Techniques of Integration Because of the Fundamental Theorem of Calculus, We Can Integrate a Function If We Know an Antiderivative, That Is, an Indeﬁnite Integral

Home , Integration by parts, Trigonometric integral

The techniques of this chapter enable us to ﬁnd the height of a rocket a minute after liftoff and to compute the escape velocity of the rocket.

Techniques of Integration Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an antiderivative, that is, an indeﬁnite integral. We summarize here the most important integrals that we have learned so far.

x n1 1 y x n dx C n 1 y dx ln x C n 1 x

a x y e x dx e x C y a x dx C ln a

y sin x dx cos x C y cos x dx sin x C

y sec2xdx tan x C y csc2xdx cot x C

y sec x tan x dx sec x C y csc x cot x dx csc x C

y sinh x dx cosh x C y cosh x dx sinh x C

y tan x dx ln sec x C y cot x dx ln sin x C

1 1 x 1 x y dx tan1 C y dx sin1 C x 2 a 2 a a sa 2 x 2 a

In this chapter we develop techniques for using these basic integration formulas to obtain indeﬁnite integrals of more complicated functions. We learned the most important method of integration, the Substitution Rule, in Section 5.5. The other general technique, integration by parts, is presented in Section 7.1. Then we learn methods that are special to particular classes of functions such as trigonometric functions and rational functions. Integration is not as straightforward as differentiation; there are no rules that absolutely guarantee obtaining an indeﬁnite integral of a function. Therefore, in Section 7.5 we discuss a strategy for integration.

|||| 7.1 Integration by Parts

Every differentiation rule has a corresponding integration rule. For instance, the Substi- tution Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for integration by parts. The Product Rule states that if f and t are differentiable functions, then

d f xtx f xtx txf x dx 476 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

In the notation for indeﬁnite integrals this equation becomes

y f xtx txf x dx f xtx

or y f xtx dx y txf x dx f xtx

We can rearrange this equation as

1 y f xtx dx f xtx y txf x dx

Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f x and v tx . Then the differentials are du f x dx and dv tx dx , so, by the Substitution Rule, the formula for integration by parts becomes

2 y udv uv y v du

EXAMPLE 1 Find y x sin x dx.

SOLUTION USING FORMULA 1 Suppose we choose f x x and tx sin x . Then f x 1 and tx cos x . (For t we can choose any antiderivative of t .) Thus, using Formula 1, we have y x sin x dx f xtx y txf x dx

xcos x y cos x dx

x cos x y cos x dx

x cos x sin x C It’s wise to check the answer by differentiating it. If we do so, we get x sin x , as expected.

SOLUTION USING FORMULA 2 Let

|||| It is helpful to use the pattern: u x dv sin x dx u dv du v Then du dx v cos x

and so u d√ u √ √ du y x sin x dx y x sin x dx x cos x y cos x dx

x cos x y cos x dx

x cos x sin x C SECTION 7.1 INTEGRATION BY PARTS ❙❙❙❙ 477

NOTE Our aim in using integration by parts is to obtain a simpler integral than the one we started with. Thus, in Example 1 we started with x x sin x dx and expressed it in terms of the simpler integral x cos x dx . If we had chosen u sin x and dv x dx , then du cos x dx and v x 22 , so integration by parts gives

x 2 1 y x sin x dx sin x y x 2 cos x dx 2 2

Although this is true,x x 2 cos x dx is a more difﬁcult integral than the one we started with. In general, when deciding on a choice for u and dv , we usually try to choose u f x to be a function that becomes simpler when differentiated (or at least not more complicated) as long as dv tx dx can be readily integrated to give v .

EXAMPLE 2 Evaluate y ln x dx.

SOLUTION Here we don’t have much choice for u and dv . Let

u ln x dv dx

1 Then du dx v x x

Integrating by parts, we get dx y ln x dx x ln x y x x

|||| It’s customary to write x 1 dx as x dx . x ln x y dx

|||| Check the answer by differentiating it. x ln x x C

Integration by parts is effective in this example because the derivative of the function f x ln x is simpler than f .

EXAMPLE 3 Find y t 2et dt.

SOLUTION Notice that t 2 becomes simpler when differentiated (whereas et is unchanged when differentiated or integrated), so we choose

u t 2 dv et dt

Then du 2tdt v et

Integration by parts gives

2 t 2 t t 3 y t e dt t e 2 y te dt

The integral that we obtained,x tet dt , is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time, this time with u t and 478 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

dv et dt. Then du dt ,v et , and

y tet dt tet y et dt

tet et C

Putting this in Equation 3, we get

y t 2et dt t 2et 2 y tet dt

t 2et 2tet et C

2 t t t t e 2te 2e C1 where C1 2C

EXAMPLE 4 Evaluate y e x sin x dx.

|||| An easier method, using complex numbers, SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing is given in Exercise 48 in Appendix G. u e x and dv sin x dx anyway. Then du e x dx and v cos x , so integration by parts gives

4 y e x sin x dx e x cos x y e x cos x dx

The integral that we have obtained,x e x cos x dx , is no simpler than the original one, but at least it’s no more difﬁcult. Having had success in the preceding example integrating by parts twice, we persevere and integrate by parts again. This time we use u e x and dv cos x dx. Then du e x dx ,v sin x , and

5 y e x cos x dx e x sin x y e x sin x dx

At ﬁrst glance, it appears as if we have accomplished nothing because we have arrived at x |||| Figure 1 illustrates Example 4 by x e sin x dx, which is where we started. However, if we put Equation 5 into Equation 4 showing the graphs of f x e x sin x and we get 1 x F x 2 e sin x cos x . As a visual check on our work, notice that f x 0 when F has x x x x a maximum or minimum. y e sin x dx e cos x e sin x y e sin x dx

12 This can be regarded as an equation to be solved for the unknown integral. Adding x ex sin x dx to both sides, we obtain F f 2 y e x sin x dx e x cos x e x sin x

_3 6 Dividing by 2 and adding the constant of integration, we get _4 x 1 x y e sin x dx 2 e sin x cos x C FIGURE 1

If we combine the formula for integration by parts with Part 2 of the Fundamental Theorem of Calculus, we can evaluate deﬁnite integrals by parts. Evaluating both sides of Formula 1 between a and b , assuming f and t are continuous, and using the Fundamental SECTION 7.1 INTEGRATION BY PARTS ❙❙❙❙ 479

