and there is no simplifi-cation or substitution that will help. We choose

u = x2 dv = sin x dx

du = 2x dx v = −cos x

With this choice, integration by parts yields∫x2︸︷︷︸ sin x dx︸︷︷︸ = −x2 cos x + 2

∫x cos x dx .

u dv

Of course, this last integral cannot be evaluated as it stands, but we could do it using afurther integration by parts. We now choose

u = x dv = cos x dx

du = dx v = sin x

Applying integration by parts to the last integral, we now have∫x2 sin x dx = −x2 cos x + 2

∫x︸︷︷︸ cos x dx︸︷︷︸u dv

= −x2 cos x + 2

(x sin x −

∫sin x dx

)= −x2 cos x + 2x sin x + 2 cos x + c.

■

Repeated Integration by PartsExample 2.4

An Integrand with a Single TermExample 2.3


In the second integration by parts inexample 2.4, if you choose u = cos xand dv = x dx, then integration byparts will fail and leave you with theless than astounding conclusion thatthe integral equals itself. (Try this asan exercise.)

Remark 2.2


Based on our work in example 2.4, try to figure out how many integrations by partswould be required to evaluate

∫xn sin x dx, for a positive integer n. (There will be more

on this in the exercises.)Repeated integration by parts sometimes takes you back to the integral you started

with. This can be bad news (see Remark 2.2), or this can give us a clever way of evaluat-ing an integral, as in the following example.

Evaluate ∫

e2x sin x dx .

Solution Again, you should verify that none of our elementary methods works onthis integral. This leaves us with integration by parts. In this case, there are two viablechoices for u and dv. We take

u = e2x dv = sin x dx

du = 2e2x dx v = −cos x

(The opposite choice also works. Try this as an exercise.) Applying integration by partsyields ∫

e2x︸︷︷︸ sin x dx︸︷︷︸ = −e2x cos x + 2∫

e2x cos x dx .

u dv

The remaining integral again requires integration by parts. We choose

u = e2x dv = cos x dx

du = 2e2x dx v = sin x

It now follows that∫e2x sin x dx = −e2x cos x + 2

∫e2x︸︷︷︸ cos x dx︸︷︷︸

u dv

= −e2x cos x + 2

(e2x sin x − 2

∫e2x sin x dx

)

= −e2x cos x + 2e2x sin x − 4∫

e2x sin x dx . (2.3)

Look carefully at the last line and you’ll see the integral that we started with! If you treatthe integral

∫e2x sin x dx as the unknown you’re trying to solve for, you can add

4∫

e2x sin x dx to both sides of equation (2.3). This leaves us with

5∫

e2x sin x dx = −e2x cos x + 2e2x sin x + K ,

where we have added the constant of integration K on the right side. Dividing both sidesby 5 then gives us ∫

e2x sin x dx = −1

5e2x cos x + 2

5e2x sin x + c,

where we have replaced the arbitrary constant of integration K5 by c.

■

Repeated Integration by Parts with a TwistExample 2.5



For integrals like ∫

e2x cos x dx (or related integrals like ∫

e−3x sin x dx ), repeated integration byparts as in example 2.5 will produce an antiderivative. The first choice of u and dv is up to you (eitherchoice will work) but your choice of u and dv in the second integration by parts must be consistentwith your first choice. For instance, in example 2.5, our initial choice of u = e2x commits us to usingu = e2x for the second integration by parts, as well. If you want to know why, rework the secondintegral taking u = cos x and see what happens!

Observe that for any positive integer n, the integral ∫

xnex dx will require integrationby parts. At this point, it should be no surprise that we take

u = xn dv = ex dx

du = nxn−1 dx v = ex

Applying integration by parts gives us∫xn︸︷︷︸ ex dx︸︷︷︸ = xnex − n

∫xn−1ex dx . (2.4)

u dv

Notice that if n − 1 > 0, we will need to perform another integration by parts. In fact, we’llneed to perform a total of n integrations by parts to complete the process. An alternative isto apply formula (2.4) (called a reduction formula) repeatedly to evaluate a given integral.We illustrate this in the following example.

Evaluate the integral ∫

x4ex dx .

Solution The prospect of performing four integrations by parts may not particu-larly appeal to you. However, we can use the reduction formula (2.4) repeatedly to eval-uate the integral with relative ease, as follows. From (2.4), with n = 4, we have∫

x4ex dx = x4ex − 4∫

x4−1ex dx = x4ex − 4∫

x3ex dx .

Applying (2.4) again, this time with n = 3, gives us∫x4ex dx = x4ex − 4

(x3ex − 3

∫x2ex dx

).

By now, you should see that we can resolve this by applying the reduction formula twomore times. By doing so, we get∫

x4ex dx = x4ex − 4x3ex + 12x2ex − 24xex + 24ex + c,

where we leave the details of the remaining calculations to you.

■

Note that to evaluate a definite integral, it is always possible to apply integration byparts to the corresponding indefinite integral and then simply evaluate the resultingantiderivative between the limits of integration. Whenever possible, however (i.e., when

Using a Reduction FormulaExample 2.6

Remark 2.3



the integration is not too involved), you should apply integration by parts directly to thedefinite integral. Observe that the integration by parts algorithm for definite integrals issimply

where we have written the limits of integration as we have to remind you that these refer tothe values of x. (Recall that we derived the integration by parts formula by taking u and vboth to be functions of x.)

Evaluate ∫ 2

1 x2 ln x dx .

Solution Again, since more elementary methods are fruitless, we try integrationby parts. Since we do not know how to integrate ln x (except via integration by parts),we choose

u = ln x dv = x2 dx

du = 1x dx v = 1

3 x3

and hence, we have∫ 2

1ln x︸︷︷︸ x2 dx︸︷︷︸ = uv

∣∣∣∣2

1

−∫ 2

1v du = 1

3x3 ln x

∣∣∣∣2

1

− 1

3

∫ 2

1x3

(1

x

)dx

u dv

= 1

3(23 ln 2 − 13 ln 1) − 1

3

∫ 2

1x2 dx

= 8 ln 2

3− 0 − 1

9x3

∣∣∣∣2

1

= 8 ln 2

3− 1

9(23 − 13)

= 8 ln 2

3− 1

9(8 − 1) = 8 ln 2

3− 7

9.

■

Integration by parts is the most powerful tool in our integration arsenal. In order tomaster its use, you will need to work through many problems. We provide a wide assort-ment of these in the exercise set that follows.

Integration by Parts for a Definite IntegralExample 2.7

∫ x=b

x=au dv = uv

∣∣∣∣x=b

x=a

−∫ x=b

x=av du,


EXERCISES 7.2

1. Discuss your best strategy for determining which part ofthe integrand should be u and which part should be dv.

2. Integration by parts comes from the product rule for de-rivatives. Which integration technique comes from the

chain rule? Briefly discuss why there is no commonly used in-tegration technique derived from the quotient rule.

In exercises 3–32, evaluate the integrals.