Theorem, we obtain

b t t b b t 6 y f x x dx f x x ]a y x f x dx a a

1 EXAMPLE 5 Calculate y tan 1xdx. 0

SOLUTION Let

u tan1x dv dx dx Then du v x 1 x 2

So Formula 6 gives

1 1 1 x 1 1 y tan xdx x tan x]0 y 2 dx 0 0 1 x

1 x 1 1 1 tan 1 0 tan 0 y 2 dx 0 1 x

|||| 1 Since tan x 0 for x 0 , the integral in 1 x Example 5 can be interpreted as the area of the y 2 dx 4 0 1 x region shown in Figure 2. To evaluate this integral we use the substitution t 1 x 2 (since u has another meaning y in this example). Then dt 2xdx , so xdx dt2 . When x 0 ,t 1 ; when x 1 , y=tan–!x t 2; so

1 x 2 dt 2 0 1 1 y 2 dx 2 y 2 ln t ]1 0 1 x 1 t 1 x 1 1 2 ln 2 ln 1 2 ln 2

1 1 x ln 2 1 Therefore y tan x dx y 2 dx FIGURE 2 0 4 0 1 x 4 2

EXAMPLE 6 Prove the reduction formula

|||| Equation 7 is called a reduction formula 1 n 1 because the exponent n has been reduced to 7 y sinnxdx cos x sinn 1x y sinn 2xdx n 1 and n 2. n n

where n 2 is an integer.

SOLUTION Let u sinn 1x dv sin x dx

Then du n 1 sinn2x cos x dx v cos x

so integration by parts gives

y sinnx dx cos x sinn1x n 1 y sinn2x cos2x dx 480 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

Since cos2x 1 sin2x, we have

y sinnxdx cos x sinn1x n 1 y sinn2x dx n 1 y sinnx dx

As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side. Thus, we have

n y sinnx dx cos x sinn1x n 1 y sinn2x dx

1 n 1 or y sinnx dx cos x sinn1x y sinn2x dx n n

The reduction formula (7) is useful because by using it repeatedly we could eventually express x sinnx dx in terms of x sin x dx (if n is odd) or x sin x0 dx x dx (if n is even).

|||| 7.1 Exercises

1–2 |||| Evaluate the integral using integration by parts with the 1 y 2 2 23.y 2y dy 24. y x csc x dx indicated choices of u and dv . 0 e 4

1. y x ln x dx;,u ln x dv x dx 1 2 1 x 25.y cos 1x dx 26. y x5 dx 0 0

y 2 v 2 2. sec d ;,u d sec d s3 27.y cos x lnsin x dx 28. y arctan1x dx 1

3 3–32 |||| Evaluate the integral. 1 r 29. y cosln x dx 30. y dr 0 s4 r 2 3.y x cos 5x dx 4. y xex dx

2 t 31.y x 4ln x2 dx 32. y e s sint s ds 5.y rer2 dr 6. y t sin 2t dt 1 0

y x 2 sin x dx y x 2 cos mx dx 7. 8. 33–36 |||| First make a substitution and then use integration by parts to evaluate the integral. 1 9.y ln 2x 1 dx 10. y sin x dx 4 33.y sin sx dx 34. y esx dx 1 11.y arctan 4t dt 12. y p5 ln p dp s 2 35. y 3 cos 2 d 36. y x 5e x dx s2 y 2 y 3 t 13. ln x dx 14. t e dt

15. y e 2 sin 3 d 16. y e cos 2 d ; 37–40 |||| Evaluate the indeﬁnite integral. Illustrate, and check that your answer is reasonable, by graphing both the function and its antiderivative (take C 0 ). 17.y y sinh y dy 18. y y cosh ay dy 37.y x cos x dx 38. y x 32 ln x dx 1 19. y t sin 3t dt 20. y x 2 1ex dx 0 0 2 39.y 2x 3e x dx 40. y x 3e x dx 2 ln x 4 s 21.y 2 dx 22. y t ln t dt 1 x 1 SECTION 7.1 INTEGRATION BY PARTS ❙❙❙❙ 481

41. (a) Use the reduction formula in Example 6 to show that ; 53–54 |||| Use a graph to ﬁnd approximate x -coordinates of the points of intersection of the given curves. Then ﬁnd (approxi- x sin 2x y sin2x dx C mately) the area of the region bounded by the curves. 2 4 53. y x sin x, y x 22 (b) Use part (a) and the reduction formula to evaluate 54. y arctan 3x, y x2 x sin4x dx.

42. (a) Prove the reduction formula 55–58 |||| Use the method of cylindrical shells to ﬁnd the volume 1 n 1 generated by rotating the region bounded by the given curves about y cosnx dx cosn 1x sin x y cosn 2x dx n n the speciﬁed axis. (b) Use part (a) to evaluate x cos2x dx . 55. y cos x 2 ,y 0 ,0 x 1 ; about the y -axis x 4 (c) Use parts (a) and (b) to evaluate cos x dx . 56. y e x,y e x ,x 1 ; about the y -axis 43. (a) Use the reduction formula in Example 6 to show that 57. y ex,y 0 ,x 1 ,x 0 ; about x 1

x 2 n 1 2 58. y e ,x 0 ,y ; about the x -axis y sinnx dx y sinn2x dx 0 0 n

where n 2 is an integer. 59. Find the average value of f x x 2 ln x on the interval 1, 3 . (b) Use part (a) to evaluate x2 sin3x dx and x2 sin5x dx . 0 0 60. A rocket accelerates by burning its onboard fuel, so its mass (c) Use part (a) to show that, for odd powers of sine, decreases with time. Suppose the initial mass of the rocket at

2 2 4 6 2n liftoff (including its fuel) is m , the fuel is consumed at rate r , 2n1 y sin x dx and the exhaust gases are ejected with constant velocity ve 0 3 5 7 2n 1 (relative to the rocket). A model for the velocity of the rocket 44. Prove that, for even powers of sine, at time t is given by the equation m rt 2 2n 1 3 5 2n 1 vt tt ve ln y sin x dx m 0 2 4 6 2n 2 where t is the acceleration due to gravity and t is not too 45–48 |||| Use integration by parts to prove the reduction formula. large. If t 9.8 ms2 ,m 30,000 kg,r 160 kgs, and ve 3000 ms, ﬁnd the height of the rocket one minute 45. y ln xn dx xln xn n y ln xn1 dx after liftoff.

61. A particle that moves along a straight line has velocity 2 t 46. y x ne x dx x ne x n y x n1e x dx vt t e meters per second after t seconds. How far will it travel during the ﬁrst t seconds?

2 2 n 62. If f 0 t0 0 and f and t are continuous, show that 47. y x a dx

a a 2 2 n 2 t t t t x x a 2na 1 y f x x dx f a a f a a y f x x dx y x 2 a 2 n 1 dx (n ) 0 0 2n 1 2n 1 2 f f f f n2 63. Suppose that 1 2 ,4 7 ,1 5 ,4 3 , and 4 n tan x sec x n 2 n2 x 48. y sec x dx y sec x dx n 1 f is continuous. Find the value of 1 xf x dx . n 1 n 1 64. (a) Use integration by parts to show that

49. Use Exercise 45 to ﬁnd x ln x3 dx. y f x dx xfx y xfx dx 50. Use Exercise 46 to ﬁnd x x 4e x dx. (b) If f and t are inverse functions and f is continuous, prove

51–52 |||| Find the area of the region bounded by the given curves. that 0.4x b f b 51. y xe ,,y 0 x 5 y f x dx bfb afa y ty dy a f a 52. y 5 ln x, y x ln x

[Hint: Use part (a) and make the substitution y f x .] 482 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

(c) In the case where f and t are positive functions and (c) Use parts (a) and (b) to show that b a 0, draw a diagram to give a geometric interpreta- 2n 1 I tion of part (b). 2n 1 1 xe 2n 2 I2n (d) Use part (b) to evaluate 1 ln x dx .

xb and deduce that limn l I2n1 I2n 1 . 65. We arrived at Formula 6.3.2,V a 2 xf x dx , by using cylindrical shells, but now we can use integration by parts to (d) Use part (c) and Exercises 43 and 44 to show that prove it using the slicing method of Section 6.2, at least for the 2 2 4 4 6 6 2n 2n case where f is one-to-one and therefore has an inverse func- lim n l 1 3 3 5 5 7 2n 1 2n 1 2 tion t . Use the ﬁgure to show that

d This formula is usually written as an inﬁnite product: V b 2d a 2c y ty2 dy c 2 2 4 4 6 6 Make the substitution y f x and then use integration by 2 1 3 3 5 5 7 xb parts on the resulting integral to prove that V a 2 xf x dx . and is called the Wallis product. y (e) We construct rectangles as follows. Start with a square of x=g(y) y=ƒ area 1 and attach rectangles of area 1 alternately beside or d on top of the previous rectangle (see the ﬁgure). Find the limit of the ratios of width to height of these rectangles.

c x=b x=a 0 abx

x2 n 66. Let In 0 sin x dx.

(a) Show that I2n2 I2n1 I2n . (b) Use Exercise 44 to show that I2n2 2n 1 I2n 2n 2

|||| 7.2 Trigonometric Integrals

In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.

EXAMPLE 1 Evaluate y cos3x dx.

SOLUTION Simply substituting u cos x isn’t helpful, since then du sin x dx . In order to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of sine would require an extra cos x factor. Thus, here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine using the identity sin2x cos2x 1: cos3x cos2x cos x 1 sin2x cos x We can then evaluate the integral by substituting u sin x , so du cos x dx and

y cos3x dx y cos2x cos x dx y 1 sin2x cos x dx

2 1 3 y 1 u du u 3 u C 1 3 sin x 3 sin x C SECTION 7.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 483

In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor (and the remainder of the expression in terms of cosine) or only one cosine factor (and the remainder of the expression in terms of sine). The identity sin2x cos2x 1 enables us to convert back and forth between even powers of sine and cosine.

EXAMPLE 2 Find y sin5x cos2x dx

SOLUTION We could convert cos2x to 1 sin2x , but we would be left with an expression in terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and rewrite the remaining sin4x factor in terms of cos x :

sin5x cos2x sin2x2 cos2x sin x 1 cos2x2 cos2x sin x

|||| Figure 1 shows the graphs of the integrand Substituting u cos x, we have du sin x dx and so sin5x cos2x in Example 2 and its indeﬁnite integral (with C 0 ). Which is which? y sin5x cos2x dx y sin2x2 cos2x sin x dx 0.2 y 1 cos2x2 cos2x sin x dx

_π π y 1 u 2 2 u 2 du y u 2 2u 4 u 6 du

u 3 u 5 u 7 2 C _0.2 3 5 7 1 3 2 5 1 7 FIGURE 1 3 cos x 5 cos x 7 cos x C

In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix D):

2 1 2 1 sin x 2 1 cos 2x and cos x 2 1 cos 2x

|||| Example 3 shows that the area of the region EXAMPLE 3 Evaluate y sin2x dx. shown in Figure 2 is 2 . 0 SOLUTION If we write sin2x 1 cos2x , the integral is no simpler to evaluate. Using the 1.5 half-angle formula for sin2x , however, we have

y=sin@ x 2 1 1 1 y sin x dx 2 y 1 cos 2x dx [ 2 (x 2 sin 2x)]0 0 0 1 1 1 1 1 2 ( 2 sin 2 ) 2 (0 2 sin 0) 2 0 π Notice that we mentally made the substitution u 2x when integrating cos 2x . Another _0.5 method for evaluating this integral was given in Exercise 41 in Section 7.1.

FIGURE 2 EXAMPLE 4 Find y sin4x dx.

SOLUTION We could evaluate this integral using the reduction formula for x sinnx dx (Equation 7.1.7) together with Example 1 (as in Exercise 41 in Section 7.1), but a better 484 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

method is to write sin4x sin2x2 and use a half-angle formula:

y sin4x dx y sin2x2 dx

1 cos 2x 2 y dx 2 1 2 4 y 1 2 cos 2x cos 2x dx Since cos2 2x occurs, we must use another half-angle formula

2 1 cos 2x 2 1 cos 4x This gives 4 1 1 y sin x dx 4 y 1 2 cos 2x 2 1 cos 4x dx

1 3 1 4 y ( 2 2 cos 2x 2 cos 4x) dx

1 3 1 4 ( 2 x sin 2x 8 sin 4x) C

To summarize, we list guidelines to follow when evaluating integrals of the form x sinmx cosnx dx, where m 0 and n 0 are integers.

Strategy for Evaluating y sinmx cosnx dx

(a) If the power of cosine is odd n 2k 1 , save one cosine factor and use cos2x 1 sin2x to express the remaining factors in terms of sine:

y sinmx cos2k1x dx y sinmx cos2xk cos x dx

y sinmx 1 sin2xk cos x dx

Then substitute u sin x . (b) If the power of sine is odd m 2k 1 , save one sine factor and use sin2x 1 cos2x to express the remaining factors in terms of cosine:

y sin2k1x cosnx dx y sin2xk cosnx sin x dx

y 1 cos2xk cosnx sin x dx

Then substitute u cos x . [Note that if the powers of both sine and cosine are odd, either (a) or (b) can be used.] (c) If the powers of both sine and cosine are even, use the half-angle identities 2 1 2 1 sin x 2 1 cos 2x cos x 2 1 cos 2x It is sometimes helpful to use the identity 1 sin x cos x 2 sin 2x SECTION 7.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 485

We can use a similar strategy to evaluate integrals of the form x tanmx secnx dx . Since ddx tan x sec2x, we can separate a sec2x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec2x 1 tan2x . Or, since ddx sec x sec x tan x , we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant.