3.∫

x cos x dx 4.∫

x sin 4x dx

5.∫

xe2x dx 6.∫

x ln x dx

Integration by Parts

for a definite integral


7.∫

x2 ln x dx 8.∫

ln x

xdx

9.∫

x2e−3x dx 10.∫

x2ex 3dx

11.∫

x2 cos 2x dx 12.∫

x2 sin 3x dx

13.∫

ex sin 4x dx 14.∫

e2x cos x dx

15.∫

cos x cos 2x dx 16.∫

sin x sin 2x dx

17.∫

x sec2 x dx 18.∫

x3ex 2dx

19.∫

(ln x)2 dx 20.∫

x2e3x dx

21.∫

cos x ln(sin x) dx 22.∫

x sin x2 dx

23.∫

x cosh 2x dx 24.∫

2x sinh 3x dx

25.∫ 1

0x sin 2x dx 26.

∫ π

02x cos x dx

27.∫ 1

0x cos πx dx 28.

∫ 1

0xe3x dx

29.∫ 1

0x sin πx dx 30.

∫ 1

0x cos 2πx dx

31.∫ 10

1ln x dx 32.

∫ 2

1x ln x dx

In exercises 33–36, evaluate the integral using integration byparts (as we recommended in the text, “Try something!’’).

33.∫

cos−1 x dx 34.∫

tan−1 x dx

35.∫

sin√

x dx 36.∫

e√

x dx

37. How many times would integration by parts need to beperformed to evaluate

∫xn sin x dx (where n is a positive

integer)?

38. How many times would integration by parts need to beperformed to evaluate

∫xn ln x dx (where n is a positive

integer)?


39. Several useful integration formulas called reduction formulasare used to automate the process of performing multiple inte-grations by parts. Prove that for any positive integer n,∫

cosn x dx = 1

ncosn−1 x sin x + n − 1

n

∫cosn−2 x dx .

(Use integration by parts with u = cosn−1 x and dv = cos x dx.)

40. Use integration by parts to prove that for any positive integer n,∫sinn x dx = − 1

nsinn−1 x cos x + n − 1

n

∫sinn−2 x dx .

In exercises 41–48, evaluate the integral using the reduction for-mulas from exercises 39 and 40 and (2.4).

41.∫

x3ex dx 42.∫

cos5 x dx

43.∫

cos3 x dx 44.∫

sin4 x dx

45.∫ 1

0x4ex dx 46.

∫ π/2

0sin4 x dx

47.∫ π/2

0sin5 x dx 48.

∫ π/2

0sin6 x dx

49. Based on exercises 46–48, conjecture a formula for∫ π/20 sinm x dx . (Note: You will need different formulas for

m odd and for m even.)

50. Conjecture a formula for ∫ π/2

0 cosm x dx .

51. The excellent movie Stand and Deliver tells the story of math-ematics teacher Jaime Escalante, who developed a remarkableAP calculus program in inner-city Los Angeles. In one scene,Escalante shows a student how to evaluate the integral∫

x2 sin x dx . He forms a chart like the following:

Multiplying across each full row, the antiderivative is−x2 cos x + 2x sin x + 2 cos x + c . Explain where each col-umn comes from and why the method works on this problem.

In exercises 52–57, use the method of exercise 51 to evaluate theintegral.

52.∫

x4 sin x dx 53.∫

x4 cos x dx

sin x

x2 −cos x +2x −sin x −2 cos x +



54.∫

x4ex dx 55.∫

x4e2x dx

56.∫

x5 cos 2x dx 57.∫

x3e−3x dx

58. You should be aware that the method of exercise 51 doesn’talways work, especially if both the derivative and antideriva-tive columns have powers of x . Show that the method doesn’twork on

∫x2 ln x dx .

59. Show that ∫ π

−πcos(mx) cos(nx) dx = 0 for positive integers

m �= n.


−πsin(mx) sin(nx) dx = 0 for positive integers

m �= n.


−πcos(mx) sin(nx) dx = 0 for positive integers m

and n.


−πcos2 nx dx = ∫ π

−πsin2 nx dx = π for any posi-

tive integer n.

63. Integration by parts can be used to compute coeffi-cients for important functions called Fourier series. We

cover Fourier series in detail in Chapter 8. Here, you will dis-cover what some of the fuss is about. Start by computingan = 2

π

∫ π

−πx sin nx dx for an unspecified positive integer n.

Write out the specific values for a1, a2 , a3 and a4 and then formthe function

f (x) = a1 sin x + a2 sin 2x + a3 sin 3x + a4 sin 4x .

Compare the graphs of y = x and y = f (x) on the interval[−π, π]. From writing out a1 through a4 , you should notice anice pattern. Use it to form the function

g(x) = f (x) + a5 sin 5x + a6 sin 6x + a7 sin 7x + a8 sin 8x .

Compare the graphs of y = x and y = g(x) on the interval[−π, π]. Is it surprising that you can add sine functionstogether and get something close to a straight line? It turnsout that Fourier series can be used to find cosine and sineapproximations to nearly any continuous function on a closedinterval.

64. Along with giving us a technique to compute antideriv-atives, integration by parts is very important theoreti-

cally. In this context, it can be thought of as a technique formoving derivatives off of one function and onto another. To seewhat we mean, suppose that f (x) and g(x) are functions withf (0) = g(0) = 0, f (1) = g(1) = 0 and with continuous sec-ond derivatives f ′′(x) and g′′(x). Use integration by partstwice to show that∫ 1

0f ′′(x)g(x) dx =

∫ 1

0f (x)g′′(x) dx .

7.3 TRIGONOMETRIC TECHNIQUES OF INTEGRATION

Integrals Involving Powers of Trigonometric FunctionsEvaluating an integral whose integrand contains powers of one or more trigonometric func-tions often involves making a clever substitution. These integrals are sufficiently commonthat we present them here as a group.

Our first aim is to evaluate integrals of the form∫sinm x cosn x dx,

where m and n are positive integers.

Case 1: m or n Is an Odd Positive IntegerIf m is odd, first isolate one factor of sin x . (You’ll need this for du.) Then, replace any fac-tors of sin2 x with 1 − cos2 x and make the substitution u = cos x . Likewise, if n is odd,first isolate one factor of cos x . (You’ll need this for du.) Then, replace any factors of cos2 xwith 1 − sin2 x and make the substitution u = sin x .


Section 7.3 Trigonometric Techniques of Integration 569

We illustrate this for the case where m is odd in the following example.

Evaluate ∫

cos4 x sin x dx .

Solution There is nothing particularly new or tricky about this integral. Since youcannot evaluate it as it stands, you might think of any available substitutions. (Hint:Look for terms that are derivatives of other terms.) You should quickly settle on lettingu = cos x, so that du = −sin x dx . This gives us∫

cos4 x sin x dx = −∫

cos4 x︸︷︷︸ (−sin x) dx︸︷︷︸ = −∫

u4 du

u4 du

= −u5

5+ c = −cos5 x

5+ c. Since u = cos x.

■

While this first example was not a particularly challenging one, it should help to giveyou an idea of what to do with the following example.

Evaluate ∫

cos4 x sin3 x dx .

Solution If you’re looking for terms that are derivatives of other terms, you shouldsee both sine and cosine terms, but for which do you substitute? Consider u = cos x, sothat du = −sin x dx . Notice that we can now rewrite the integral as∫

cos4 x sin3 x dx =∫

cos4 x sin2 x sin x dx = −∫

cos4 x sin2 x(−sin x) dx

= −∫

cos4 x(1 − cos2 x)︸︷︷︸ (−sin x) dx︸︷︷︸ = −∫

u4(1 − u2) du

u4(1 − u2) du

= −∫

(u4 − u6) du = −(

u5

5− u7

7

)+ c

= −cos5 x

5+ cos7 x

7+ c. Since u = cos x.

Pay close attention to how we did this. We took the odd power (in this case, sin3 x) andfactored out one power of sin x (to use for du). The remaining (even) powers of sin xwere rewritten in terms of cos x using the Pythagorean identity

sin2 x + cos2 x = 1.

■

The ideas used in example 3.2 can be applied to any integral of the specified form.

An Integrand with an Odd Power of SineExample 3.2

A Typical SubstitutionExample 3.1


Evaluate ∫ √

sin x cos5 x dx .

Solution Observe that we can rewrite this as∫ √sin x cos5 x dx =

∫ √sin x cos4 x cos x dx =

∫ √sin x(1 − sin2 x)2 cos x dx .

Substitute u = sin x, so that du = cos x dx and we have∫ √sin x cos5 x dx =

∫ √sin x(1 − sin2 x)2︸︷︷︸ cos x dx︸︷︷︸√

u(1 − u2)2 du

=∫ √

u(1 − u2)2 du =∫

u1/2(1 − 2u2 + u4) du

=∫

(u1/2 − 2u5/2 + u9/2) du

= 2

3u3/2 − 2

(2

7

)u7/2 + 2

11u11/2 + c

= 2

3sin3/2 x − 4

7sin7/2 x + 2

11sin11/2 x + c. Since u = sin x .

Although the details of calculation here got a bit messy, you should see through the detailsto the main point: all integrals of this form are calculated in essentially the same way.

■

Case 2: m and n Are Both Even Positive IntegersIn this case, we can use the half-angle formulas for sine and cosine (shown in the margin)to reduce the powers in the integrand.

We illustrate this case in the following example.

Evaluate ∫

sin2 x dx .

Solution Using the half-angle formula, we can rewrite the integral as∫sin2 x dx = 1

2

∫(1 − cos 2x) dx .

We can evaluate this last integral by using the substitution u = 2x, so that du = 2 dx .

This gives us∫sin2 x dx = 1

2

(1

2

)∫(1 − cos 2x)︸︷︷︸ 2 dx︸︷︷︸ = 1

4

∫(1 − cos u) du

1 − cos u du

= 1

4(u − sin u) + c = 1

4(2x − sin 2x) + c. Since u = 2x .

■

An Integrand with an Even Power of SineExample 3.4

An Integrand with an Odd Power of CosineExample 3.3


N O T E S

Half-angle formulas

sin2 x = 12 (1 − cos 2x )

cos2 x = 12 (1 + cos 2x )


With some integrals, you may need to apply the half-angle formulas several times, asin the following example.

Evaluate ∫

cos4 x dx .

Solution Using the half-angle formula for cosine, we have∫cos4 x dx =

∫(cos2 x)2 dx = 1

4

∫(1 + cos 2x)2 dx

= 1

4

∫(1 + 2 cos 2x + cos2 2x) dx .

Using the half-angle formula again, on the last term in the integrand, we get∫cos4 x dx = 1

4

∫ [1 + 2 cos 2x + 1

2(1 + cos 4x)

]dx

= 3

8x + 1

4sin 2x + 1

32sin 4x + c.

We leave it as an (easy) exercise to obtain this last expression.

■

Our next aim is to devise a strategy for evaluating integrals of the form∫tanm x secn x dx,

where m and n are integers.

Case 1: m Is an Odd Positive IntegerFirst, isolate one factor of sec x tan x . (You’ll need this for du.) Then, replace any factorsof tan2 x with sec2 x − 1 and make the substitution u = sec x .

We illustrate this in the following example.

Evaluate ∫

tan3 x sec3 x dx .

Solution Looking for terms that are derivatives of other terms, we rewrite theintegral as ∫

tan3 x sec3 x dx =∫

tan2 x sec2 x (sec x tan x) dx

=∫

(sec2 x − 1) sec2 x (sec x tan x) dx,

where we have used the Pythagorean identity

tan2 x = sec2 x − 1.

An Integrand with an Odd Power of TangentExample 3.6

An Integrand with an Even Power of CosineExample 3.5



You should see the substitution now. We let u = sec x, so that du = sec x tan x dx andhence,∫


(sec2 x − 1) sec2 x︸︷︷︸ (sec x tan x) dx︸︷︷︸(u2 − 1) u2 du

=∫

(u2 − 1)u2 du =∫

(u4 − u2) du

= 1

5u5 − 1

3u3 + c = 1

5sec5 x − 1

3sec3 x + c. Since u = sec x .

■

Case 2: n Is an Even Positive IntegerFirst, isolate one factor of sec2 x . (You’ll need this for du.) Then, replace any remainingfactors of sec2 x with 1 + tan2 x and make the substitution u = tan x .

We illustrate this in the following example.

Evaluate ∫

tan2 x sec4 x dx .

Solution Since ddx tan x = sec2 x, we rewrite the integral as∫


tan2 x sec2 x sec2 x dx =∫

tan2 x(1 + tan2 x) sec2 x dx .

Now, we let u = tan x, so that du = sec2 x dx and∫tan2 x sec4 x dx =

∫tan2 x(1 + tan2 x)︸︷︷︸ sec2 x dx︸︷︷︸

u2(1 + u2) du

=∫

u2(1 + u2) du =∫

(u2 + u4) du

= 1

3u3 + 1

5u5 + c

= 1

3tan3 x + 1

5tan5 x + c. Since u = tan x .

■

Case 3: m Is an Even Positive Integer and n Is an Odd Positive IntegerReplace any factors of tan2 x with sec2 x − 1 and then use a special reduction formula(given in the exercises) to evaluate integrals of the form

∫secn x dx . This complicated case

will be covered briefly in the exercises. Much of this depends on the following example.


sec x dx .

Solution Think about this integral some and you’ll probably realize that you don’thave a clue as to how to evaluate it. Finding an antiderivative depends on an unusual

observation. Notice that if we multiply the integrand by the fraction sec x + tan xsec x + tan x

An Unusual IntegralExample 3.8

An Integrand with an Even Power of SecantExample 3.7



(which is of course equal to 1), we get∫sec x dx =

∫sec x

(sec x + tan x

sec x + tan x

)dx

=∫

sec2 x + sec x tan x

sec x + tan xdx .

Now, observe that the numerator is exactly the derivative of the denominator. That is,

d

dx(sec x + tan x) = sec x tan x + sec2 x,

so that taking u = sec x + tan x gives us∫sec x dx =

∫sec2 x + sec x tan x

sec x + tan xdx

=∫

1

udu = ln |u| + c

= ln |sec x + tan x | + c. Since u = sec x + tan x .

■

Trigonometric SubstitutionIf an integral contains a term of the form

√a2 − x2,

√a2 + x2 or

√x2 − a2, for some

a > 0, you can often evaluate the integral by making a substitution involving a trig func-tion (hence, the name trigonometric substitution).

First, suppose that an integrand contains a term of the form √

a2 − x2, for some a > 0.