EXAMPLE 5 Evaluate y tan6x sec4x dx.

SOLUTION If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x 1 tan2x . We can then evaluate the integral by substituting u tan x with du sec2x dx :

y tan6x sec4x dx y tan6x sec2x sec2x dx

y tan6x 1 tan2x sec2x dx

y u 61 u 2 du y u 6 u 8 du

u 7 u 9 C 7 9

1 7 1 9 7 tan x 9 tan x C

EXAMPLE 6 Find y tan5 sec7 d.

SOLUTION If we separate a sec2 factor, as in the preceding example, we are left with a sec5 factor, which isn’t easily converted to tangent. However, if we separate a sec tan factor, we can convert the remaining power of tangent to an expression involving only secant using the identity tan2 sec2 1 . We can then evaluate the integral by substituting u sec , so du sec tan d :

y tan5 sec7 d y tan4 sec6 sec tan d

y sec2 12 sec6 sec tan d

y u 2 12 u 6 du y u 10 2u 8 u 6 du

u 11 u 9 u 7 2 C 11 9 7

1 11 2 9 1 7 11 sec 9 sec 7 sec C

The preceding examples demonstrate strategies for evaluating integrals of the form x tanmx secnx dx for two cases, which we summarize here. 486 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

Strategy for Evaluating y tanmx secnx dx

(a) If the power of secant is even n 2k, k 2 , save a factor of sec2x and use sec2x 1 tan2x to express the remaining factors in terms of tan x :

y tanmx sec2kx dx y tanmx sec2xk1 sec2x dx

y tanmx 1 tan2xk1 sec2x dx

Then substitute u tan x . (b) If the power of tangent is odd m 2k 1 , save a factor of sec x tan x and use tan2x sec2x 1 to express the remaining factors in terms of sec x :

y tan2k1x secnx dx y tan2xk secn1x sec x tan x dx

y sec2x 1k secn1x sec x tan x dx

Then substitute u sec x .

For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to integrate tan x by using the formula established in (5.5.5):

y tan x dx ln sec x C

We will also need the indeﬁnite integral of secant:

1 y sec x dx ln sec x tan x C

We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x :

sec x tan x y sec x dx y sec x dx sec x tan x sec2x sec x tan x y dx sec x tan x

If we substitute u sec x tan x , then du sec x tan x sec2x dx , so the integral becomes x 1u du ln u C. Thus,we have

y sec x dx ln sec x tan x C SECTION 7.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 487

EXAMPLE 7 Find y tan3x dx.

SOLUTION Here only tan x occurs, so we use tan2x sec2x 1 to rewrite a tan2x factor in terms of sec2x : y tan3x dx y tan x tan2x dx

y tan x sec2x 1 dx

y tan x sec2x dx y tan x dx

tan2x ln sec x C 2 In the ﬁrst integral we mentally substituted u tan x so that du sec2x dx .

If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x . Powers of sec x may require integration by parts, as shown in the following example.

EXAMPLE 8 Find y sec3x dx.

SOLUTION Here we integrate by parts with u sec x dv sec2x dx du sec x tan x dx v tan x

Then y sec3x dx sec x tan x y sec x tan2x dx

sec x tan x y sec x sec2x 1 dx

sec x tan x y sec3x dx y sec x dx

Using Formula 1 and solving for the required integral, we get

3 1 y sec x dx 2 (sec x tan x ln sec x tan x ) C

Integrals such as the one in the preceding example may seem very special but they occur frequently in applications of integration, as we will see in Chapter 8. Integrals of the form x cotmx cscnx dx can be found by similar methods because of the identity 1 cot2x csc2x. Finally, we can make use of another set of trigonometric identities:

2 To evaluate the integrals (a) x sin mx cos nx dx , (b) x sin mx sin nx dx , or (c)x cos mx cos nx dx , use the corresponding identity:

|||| 1 These product identities are discussed in (a) sin A cos B 2 sin A B sin A B Appendix D. 1 (b) sin A sin B 2 cos A B cos A B 1 (c)cos A cos B 2 cos A B cos A B 488 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

EXAMPLE 9 Evaluate y sin 4x cos 5x dx.

SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use the identity in Equation 2(a) as follows:

1 y sin 4x cos 5x dx y 2 sin x sin 9x dx

1 2 y sin x sin 9x dx

1 1 2 (cos x 9 cos 9x C

|||| 7.2 Exercises

1–47 |||| Evaluate the integral. 31.y tan5x dx 32. y tan6ay dy

1.y sin3x cos2x dx 2. y sin6x cos3x dx tan3 33.y d 34. y tan2x sec x dx cos4 34 2 3. y sin5x cos3x dx 4. y cos5x dx 2 0 2 2 35.y cot2x dx 36. y cot3x dx 6 4 5.y cos5x sin4x dx 6. y sin3mx dx 37.y cot 3 csc3 d 38. y csc 4x cot 6x dx 2 2 7. y cos2 d 8. y sin22 d 0 0 3 39.y csc x dx 40. y csc3x dx 6 9.y sin43t dt 10. y cos6 d 0 0 41. y sin 5x sin 2x dx 42. y sin 3x cos x dx 11.y 1 cos 2 d 12. y x cos2x dx cos x sin x 43.y cos 7 cos 5 d 44. y dx 4 2 sin 2x 13. y sin4x cos2x dx 14. y sin2x cos2x dx 0 0 1 tan2x dx 45.y dx 46. y sec2x cos x 1 15.y sin3x scos x dx 16. y cos cos5sin d 47. y t sec2t 2 tan4t 2 dt 2 3 5 4 17.y cos x tan x dx 18. y cot sin d

1 sin x 48. If x 4 tan6x sec x dx I , express the value of 19.y dx 20. y cos2x sin 2x dx 0 cos x x4 8 0 tan x sec x dx in terms of I . 2 2 4 21.y sec x tan x dx 22. y sec t2 dt ; 49–52 |||| Evaluate the indeﬁnite integral. Illustrate, and check that 0 your answer is reasonable, by graphing both the integrand and its antiderivative (taking C 0 . 23. y tan2x dx 24. y tan4x dx 49.y sin5x dx 50. y sin4x cos4x dx 4 25.y sec6t dt 26. y sec4 tan4 d 0 x 51.y sin 3x sin 6x dx 52. y sec4 dx 3 2 27.y tan5x sec4x dx 28. y tan32x sec52x dx 0

2 3 3 53. Find the average value of the function f x sin x cos x on 29. y tan3x sec x dx 30. y tan5x sec6x dx 0 the interval , . SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 489