If we let x = a sin θ, where −π2 ≤ θ ≤ π

2 , then we can eliminate the square root, asfollows. Notice that we now have√

a2 − x2 =√

a2 − (a sin θ)2 =√

a2 − a2 sin2 θ

= a√

1 − sin2 θ = a√

cos2 θ = a cos θ,

since for −π2 ≤ θ ≤ π

2 , cos θ ≥ 0. The following example is typical of how these substitu-tions are used.

Evaluate ∫

1

x2√

4 − x2dx .

Solution You should always first consider whether an integral can be done directly,with a simple substitution or by parts. Since none of these methods help here, we con-sider trigonometric substitution. Keep in mind that the immediate objective here is toeliminate the square root term. So, make a substitution that will accomplish this. We let

x = 2 sin θ, −π

2< θ <

π

2.

(Why do we need strict inequalities here?) This gives us

dx = 2 cos θ dθ

An Integral Involving √

a2 − x2Example 3.9



and hence,∫1

x2√

4 − x2dx =

∫2 cos θ

(2 sin θ)2√

4 − (2 sin θ)2dθ

=∫

2 cos θ

4 sin2 θ√

4 − 4 sin2 θdθ

=∫

cos θ

(2 sin2 θ) 2√

1 − sin2 θdθ

=∫

cos θ

4 sin2 θ cos θdθ Since 1 − sin2 θ = cos2 θ.

= 1

4

∫csc2 θ dθ = −1

4cot θ + c. Since

1

sin2 θ= csc2 θ.

The only remaining problem is that the antiderivative is presently written in terms ofthe variable θ. When converting back to the original variable x = 2 sin θ, we urgeyou to draw a diagram, as in Figure 7.1. Since the substitution was x = 2 sin θ, sin θ =x

2= opposite

hypotenuse, so we label the hypotenuse as 2. The side opposite the angle θ is then

2 sin θ. By the Pythagorean Theorem, we get that the adjacent side is √

4 − x2, as indi-cated. So, we have

cot θ = cos θ

sin θ=

√4 − x2

x.

It now follows that∫1

x2√

4 − x2dx = −1

4cot θ + c = −1

4

√4 − x2

x+ c.

■

Next, suppose that an integrand contains a term of the form √

a2 + x2, for somea > 0. If we let x = a tan θ, where −π

2 < θ < π2 , then we can eliminate the square root,

as follows. Notice that in this case, we have√a2 + x2 =

√a2 + (a tan θ)2 =

√a2 + a2 tan2 θ

= a√

1 + tan2 θ = a√

sec2 θ = a sec θ,

since for −π2 < θ < π

2 , sec θ ≥ 0. The following example is typical of how these substitu-tions are used.


1√9 + x2

dx .

Solution First, you should recognize that you do not know how to evaluate thisintegral directly. Of course, the big problem here is the square root of the sum. You caneliminate this by letting x = 3 tan θ, for −π

2 < θ < π2 . This gives us dx = 3 sec2 θ dθ,


a2 + x2Example 3.10


u

2 cos u � �4 � x2

2 2 sin u � x

Figure 7.1


so that ∫1√

9 + x2dx =

∫1√

9 + (3 tan θ)23 sec2 θ dθ

=∫

3 sec2 θ√9 + 9 tan2 θ

dθ

=∫

3 sec2 θ

3√

1 + tan2 θdθ

=∫

sec2 θ

sec θdθ Since 1 + tan2 θ = sec2 θ.

=∫

sec θ dθ

= ln |sec θ + tan θ | + c,

from example 3.8. We’re not done here, though, since we first need to express theintegral in terms of the original variable x. Observe that we had x = 3 tan θ, so thattan θ = x

3 . It remains only to solve for sec θ. Although you can do this with a triangle,as in example 3.9, the simplest way to do this is to recognize that for −π

2 < θ < π2 ,

sec θ =√

1 + tan2 θ =√

1 +(

x

3

)2

.

This leaves us with ∫1√

9 + x2dx = ln |sec θ + tan θ | + c

= ln

∣∣∣∣∣√

1 +(

x

3

)2

+ x

3

∣∣∣∣∣ + c.

■

Finally, suppose that an integrand contains a term of the form √

x2 − a2, for somea > 0. If we let x = a sec θ, where θ ∈ [

0, π2

) ∪ (π2 , π

], then we can eliminate the square

root, as follows. Notice that in this case, we have√x2 − a2 =

√(a sec θ)2 − a2 =

√a2 sec2 θ − a2

= a√

sec2 θ − 1 = a√

tan2 θ = a |tan θ |.Notice that the absolute values are needed, as tan θ can be both positive and negative on[0, π

2

) ∪ (π2 , π

]. The following example is typical of how these substitutions are used.

Evaluate the integral ∫ √

x2 − 25

xdx, for x > 5.

Solution Here, we make the substitution x = 5 sec θ, for θ ∈ [0, π

2

). Notice that

we chose the first half of the domain [0, π

2

) ∪ (π2 , π

], so that we would have

x = 5 sec θ > 5. (If we had x < −5, we would have chosen θ ∈ (π2 , π

].) Note that this


x2 − a2Example 3.11



gives us dx = 5 sec θ tan θ dθ and the integral then becomes:∫ √x2 − 25

xdx =

∫ √(5 sec θ)2 − 25

5 sec θ(5 sec θ tan θ) dθ

=∫ √

25 sec2 θ − 25 tan θ dθ

=∫

5√

sec2 θ − 1 tan θ dθ

= 5∫

tan2 θ dθ Since sec2 θ − 1 = tan2 θ.

= 5∫

(sec2 θ − 1) dθ

= 5(tan θ − θ) + c.

Finally, observe that since x = 5 sec θ, for θ ∈ [0, π

2

), we have that

tan θ =√

sec2 θ − 1 =√(

x

5

)2

− 1 = 1

5

√x2 − 25

and θ = sec−1(

x5

). We now have∫ √

x2 − 25

xdx = 5(tan θ − θ) + c

=√

x2 − 25 − 5 sec−1

(x

5

)+ c.

■

You will find a number of additional integrals requiring trigonometric substitution inthe exercises. The principal idea here is to see that you can eliminate certain square rootterms in an integrand by making use of a carefully chosen trigonometric substitution.

We summarize the three trigonometric substitutions presented here in the following table.


TrigonometricExpression Substitution Interval Identity

EXERCISES 7.3

1. Suppose a friend in your calculus class tells you thatthis section just has too many rules to memorize. (By

the way, the authors would agree.) Help your friend out bymaking it clear that each rule indicates when certain substitu-tions will work. In turn, a substitution u(x) works if the ex-pression u′(x) appears in the integrand and the resulting inte-

gral is easier to integrate. In each of the rules covered in thetext, identify u′(x) and point out why n has to be odd (or what-ever the rule says) for the remaining integrand to be workable.Without memorizing rules, you can remember a small numberof potential substitutions and see which one works for a givenproblem.

√a2 − x2 x = a sin θ − π

2 ≤ θ ≤ π

2 1 − sin2 θ = cos2 θ√

a2 + x2 x = a tan θ − π

2 < θ < π

2 1 + tan2 θ = sec2 θ√

x2 − a2 x = a sec θ θ ∈ [0, π

2

) ∪ (π

2 , π]

sec2 θ − 1 = tan2 θ


2. In the text, we suggested that when the integrand con-tains a term of the form

√4 − x2 , you might try the

trigonometric substitution x = 2 sin θ . We should admit nowthat this does not always work. How can you tell whether ornot this substitution will work?