54. Evaluate x sin x cos x dx by four methods: (a) the substitution 64. Household electricity is supplied in the form of alternating u cos x, (b) the substitution u sin x , (c) the identity current that varies from 155 V to 155 V with a frequency sin 2x 2 sin x cos x, and (d) integration by parts. Explain the of 60 cycles per second (Hz). The voltage is thus given by different appearances of the answers. the equation Et 155 sin120t 55–56 |||| Find the area of the region bounded by the given curves. where t is the time in seconds. Voltmeters read the RMS (root- 3 55. y sin x, y sin x, x 0, x 2 mean-square) voltage, which is the square root of the average 2 56. y sin x, y 2 sin2x, x 0, x 2 value of E t over one cycle. (a) Calculate the RMS voltage of household current. (b) Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude A needed for the voltage ; 57–58 |||| Use a graph of the integrand to guess the value of the Et A sin120t. integral. Then use the methods of this section to prove that your guess is correct. 65–67 |||| Prove the formula, where m and n are positive integers. 2 2 65. y sin mx cos nx dx 0 57.y cos3x dx 58. y sin 2x cos 5x dx 0 0 0 if m n 66. y sin mx sin nx dx if m n 59–62 |||| Find the volume obtained by rotating the region bounded 0 if m n by the given curves about the specified axis. 67. y cos mx cos nx dx if m n 59. y sin x,x 2 ,x ,y 0 ; about the x -axis 60. y tan2x,y 0 ,x 0 ,x 4 ; about the x -axis 68. A finite Fourier series is given by the sum 61. y cos x,y 0 ,x 0 ,x 2 ; about y 1 N f x an sin nx 62. y cos x,y 0 ,x 0 ,x 2 ; about y 1 n1 a1 sin x a2 sin 2x aN sin Nx Show that the m th coefficient a is given by the formula 63. A particle moves on a straight line with velocity function m vt sin t cos2t. Find its position function s f t if 1 am y f x sin mx dx f 0 0.

|||| 7.3 Trigonometric Substitution

In ﬁnding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 dx arises, where a 0 . If it were x xsa 2 x 2 dx , the substitution u a 2 x 2 would be effective but, as it stands,x sa 2 x 2 dx is more difﬁcult. If we change the variable from x to by the substitution x a sin , then the identity 1 sin2 cos2 allows us to get rid of the root sign because

sa 2 x 2 sa 2 a 2 sin2 sa 21 sin 2 sa 2 cos2 a cos

Notice the difference between the substitution u a 2 x 2 (in which the new variable is a function of the old one) and the substitution x a sin (the old variable is a function of the new one). In general we can make a substitution of the form x tt by using the Substitution Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is,t is one-to-one. In this case, if we replace u by x and x by t in the Substitution Rule (Equation 5.5.4), we obtain

y f x dx y f tttt dt

This kind of substitution is called inverse substitution. 490 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

We can make the inverse substitution x a sin provided that it defines a one-to-one function. This can be accomplished by restricting to lie in the interval 2, 2 . In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities. In each case the restric- tion on is imposed to ensure that the function that defines the substitution is one-to-one. (These are the same intervals used in Appendix D in defining the inverse functions.)

Table of Trigonometric Substitutions

Expression Substitution Identity sa 2 x 2 x a sin , 1 sin2 cos2 2 2 sa 2 x 2 x a tan , 1 tan2 sec2 2 2 3 sx 2 a 2 x a sec ,0 or sec2 1 tan2 2 2

s9 x 2 EXAMPLE 1 Evaluate y dx. x 2

SOLUTION Let x 3 sin , where 2 2 . Then dx 3 cos d and

s9 x 2 s9 9 sin2 s9 cos2 3 cos 3 cos

(Note that cos 0 because 2 2 .) Thus, the Inverse Substitution Rule gives s9 x 2 3 cos y dx y 3 cos d x 2 9 sin2 cos2 y d y cot2 d sin2

y csc2 1 d

cot C

Since this is an indeﬁnite integral, we must return to the original variable x . This can be 3 done either by using trigonometric identities to express cot in terms of sin x 3 or x by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle. Sincesin x3 , we label the opposite side and the hypotenuse as having lengthsx and 3. ¨ Then the Pythagorean Theorem gives the length of the adjacent side as s9 x 2 , so we œ„„„„9-≈„ can simply read the value of cot from the ﬁgure:

FIGURE 1 s9 x 2 x cot sin ¨= 3 x (Although 0 in the diagram, this expression for cot is valid even when 0 .) Since sin x3, we have sin1x3 and so

s9 x 2 s9 x 2 x y dx sin1 C x 2 x 3 SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 491

EXAMPLE 2 Find the area enclosed by the ellipse

x 2 y 2 1 a 2 b 2

y SOLUTION Solving the equation of the ellipse for y , we get

(0, b) y 2 x 2 a 2 x 2 b 1 or y sa 2 x 2 b 2 a 2 a 2 a (a, 0) 0 x Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the ﬁrst quadrant (see Figure 2). The part of the ellipse in the ﬁrst quadrant is given by the function b y sa 2 x 2 0 x a a FIGURE 2

≈ ¥ a b 1 s 2 2 +=1 and so 4 A y a x dx a@ b@ 0 a

To evaluate this integral we substitute x a sin . Then dx a cos d . To change the limits of integration we note that when x 0 ,sin 0 , so 0 ; when x a , sin 1, so 2. Also

sa 2 x 2 sa 2 a 2 sin2 sa 2 cos2 a cos a cos

since 0 2. Therefore

b a b 2 A 4 y sa 2 x 2 dx 4 y a cos a cos d a 0 a 0

2 2 2 1 4ab y cos d 4ab y 2 1 cos 2 d 0 0

1 2 2ab[ 2 sin 2 ]0 2ab 0 0 2 ab

We have shown that the area of an ellipse with semiaxes a and b is ab . In particular, taking a b r , we have proved the famous formula that the area of a circle with radius r is r 2 .

NOTE Since the integral in Example 2 was a deﬁnite integral, we changed the limits of integration and did not have to convert back to the original variable x .

1 EXAMPLE 3 Find y dx. x 2sx 2 4

SOLUTION Let x 2 tan , 2 2. Then dx 2 sec2 d and

sx 2 4 s4tan 2 1 s4 sec 2 2 sec 2 sec

Thus, we have dx 2 sec2 d 1 sec y y y d x 2sx 2 4 4tan2 2sec 4 tan2 492 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

To evaluate this trigonometric integral we put everything in terms of sin and cos :

2 sec 1 cos cos tan2 cos sin2 sin2

Therefore, making the substitution u sin , we have

dx 1 cos 1 du y y d y x 2sx 2 4 4 sin2 4 u 2

1 1 1 C C œ„„„„„≈+4 4 u 4 sin x csc C ¨ 4 2 We use Figure 3 to determine that csc sx 2 4x and so FIGURE 3 s 2 x dx x 4 tan ¨= y C 2 x 2sx 2 4 4x x EXAMPLE 4 Find y dx. sx 2 4

SOLUTION It would be possible to use the trigonometric substitution x 2tan here (as in Example 3). But the direct substitution u x 2 4 is simpler, because then du 2x dx and x 1 du y dx y su C sx 2 4 C sx 2 4 2 su

NOTE Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method ﬁrst.

dx EXAMPLE 5 Evaluate y , where a 0. sx 2 a 2

SOLUTION 1 We let x a sec , where 0 2 or 32 . Then dx a sec tan d and

sx 2 a 2 sa 2sec 2 1 sa 2 tan2 a tan a tan

Therefore dx a sec tan y y d sx 2 a 2 a tan

x y sec d ln sec tan C œ„„„„„≈-a@

¨ The triangle in Figure 4 gives tan sx 2 a 2a , so we have a dx x sx 2 a 2 y ln C FIGURE 4 sx 2 a 2 a a x sec ¨= a ln x sx 2 a 2 ln a C SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 493

Writing C1 C ln a, we have

dx 2 2 1 y ln x sx a C1 sx 2 a 2

SOLUTION 2 For x 0 the hyperbolic substitution x a cosh t can also be used. Using the identity cosh2 y sinh2 y 1, we have

sx 2 a 2 sa 2 cosh2 t 1 sa 2 sinh2 t a sinh t

Since dx a sinh t dt, we obtain

dx a sinh t dt y y y dt t C sx 2 a 2 a sinh t

Since cosh t xa, we have t cosh1xa and

dx x 2 y cosh1 C sx 2 a 2 a

Although Formulas 1 and 2 look quite different, they are actually equivalent by Formula 3.9.4.

NOTE As Example 5 illustrates, hyperbolic substitutions can be used in place of trigonometric substitutions and sometimes they lead to simpler answers. But we usually use trigonometric substitutions because trigonometric identities are more familiar than hyperbolic identities.

3 3 s32 x EXAMPLE 6 Find y 2 32 dx. 0 4x 9 SOLUTION First we note that 4x 2 93 2 s4x 2 9)3 so trigonometric substitution is appropriate. Although s4x 2 9 is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitu- 3 tion u 2x . When we combine this with the tangent substitution, we have x 2 tan , 3 2 which gives dx 2 sec d and

s4x 2 9 s9 tan2 9 3 sec

When x 0,tan 0 , so 0; when x 3s32,tan s3 , so 3 .

3 27 3 3 s32 x 3 tan 8 3 2 y 2 32 dx y 3 2 sec d 0 4x 9 0 27 sec

3 3 3 3 tan 3 3 sin 16 y d 16 y 2 d 0 sec 0 cos

2 3 3 1 cos 16 y 2 sin d 0 cos

Now we substitute u cos so that du sin d . When 0 ,u 1 ; when 1 3, u 2. 494 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

Therefore

3 2 3 s32 x 12 1 u 12 3 3 2 y 2 32 dx 16 y 2 du 16 y 1 u du 0 4x 9 1 u 1 1 1 2 3 3 1 3 16 u 16 [( 2 2) 1 1 ] 32 u 1 x EXAMPLE 7 Evaluate y dx. s3 2x x 2

SOLUTION We can transform the integrand into a function for which trigonometric substitution is appropriate by ﬁrst completing the square under the root sign:

3 2x x 2 3 x 2 2x 3 1 x 2 2x 1 4 x 12

This suggests that we make the substitutionu x 1 . Thendu dx andx u 1 , so

x u 1 y dx y du s3 2x x 2 s4 u 2

We now substitute u 2 sin , giving du 2 cos d and s4 u 2 2 cos , so |||| Figure 5 shows the graphs of the integrand in Example 7 and its indeﬁnite integral (with x 2 sin 1 C 0). Which is which? y dx y 2 cos d s3 2x x 2 2 cos 3 y 2 sin 1 d

_4 2 2 cos C u s4 u 2 sin1 C 2 _5 x 1 s3 2x x 2 sin1 C FIGURE 5 2

|||| 7.3 Exercises

1–3 |||| Evaluate the integral using the indicated trigonometric 4–30 |||| Evaluate the integral. substitution. Sketch and label the associated right triangle. 3 2 s3 x 4. y dx 1 s 2 1. y dx; x 3 sec 0 16 x x 2 sx 2 9 2 1 2 3 s 2 5.y 3 dt 6. y x x 4 dx s2 t st 2 1 0 2. y x 3 s9 x 2 dx; x 3 sin 1 sx 2 a 2 7. y dx 8. y dx 2 s 2 4 x 3 x 25 x x 3. y dx; x 3 tan sx 2 9 dx t 5 9.y 10. y dt s 2 s 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ x 16 t 2 SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 495

1 2 2 2 11.y s1 4x 2 dx 12. y xsx 2 4 dx equation x y r . Then A is the sum of the area of the 0 triangle POQ and the area of the region PQR in the ﬁgure.] sx 2 9 du 13. y dx 14. y y x 3 us5 u 2 P x 2 dx 15.y dx 16. y a 2 x 2 32 x 2 s16x 2 9 x dx 17. y dx 18. y ¨ sx 2 7 ax2 b 2 32 O Q R x s1 x 2 t 19.y dx 20. y dt x s25 t 2 ; 36. Evaluate the integral

23 1 dx 3s 2 s 2 y 21. y x 4 9x dx 22. y x 1 dx 4 2 0 0 x sx 2

dt Graph the integrand and its indeﬁnite integral on the same 23.y s5 4x x 2 dx 24. y st 2 6t 13 screen and check that your answer is reasonable. 1 x 2 ; 37. Use a graph to approximate the roots of the equation 25.y dx 26. y dx s9x 2 6x 8 s4x x 2 x 2 s4 x 2 2 x. Then approximate the area bounded by the curve y x 2 s4 x 2 and the line y 2 x . dx dx 27.y 28. y x 2 2x 22 5 4x x 2 52 38. A charged rod of length L produces an electric ﬁeld at point Pa, b given by 2 cos t 29.y xs1 x 4 dx 30. y dt 0 s1 sin2t La b EP y dx 2 2 3 2 a 40x b ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

where is the charge density per unit length on the rod and 0 31. (a) Use trigonometric substitution to show that is the free space permittivity (see the ﬁgure). Evaluate the integral to determine an expression for the electric ﬁeld EP. dx y ln(x sx 2 a 2 ) C y sx 2 a 2 P (a, b) (b) Use the hyperbolic substitution x a sinh t to show that 0 L x dx x y sinh1 C sx 2 a 2 a

These formulas are connected by Formula 3.9.3. 39. Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii r and R . (See the ﬁgure.) 32. Evaluate x 2 y dx x 2 a 2 32 r

(a) by trigonometric substitution. R (b) by the hyperbolic substitution x a sinh t .