In exercises 3–40, evaluate the integrals.

3.∫

cos x sin4 x dx 4.∫

cos3 x sin4 x dx

5.∫ π/4

0cos x sin3 x dx 6.

∫ π/3

π/4cos3 x sin3 x dx

7.∫

cos2 x sin x dx 8.∫

cos x sin2 x dx

9.∫ π/2

0cos x sin x dx 10.

∫ 0

−π/2cos3 x sin x dx

11.∫

cos2 x dx 12.∫

sin4 x dx

13.∫

tan x sec3 x dx 14.∫

tan3 x sec x dx

15.∫ π/4

0tan4 x sec4 x dx 16.

∫ π/4

−π/4tan4 x sec2 x dx

17.∫

cos2 x sin2 x dx 18.∫

(cos2 x + sin2 x) dx

19.∫

cot3 x csc3 x dx 20.∫

cot x csc4 x dx

21.∫ 0

−π/3

√cos x sin3 x dx 22.

∫ π/2

π/6cos x/

√sin x dx

23.∫

cot2 x csc4 x dx 24.∫

cot4 x csc4 x dx

25.∫

sin x

tan xdx 26.

∫sec 2x

csc 2xdx

27.∫

1

x2√

9 − x2dx 28.

∫1

x2√

16 − x2dx

29.∫ 2

0

√4 − x2 dx 30.

∫ 1

0

x√4 − x2

dx

31.∫

x2

√x2 − 9

dx 32.∫

x3√

x2 − 1 dx

33.∫

2√x2 − 4

dx 34.∫

x√x2 − 4

dx

35.∫

x2

√x2 + 9

dx 36.∫

x3√

x2 + 8 dx


37.∫ √

x2 + 16 dx 38.∫

1√x2 + 4

dx

39.∫ 1

0x√

x2 + 8 dx 40.∫ 2

0x2

√x2 + 9 dx

In exercises 41 and 42, evaluate the integral using both substitu-tions u = tan x and u = sec x and compare the results.

41.∫

tan x sec4 x dx 42.∫

tan3 x sec4 x dx

43. Show that for any integer n > 1, we have the reduction formula∫secn x dx = 1

n − 1secn−2 x tan x + n − 2

n − 1

∫secn−2 x dx .

44. Evaluate (a) ∫

sec3 x dx , (b) ∫

sec4 x dx and (c) ∫

sec5 x dx .

Hint: Use the result of exercise 43.

45. In an AC circuit, the current has the form i(t) = I cos(ωt) forconstants I and ω. The power is defined as Ri2 for a constantR. Find the average value of the power by integrating over theinterval [0, 2π/ω].

46. The area of the ellipse x2

a2+ y2

b2= 1 is given by

4b

a

∫ a

0

√a2 − x2 dx . Compute this integral.

47. Evaluate the antiderivatives in examples 3.2, 3.3, 3.5, 3.6 and3.7 using your CAS. Based on these examples, speculatewhether or not your CAS uses the same techniques that we do.In the cases where your CAS gives a different antiderivativethan we do, comment on which antiderivative looks simpler.

48. Repeat exercise 47 for examples 3.9, 3.10 and 3.11.

49. One CAS produces − 17 sin2 x cos5 x − 2

35 cos5 x as an anti-derivative in example 3.2. Find c such that this equals our anti-derivative of − 1

5 cos5 x + 17 cos7 x + c.

50. One CAS produces − 215 tan x − 1

15 sec2 x tan x + 15 sec4 x tan x

as an antiderivative in example 3.7. Find c such that this equalsour antiderivative of 1

3 tan3 x + 15 tan5 x + c.

51. In section 7.2, you were asked to show that forpositive integers m and n with m �= n, each of the

following integrals equals 0. ∫ π

−πcos mx cos nx dx = 0,∫ π

−πsin mx sin nx dx = 0. Also,

∫ π

−πcos2 nx dx =∫ π

−πsin2 nx dx = π. Finally,

∫ π

−πcos mx sin nx dx = 0 for any

positive integers m and n. We will use these formulas toexplain how a radio can tune in an AM station.

Amplitude modulation (or AM) radio sends a signal (e.g.,music) that modulates the carrier frequency. For example, if thesignal is 2 sin t and the carrier frequency is 16, then the radio


sends out the modulated signal 2 sin t sin 16t . The graphs ofy = 2 sin t , y = −2 sin t and y = 2 sin t sin 16t are shown inthe figure.

The graph of y = 2 sin t sin 16t oscillates as rapidly as the car-rier sin 16t , but the amplitude varies between 2 sin t and −2 sin t(hence the term amplitude modulation). The radio’s problem isto tune in the frequency 16 and recover the signal 2 sin t . The dif-

y

t

�2

�1

1

2

3p


ficulty is that other radio stations are broadcasting simultane-ously. A radio receives all the signals mixed together. To see howthis works, suppose a second station broadcasts the signal 3 sin tat frequency 32. The combined signal that the radio receives is2 sin t sin 16t + 3 sin t sin 32t. We will decompose this signal.The first step is to rewrite the signal using the identity

sin A sin B = 1

2cos(B − A) − 1

2cos(B + A).

The signal then equals

f (t) = cos 15t − cos 17t + 3

2cos 31t − 3

2cos 33t.

If the radio “knows” that the signal has the form c sin t forsome constant c, it can determine the constant c at frequency 16by computing the integral

∫ π

−πf (t) cos 15t dt and multiplying

by 2/π . Show that ∫ π

−πf (t) cos 15t dt = π, so that the correct

constant is c = π(2/π) = 2. The signal is then 2 sin t. Torecover the signal sent out by the second station, compute∫ π

−πf (t) cos 31t dt and multiply by 2/π . Show that you cor-

rectly recover the signal 3 sin t .

7.4 INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS

The notion we introduce in this section is less a technique of integration than it is an alge-braic method for rewriting certain rational functions. As we will see, it is very useful inintegration as well as in other applications. We begin with a simple observation. Note that

3

x + 2− 2

x − 5= 3(x − 5) − 2(x + 2)

(x + 2)(x − 5)= x − 19

x2 − 3x − 10. (4.1)

So, suppose that you wanted to evaluate the integral of the function on the right-hand sideof (4.1). While it’s not clear how to evaluate this integral, the integral of the (equivalent)function on the left-hand side of (4.1) is easy to evaluate. From (4.1), we now have∫

x − 19

x2 − 3x − 10dx =

∫ (3

x + 2− 2

x − 5

)dx = 3 ln |x + 2| − 2 ln |x − 5| + c.

The second integrand,

3

x + 2− 2

x − 5

is called a partial fractions decomposition of the first integrand. You may be wonderingif this is a fluke or if this kind of expansion can be done more generally. It turns out that wecan decompose any rational function (i.e., any quotient of two polynomials) into a sum ofsimpler rational functions. We develop this method in this section. Equation (4.1) is anexample of a special case of partial fractions. More generally, if the three factors a1x + b1,a2x + b2 and a3x + b3 are all distinct (i.e., no one is a constant multiple of another), then


we can write

for some choice of constants A and B to be determined. Notice that if you wanted to inte-grate this expression, the partial fractions on the right-hand side are very easy to integrate,just as they were in the introductory example above.

Evaluate ∫

1

x2 + x − 2dx .

Solution First, note that you can’t evaluate this as it stands and all of our earliermethods fail to help (consider each of these for this problem). However, we can make apartial fractions decomposition, as follows.

1

x2 + x − 2= 1

(x − 1)(x + 2)= A

x − 1+ B

x + 2= A(x + 2) + B(x − 1)

(x − 1)(x + 2).

Looking at the first and last of the fractions above, notice that their denominators are thesame. So, to be equal, their numerators must be the same, too. Thus,

1 = A(x + 2) + B(x − 1). (4.2)

We would like to solve this equation for A and B. The key is to realize that this equationmust be true for all x , including x = 1 and x = −2. [We single out these two valuesbecause they will make one or the other of the terms in (4.2) zero and thereby allow usto easily solve for the unknowns A and B.] In particular, for x = 1, notice that from(4.2), we have

1 = A(1 + 2) + B(1 − 1) = 3A,

so that A = 13 . Likewise, taking x = −2, we have

1 = A(−2 + 2) + B(−2 − 1) = −3B,

so that B = − 13 . Thus, we have ∫

1

x2 + x − 2dx =

∫ [1

3

(1

x − 1

)− 1

3

(1

x + 2

)]dx

= 1

3ln |x − 1| − 1

3ln |x + 2| + c.

■

We can do the same as we did above whenever a rational expression has a denomina-tor that factors into n distinct linear factors, as follows. If the degree of P(x) < n and thefactors (ai x + bi ), for i = 1, 2, . . . , n are all distinct, then we can write

for some constants c1, c2, . . . , cn.

P(x)

(a1x + b1)(a2x + b2) · · · (an x + bn)= c1

a1x + b1+ c2

a2x + b2+ · · · + cn

an x + bn,

Partial Fractions: Distinct Linear FactorsExample 4.1

a1x + b1

(a2x + b2)(a3x + b3)= A

a2x + b2+ B

a3x + b3,

Section 7.4 Integration of Rational Functions Using Partial Fractions 579

Partial fractions:

distinct linear factors


Evaluate ∫

3x2 − 7x − 2

x3 − xdx .

Solution Once again, our earlier methods fail us, but we can rewrite the integrandusing partial fractions. We have

3x2 − 7x − 2

x3 − x= 3x2 − 7x − 2

x(x − 1)(x + 1)= A

x+ B

x − 1+ C

x + 1

= A(x − 1)(x + 1) + Bx(x + 1) + Cx(x − 1)

x(x − 1)(x + 1).

Again, we equate the numerators of the fractions on the far left and on the far right, to get

3x2 − 7x − 2 = A(x − 1)(x + 1) + Bx(x + 1) + Cx(x − 1). (4.3)

In this case, notice that taking x = 0, x = 1 or x = −1 will make two of the three termson the right side of (4.3) zero. Specifically, for x = 0, we get

−2 = A(−1)(1) = −A,

so that A = 2. Likewise, taking x = 1, we find B = −3 and taking x = −1, we findC = 4. Thus, we have∫

3x2 − 7x − 2

x3 − xdx =

∫ (2

x− 3

x − 1+ 4

x + 1

)dx

= 2 ln |x | − 3 ln |x − 1| + 4 ln |x + 1| + c.

■

Find the indefinite integral of f (x) = 2x3 − 4x2 − 15x + 5

x2 − 2x − 8using a partial fractions

decomposition.

Solution Since the degree of the numerator exceeds that of the denominator, firstdivide. We indicate the long division below. (You should perform the division howeveryou are most comfortable.)

2x

x2 − 2x − 8)2x3 − 4x2 − 15x + 5

−−−−−−−−−−−−−−−2x3 − 4x2 − 16x

x + 5

Thus, we have

f (x) = 2x3 − 4x2 − 15x + 5

x2 − 2x − 8= 2x + x + 5

x2 − 2x − 8.

The remaining proper fraction can be expanded as

x + 5

x2 − 2x − 8= x + 5

(x − 4)(x + 2)= A

x − 4+ B

x + 2.

Partial Fractions Where Long Division Is RequiredExample 4.3

Partial Fractions: Three Distinct Linear FactorsExample 4.2


N O T E S

If the numerator of a rationalexpression has the same or higherdegree than the denominator, youmust first perform a long divisionand follow this with a partialfractions decomposition of theremaining (proper) fraction.


It is a simple matter to solve for the constants: A = 32 and B = − 1

2 . (This is left as anexercise.) We now have∫

2x3 − 4x2 − 15x + 5

x2 − 2x − 8dx =

∫ [2x + 3

2

(1

x − 4

)− 1

2

(1

x + 2

)]dx

= x2 + 3

2ln |x − 4| − 1

2ln |x + 2| + c.

■

You may already have begun to wonder what happens when the denominator of arational expression contains repeated linear factors. For example, is there a partial fractionsdecomposition of an expression like

2x + 3

(x − 1)2?

The answer is yes and the decomposition looks like the following. If the degree of P(x) < n,then we can write

for constants c1, c2, . . . , cn to be determined. The following example is typical.

Use a partial fractions decomposition to find an antiderivative of

f (x) = 5x2 + 20x + 6

x3 + 2x2 + x.

Solution First, note that there is a repeated linear factor in the denominator.We have

5x2 + 20x + 6

x3 + 2x2 + x= 5x2 + 20x + 6

x(x + 1)2= A

x+ B

x + 1+ C

(x + 1)2

= A(x + 1)2 + Bx(x + 1) + Cx

x(x + 1)2.

Comparing numerators, we have

5x2 + 20x + 6 = A(x + 1)2 + Bx(x + 1) + Cx .

Taking x = 0, we find A = 6. Likewise, taking x = −1, we find that C = 9. To deter-mine B, substitute any convenient value for x, say x = 1. (Unfortunately, notice thatthere is no choice of x that will make the two terms containing A and C both zero, with-out also making the term containing B zero.) You should find that B = −1. So, we have∫

5x2 + 20x + 6

x3 + 2x2 + xdx =

∫ [6

x− 1

x + 1+ 9

(x + 1)2

]dx

= 6 ln |x | − ln |x + 1| − 9(x + 1)−1 + c.

■

Partial Fractions with a Repeated Linear FactorExample 4.4

P(x)

(ax + b)n= c1

ax + b+ c2

(ax + b)2+ · · · + cn

(ax + b)n,

Partial fractions: repeated

linear factors



We can extend the notion of partial fractions decomposition to rational expressionswith denominators containing irreducible quadratic factors (i.e., quadratic factors that haveno real factorization). If the degree of P(x) < 2n (the degree of the denominator) and eachof the factors in the denominator are distinct, then we can write

(4.4)

You can think of this in terms of irreducible quadratic denominators in a partial fractions de-composition getting linear numerators, while linear denominators get constant numerators.If you think this looks messy, you’re right; it is. But only the algebra is messy (and you canalways use a CAS to do the algebra for you). You should note that the partial fractions onthe right-hand side of (4.4) are integrated comparatively easily using substitution togetherwith possibly completing the square.

Use a partial fractions decomposition to find an antiderivative of f (x) = 2x2 − 5x + 2

x3 + x.