33. Find the average value of f x sx 2 1x ,1 x 7 . 40. A water storage tank has the shape of a cylinder with diameter 34. Find the area of the region bounded by the hyperbola 10 ft. It is mounted so that the circular cross-sections are verti- 2 2 9x 4y 36 and the line x 3 . cal. If the depth of the water is 7 ft, what percentage of the 1 2 total capacity is being used? 35. Prove the formula A 2 r for the area of a sector of a circle with radius r and central angle . [Hint: Assume 0 2 41. A torus is generated by rotating the circle x 2 y R2 r 2 and place the center of the circle at the origin so it has the about the x -axis. Find the volume enclosed by the torus. 496 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| 7.4 Integration of Rational Functions by Partial Fractions

In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions 2x 1 and 1x 2 to a common denominator we obtain 2 1 2 x 2 x 1 x 5 x 1 x 2 x 1x 2 x 2 x 2

If we now reverse the procedure, we see how to integrate the function on the right side of this equation:

x 5 2 1 y dx y dx x 2 x 2 x 1 x 2 2 ln x 1 ln x 2 C

To see how the method of partial fractions works in general, let’s consider a rational function Px f x Qx

where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q . Such a rational function is called proper. Recall that if

n n1 Px an x an1x a1 x a0

where an 0 , then the degree of P is n and we write degP n . If f is improper, that is,degP degQ , then we must take the preliminary step of dividing Q into P (by long division) until a remainder Rx is obtained such that degR degQ. The division statement is

Px Rx 1 f x Sx Qx Qx

where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required.

x 3 x EXAMPLE 1 Find y dx. x 1

SOLUTION Since the degree of the numerator is greater than the degree of the denominator, ≈+x +2 we ﬁrst perform the long division. This enables us to write x-1)˛ +x ˛-≈ x 3 x 2 ≈+x y dx y x 2 x 2 dx ≈-x x 1 x 1 2x x 3 x 2 2x-2 2x 2 ln x 1 C 2 3 2 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 497

The next step is to factor the denominator Qx as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax b ) and irreducible quadratic factors (of the form ax2 bx c , where b 2 4ac 0 ). For instance, if Qx x 4 16 , we could factor it as

Qx x 2 4x 2 4 x 2x 2x 2 4

The third step is to express the proper rational function RxQx (from Equation 1) as a sum of partial fractions of the form

A Ax B or ax bi ax2 bx c j

A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur.

CASE I ■ The denominator Qx is a product of distinct linear factors. This means that we can write

Qx a1 x b1 a2 x b2 ak x bk

where no factor is repeated (and no factor is a constant multiple of another). In this case

the partial fraction theorem states that there exist constants A1, A2, ..., Ak such that

Rx A1 A2 Ak 2 Qx a1 x b1 a2 x b2 ak x bk

These constants can be determined as in the following example.

x 2 2x 1 EXAMPLE 2 Evaluate y dx. 2x 3 3x 2 2x

SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide. We factor the denominator as

2x 3 3x 2 2x x2x 2 3x 2 x2x 1x 2

Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form

x 2 2x 1 A B C 3 x2x 1x 2 x 2x 1 x 2

|||| Another method for ﬁnding A , B , and C To determine the values of A ,B , and C , we multiply both sides of this equation by the is given in the note after this example. product of the denominators,x2x 1x 2 , obtaining

4 x 2 2x 1 A2x 1x 2 Bxx 2 Cx2x 1

Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get

5 x 2 2x 1 2A B 2Cx 2 3A 2B Cx 2A 498 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| Figure 1 shows the graphs of the integrand The polynomials in Equation 5 are identical, so their coefficients must be equal. The in Example 2 and its indefinite integral (with coefficient of x 2 on the right side,2A B 2C , must equal the coefficient of x 2 on the K 0). Which is which? left side—namely, 1. Likewise, the coefficients of x are equal and the constant terms are 2 equal. This gives the following system of equations for A ,B , and C :

2 A B 2 C 1 _3 3 3 A 2 B C 2

2A 2 B 2 C 1 _2 Solving, we get A 1 ,B 1 , and C 1 , and so FIGURE 1 2 5 10 x 2 2x 1 1 1 1 1 1 1 y dx y dx 2x 3 3x 2 2x 2 x 5 2x 1 10 x 2

1 1 1 2 ln x 10 ln 2x 1 10 ln x 2 K

|||| We could check our work by taking the terms In integrating the middle term we have made the mental substitution u 2x 1 , which to a common denominator and adding them. gives du 2 dx and dx du2.

NOTE ■ We can use an alternative method to ﬁnd the coefﬁcients A,B , and C in Example 2. Equation 4 is an identity; it is true for every value of x . Let’s choose values of x that simplify the equation. If we put x 0 in Equation 4, then the second and third terms 1 on the right side vanish and the equation then becomes 2A 1 , or A 2 . Likewise, 1 1 1 1 x 2 gives 5B 4 4 and x 2 gives 10C 1 , so B 5 and C 10 . (You may object 1 that Equation 3 is not valid for x 0 ,2 , or 2 , so why should Equation 4 be valid for those 1 values? In fact, Equation 4 is true for all values of x , even x 0 ,2 , and 2 . See Exercise 67 for the reason.)

dx EXAMPLE 3 Find y , where a 0. x 2 a 2

SOLUTION The method of partial fractions gives

1 1 A B x 2 a 2 x ax a x a x a

and therefore

Ax a Bx a 1

Using the method of the preceding note, we put x a in this equation and get A2a 1, so A 12a . If we put x a , we get B2a 1 , so B 12a . Thus

dx 1 1 1 y y dx x 2 a 2 2a x a x a

1 (ln x a ln x a ) C 2a SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 499

Since ln x ln y lnxy , we can write the integral as

dx 1 x a 6 y ln C x 2 a 2 2a x a

See Exercises 53Ð54 for ways of using Formula 6.