Solution First, note that

2x2 − 5x + 2

x3 + x= 2x2 − 5x + 2

x(x2 + 1)= A

x+ Bx + C

x2 + 1= A(x2 + 1) + (Bx + C)x

x(x2 + 1).

Equating the numerators gives us

2x2 − 5x + 2 = A(x2 + 1) + (Bx + C)x

= (A + B)x2 + Cx + A.

Rather than substitute numbers for x (notice that there are no convenient values to plugin, except for x = 0), we instead match up the coefficients of like powers of x :

2 = A + B

−5 = C

2 = A.

This leaves us with B = 0 and so,∫2x2 − 5x + 2

x3 + xdx =

∫ (2

x− 5

x2 + 1

)dx = 2 ln |x | − 5 tan−1 x + c.

■

Often, partial fractions decompositions involving irreducible quadratic terms lead toexpressions that require further massaging (such as completing the square) before we canfind an antiderivative. We illustrate this in the following example.

Use a partial fractions decomposition to find an antiderivative for

f (x) = 5x2 + 6x + 2

(x + 2)(x2 + 2x + 5).

Partial Fractions with a Quadratic FactorExample 4.6

Partial Fractions with a Quadratic FactorExample 4.5

P(x)

(a1x2 + b1x + c1)(a2x2 + b2x + c2) · · · (an x2 + bn x + cn)

= A1x + B1

a1x2 + b1x + c1+ A2x + B2

a2x2 + b2x + c2+ · · · + An x + Bn

an x2 + bn x + cn.


Partial fractions: irreducible

quadratic factors


Solution First, notice that the quadratic factor in the denominator does not factorand so, the correct decomposition is

5x2 + 6x + 2

(x + 2)(x2 + 2x + 5)= A

x + 2+ Bx + C

x2 + 2x + 5

= A(x2 + 2x + 5) + (Bx + C)(x + 2)

(x + 2)(x2 + 2x + 5).

If we equate the numerators, we get

5x2 + 6x + 2 = A(x2 + 2x + 5) + (Bx + C)(x + 2).

Matching up the coefficients of like powers of x, we get

5 = A + B

6 = 2A + 2B + C

2 = 5A + 2C.

You’ll need to solve this by elimination. We leave it as an exercise to show that A = 2,

B = 3 and C = −4. Integrating, we have∫5x2 + 6x + 2

(x + 2)(x2 + 2x + 5)dx =

∫ (2

x + 2+ 3x − 4

x2 + 2x + 5

)dx . (4.5)

The integral of the first term is easy, but what about the second term? Since the denom-inator doesn’t factor, you have very few choices. Try substituting for the denominator:let u = x2 + 2x + 5, so that du = (2x + 2) dx . Notice that we can now write theintegral of the second term as∫

3x − 4

x2 + 2x + 5dx =

∫3(x + 1) − 7

x2 + 2x + 5dx =

∫ [(3

2

)2(x + 1)

x2 + 2x + 5− 7

x2 + 2x + 5

]dx

= 3

2

∫2(x + 1)

x2 + 2x + 5dx −

∫7

x2 + 2x + 5dx

= 3

2ln(x2 + 2x + 5) −

∫7

x2 + 2x + 5dx . (4.6)

To see how to evaluate the remaining integral in (4.6), you’ll need to complete thesquare in the denominator. Notice that we can write∫

7

x2 + 2x + 5dx =

∫7

(x + 1)2 + 4dx = 7

2tan−1

(x + 1

2

)+ c.

(We leave the details of this last integration as an exercise.) Putting this together with(4.5) and (4.6), we now have∫

5x2 +6x + 2

(x +2)(x2 +2x + 5)dx = 2 ln |x +2|+ 3

2ln(x2 +2x +5)− 7

2tan−1

(x + 1

2

)+c.

■

Rational expressions with repeated irreducible quadratic factors in the denominatorare explored in the exercises. The idea of these is the same as the preceding decomposi-tions, but the algebra (without a CAS) is even messier.



After mastering decompositions involving repeated irreducible quadratic factors, you will be able tofind the partial fractions decomposition of any rational function. The denominator of such a function(a polynomial) can always be factored into linear and quadratic factors, some of which may be re-peated. Then, use the techniques covered in this section! You should recognize that while this obser-vation is certainly true, in practice you may require a CAS to accurately complete the calculations.

In section 6.5, we introduced the logistic model of population growth. This model de-scribes the growth of a population under the assumption that the rate of change of the pop-ulation is proportional to the current population and to the difference between the currentpopulation and the maximum sustainable population M. In this case, the population func-tion p(t) satisfies the differential equation p′(t) = kp(M − p), for some constant k. As wesee in the following example, a partial fractions decomposition can be used to solve differ-ential equations of this type.

Solve the initial value problem p′(t) = 3p(10 − p) with p(0) = 3.

Solution This is a separable differential equation (see section 6.5), which werewrite in the separable form

1

p(10 − p)p′(t) = 3.

Integrating both sides of this equation with respect to t , we get∫1

p(10 − p)p′(t) dt =

∫3 dt

or∫

1

p(10 − p)dp =

∫3 dt. (4.7)

Notice that we’ll need partial fractions in order to evaluate the integral on the left-handside of this equation. In this case, we have

1

p(10 − p)= A

p+ B

10 − p= A(10 − p) + Bp

p(10 − p),

so that

1 = A(10 − p) + Bp.

Taking p = 0, we get 1 = 10A or A = 110 ; taking p = 10, we get 1 = 10B or B = 1

10 .

Thus, we have the partial fractions decomposition

1

p(10 − p)= 1/10

p+ 1/10

10 − p.

Returning to equation (4.7), we now have∫ (1/10

p+ 1/10

10 − p

)dp =

∫3 dt

or1

10

∫1

pdp − 1

10

∫ −1

10 − pdp =

∫3 dt. Separating terms.

Using Partial Fractions to Solve a Differential EquationExample 4.7

Remark 4.1


N O T E S

Most CASs include commands forperforming partial fractionsdecomposition. Even so, we urgeyou to work through the exercisesin this section by hand. Once youhave the idea of how thesedecompositions work, by all means,use your CAS to do the drudgework for you. Until that time, bepatient and work carefully by hand.


Finally, integrating gives us

1

10ln |p| − 1

10ln |10 − p| = 3t + c. Evaluating the integrals.

It’s a fairly simple matter to solve this equation for p(t) and the details are found insection 6.5.

■


EXERCISES 7.4

1. There is a shortcut for determining the constants forlinear terms in a partial fractions decomposition. For

example, take

x − 1

(x + 1)(x − 2)= A

x + 1+ B

x − 2.

To compute A, take the original fraction on the left, cover upthe x + 1 in the denominator and replace x with −1:

A = −1 − 1−1 − 2

= 23. Similarly, to solve for B, cover up the

x − 2 and replace x with 2: B = 2 − 12 + 1

= 13

. Explain why this

works and practice it for yourself.

2. For partial fractions, there is a big distinction betweenquadratic functions that factor into linear terms and qua-

dratic functions that are irreducible. Recall that a quadraticfunction factors as (x − a)(x − b) if and only if a and b arezeros of the function. Explain how you can use the quadraticformula to determine whether or not if a given quadratic func-tion is irreducible.

In exercises 3–34, find the partial fractions decomposition and anantiderivative. If you have a CAS available, use it to check youranswer.

3.x − 5

x2 − 14.

5x − 2

x2 − 4

5.6x

x2 − x − 26.

4

x2 − 1

7.x + 1

x2 − x − 68.

3x

x2 − 3x − 4

9.−x + 5

x3 − x2 − 2x10.

3x + 8

x3 + 5x2 + 6x

11.x3 + x + 2

x2 + 2x − 812.

x2 + 1

x2 − 5x − 6

13.−3x − 1

x3 − x14.

x2 + 3

x3 − x

15.2x + 3

(x + 2)216.

3x − 5

(x − 1)2

17.x − 1

x3 + 4x2 + 4x18.

4x − 5

x3 − 3x2

19.x + 4

x3 + 3x2 + 2x20.

−2x2 + 4

x3 + 3x2 + 2x

21.x + 2

x3 + x22.

1

x3 + 4x

23.4x − 2

x4 − 124.

3x + 7

x4 − 16

25.3x2 − 6

x2 − x − 226.

x3 + x

x2 − 1

27.2x + 3

x2 + 2x + 128.

2x

x2 − 6x + 9

29.x2 + 2x + 1

x3 + x30.

2x2 − 3x + 2

x3 + x

31.4x2 + 3

x3 + x2 + x32.

4x + 4

x4 + x3 + 2x2

33.3x3 + 1

x3 − x2 + x − 134.

2x4 + 9x2 + x − 4

x3 + 4x

35. In this exercise, we find the partial fractions decomposition of4x2 + 2(x2 + 1)2

. Consistent with the form for repeated linear factors,

the form for the decomposition is Ax + Bx2 + 1

+ Cx + D

(x2 + 1)2. We

set

4x2 + 2

(x2 + 1)2= Ax + B

x2 + 1+ Cx + D

(x2 + 1)2


Multiplying through by (x2 + 1)2 , we get

4x2 + 2 = (Ax + B)(x2 + 1) + Cx + D

= Ax3 + Bx2 + Ax + B + Cx + D

As in example 4.5, we match up coefficients of like powers ofx . For x3 , we have 0 = A. For x2 , we have 4 = B . Match thecoefficients of x and the constants to finish the decomposition.

In exercises 36–40, find the partial fractions decomposition.(Refer to exercise 35.)

36.x3 + 2

(x2 + 1)237.

2x2 + 4

(x2 + 4)2

38.2x3 − x2

(x2 + 1)239.

4x2 + 3

(x2 + x + 1)2

40.x4 + x3

(x2 + 4)2

41. To solve the logistic differential equation y′ = ay(1 − y) in

section 6.5, we wrote it in the separable form 1y(1 − y)

y′ = a

and integrated. Perform the partial fractions decomposition of

the integrand, integrate ∫

1y(1 − y)

dy and then solve the

differential equation.

42. In certain chemical reactions, two molecules of substance Areact with one molecule of substance B to form one moleculeof a third substance X. This type of reaction is found in thebreakdown of the ozone layer. If there are a total of a moleculesof A available and a total of b molecules of B available, the


amount x(t) of X formed obeys the equation

x ′(t) = k[a − 2x(t)]2[b − x(t)]

for some rate constant k. Write this as a separable equation anduse partial fractions to find an implicit representation of thesolution.

43. In this exercise, you explore a more general form ofthe logistic equation than that analyzed in section 6.5.

Suppose that y′ = ay − by2 . Note that if a = b, this is thesame as the logistic equation previously studied. Assuming

that 0 < y < a/b, show that y(t) = aCe−at + b

, where the

constant C equals ay(0)

− b. What happens to y(t) as t → ∞?

How does this compare to the logistic solutions found previ-ously? Solve the equation if y > a/b.

44. In developing the definite integral, we looked at sums

such as n∑

i=1

2i2 + i

. As with Riemann sums, we are es-

pecially interested in the limit as n → ∞. Write out severalterms of the sum and try to guess what the limit is. It turns outthat this is one of the few sums for which a precise formula ex-ists, because this is a telescoping sum. To find out what this

means, write out the partial fractions decomposition for 2i2 + i

.

Using the partial fractions form, write out several terms of thesum and notice how much cancellation there is. Briefly de-scribe why the term telescoping is appropriate, and determine

n∑i=1

2i2 + i

. Then find the limit as n → ∞. Repeat this process

for the telescoping sum n∑

i=1

4ii2 − 1

.

7.5 INTEGRATION TABLES AND COMPUTERALGEBRA SYSTEMS

Ask anyone who has ever needed to evaluate a large number of integrals as part of theirwork (this includes innumerable engineers, mathematicians, physicists and others) andthey will tell you that they have made extensive use of integral tables and/or a computeralgebra system. These are extremely powerful tools for the professional user of mathemat-ics. However, they do not take the place of learning all the basic techniques of integrationthat we’ve been discussing. To use a table, you often must first rewrite the integral in theform of one of the integrals in the table. This may require you to perform some algebraicmanipulation or to make a substitution. While a CAS will report an antiderivative, it willoccasionally report it in an inconvenient form. More significantly, a CAS will from time totime report an answer that is (at least technically) incorrect. We will point out some of theseshortcomings in the examples that follow.


Using Tables of IntegralsWe include a small table of indefinite integrals at the back of the book. A larger table canbe found in the CRC Standard Mathematical Tables. An amazingly extensive table is foundin the book Table of Integrals, Series and Products, compiled by Gradshteyn and Ryzhik.If you can’t find an integral there, you’re not likely to find it anywhere.

Use a table to evaluate ∫ √

3 + 4x2

xdx .

Solution Certainly, you could evaluate this integral using trigonometric substitu-tion. However, if you look in our integral table, you will find∫ √

a2 + u2

udu =

√a2 + u2 − a ln

∣∣∣∣∣a + √a2 + u2

u

∣∣∣∣∣ + c. (5.1)

Unfortunately, the integral in question is not quite in the form of (5.1). However, we canfix this with the substitution u = 2x, so that du = 2 dx . This gives us∫ √

3 + 4x2

xdx =

∫ √3 + (2x)2

2x(2) dx =

∫ √3 + u2

udu

=√

3 + u2 −√

3 ln

∣∣∣∣∣√

3 + √3 + u2

u

∣∣∣∣∣ + c

=√

3 + 4x2 −√

3 ln

∣∣∣∣∣√

3 + √3 + 4x2

2x

∣∣∣∣∣ + c.

■

A number of the formulas in the table are called reduction formulas. These are of the form ∫f (u) du = g(u) +

∫h(u) du,

where the second integral is simpler than the first. These are often applied repeatedly, as inthe following example.

Use a reduction formula to evaluate ∫

sin6 x dx .

Solution You should recognize that this integral can be evaluated using tech-niques you already know. (How?) However, for any integer n ≥ 1, we have the reduc-tion formula ∫

sinn u du = −1

nsinn−1 u cos u + n − 1

n

∫sinn−2 u du. (5.2)

If we apply (5.2) with n = 6, we get∫sin6 x dx = −1

6sin5 x cos x + 5

6

∫sin4 x dx .

Using a Reduction FormulaExample 5.2

Using an Integral TableExample 5.1

Section 7.5 Integration Tables and Computer Algebra Systems 587


CHAPTERINTEGRATION 7 TECHNIQUES should recall from our brief intro-duction to differential equations...

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