CASE II ■ Qx is a product of linear factors, some of which are repeated. r Suppose the ﬁrst linear factor a1 x b1 is repeated r times; that is,a1 x b1 occurs in the factorization of Qx . Then instead of the single term A1a1 x b1 in Equation 2, we would use

A1 A2 Ar 7 2 r a1 x b1 a1 x b1 a1 x b1

By way of illustration, we could write

3 x x 1 A B C D E x 2x 13 x x 2 x 1 x 12 x 13

but we prefer to work out in detail a simpler example.

x 4 2x 2 4x 1 EXAMPLE 4 Find y dx. x 3 x 2 x 1

SOLUTION The ﬁrst step is to divide. The result of long division is

x 4 2x 2 4x 1 4x x 1 x 3 x 2 x 1 x 3 x 2 x 1

The second step is to factor the denominator Qx x 3 x 2 x 1 . Since Q1 0 , we know that x 1 is a factor and we obtain

x 3 x 2 x 1 x 1x 2 1 x 1x 1x 1 x 12x 1

Since the linear factor x 1 occurs twice, the partial fraction decomposition is

4x A B C x 12x 1 x 1 x 12 x 1

Multiplying by the least common denominator,x 12x 1 , we get

8 4 x Ax 1x 1 Bx 1 Cx 12 A Cx 2 B 2Cx A B C

|||| Another method for finding the coefficients: Now we equate coefficients: Put x 1 in (8):B 2 . Put x 1:C 1 . A B C 0 Put x 0:A B C 1 . A B 2 C 4

A B C 0 500 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

Solving, we obtain A 1 ,B 2 , and C 1 , so

x 4 2x 2 4x 1 1 2 1 y dx y x 1 dx x 3 x 2 x 1 x 1 x 12 x 1 x 2 2 x ln x 1 ln x 1 K 2 x 1 x 2 2 x 1 x ln K 2 x 1 x 1

CASE III ■ Qx contains irreducible quadratic factors, none of which is repeated. If Qx has the factor ax2 bx c , where b 2 4ac 0 , then, in addition to the partial fractions in Equations 2 and 7, the expression for RxQx will have a term of the form

Ax B 9 ax2 bx c

where A and B are constants to be determined. For instance, the function given by f x xx 2x 2 1x 2 4 has a partial fraction decomposition of the form x A Bx C Dx E x 2x 2 1x 2 4 x 2 x 2 1 x 2 4

The term given in (9) can be integrated by completing the square and using the formula

dx 1 x 10 y tan 1 C x 2 a 2 a a

2x 2 x 4 EXAMPLE 5 Evaluate y dx. x 3 4x

SOLUTION Since x 3 4x xx 2 4 can’t be factored further, we write

2 2x x 4 A Bx C xx 2 4 x x 2 4

Multiplying by xx 2 4 , we have

2 x 2 x 4 Ax 2 4 Bx Cx A Bx 2 Cx 4A

Equating coefﬁcients, we obtain

A B 2 C 1 4A 4

Thus A 1,B 1 , and C 1 and so

2x 2 x 4 1 x 1 y dx y dx x 3 4x x x 2 4 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 501

In order to integrate the second term we split it into two parts:

x 1 x 1 y dx y dx y dx x 2 4 x 2 4 x 2 4

We make the substitution u x 2 4 in the ﬁrst of these integrals so that du 2x dx . We evaluate the second integral by means of Formula 10 with a 2 :

2x 2 x 4 1 x 1 y dx y dx y dx y dx xx 2 4 x x 2 4 x 2 4

1 2 1 1 ln x 2 ln x 4 2 tan x 2 K

4x 2 3x 2 EXAMPLE 6 Evaluate y dx. 4x 2 4x 3

SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we ﬁrst divide and obtain

4x 2 3x 2 x 1 1 4x 2 4x 3 4x 2 4x 3

Notice that the quadratic 4x 2 4x 3 is irreducible because its discriminant is b 2 4ac 32 0. This means it can’t be factored, so we don’t need to use the partial fraction technique. To integrate the given function we complete the square in the denominator:

4x 2 4x 3 2x 12 2

This suggests that we make the substitution u 2x 1 . Then,du 2 dx and x u 12,so

4x 2 3x 2 x 1 y dx y 1 dx 4x 2 4x 3 4x 2 4x 3 1 u 1 1 u 1 x 1 y 2 du x 1 y du 2 u 2 2 4 u 2 2 u 1 x 1 y du 1 y du 4 u 2 2 4 u 2 2 1 1 u x 1 lnu 2 2 tan1 C 8 4 s2 s2 1 2x 1 x 1 ln4x 2 4x 3 tan1 C 8 4s2 s2

NOTE ■ Example 6 illustrates the general procedure for integrating a partial fraction of the form

Ax B where b 2 4ac 0 ax2 bx c 502 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

We complete the square in the denominator and then make a substitution that brings the integral into the form

Cu D u 1 y du C y du D y du u 2 a 2 u 2 a 2 u 2 a 2

Then the ﬁrst integral is a logarithm and the second is expressed in terms of tan1 .

CASE IV ■ Qx contains a repeated irreducible quadratic factor. If Qx has the factor ax2 bx cr , where b 2 4ac 0 , then instead of the single partial fraction (9), the sum

A1 x B1 A2 x B2 Ar x Br 11 ax2 bx c ax2 bx c2 ax2 bx cr

occurs in the partial fraction decomposition of RxQx . Each of the terms in (11) can be integrated by ﬁrst completing the square.

|||| It would be extremely tedious to work out by EXAMPLE 7 Write out the form of the partial fraction decomposition of the function hand the numerical values of the coefﬁcients in Example 7. Most computer algebra systems, x 3 x 2 1 however, can ﬁnd the numerical values very 2 2 3 quickly. For instance, the Maple command xx 1x x 1x 1 convertf, parfrac, x SOLUTION or the Mathematica command Apart[f] x 3 x 2 1 gives the following values: xx 1x 2 x 1x 2 13 A 1, B 1 , C D 1, 8 15 1 3 A B Cx D Ex F Gx H Ix J E 8 , F 8 , G H 4 , x x 1 x 2 x 1 x 2 1 x 2 12 x 2 13 1 1 I 2 , J 2

1 x 2x 2 x 3 EXAMPLE 8 Evaluate y dx. xx 2 12

SOLUTION The form of the partial fraction decomposition is

2 3 1 x 2x x A Bx C Dx E xx 2 12 x x 2 1 x 2 12

Multiplying by xx 2 12 , we have

x 3 2x 2 x 1 Ax 2 12 Bx Cxx 2 1 Dx Ex Ax 4 2x 2 1 Bx 4 x 2 Cx 3 x Dx 2 Ex A Bx 4 Cx3 2A B Dx 2 C Ex A

If we equate coefﬁcients, we get the system

A B 0 C 1 2A B D 2 C E 1 A 1

which has the solution A 1 ,B 1 ,C 1 ,D 1 , and E 0 . Thus SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 503

1 x 2x 2 x 3 1 x 1 x y dx y dx xx 2 12 x x 2 1 x 2 12 dx x dx x dx y y dx y y x x 2 1 x 2 1 x 2 12

1 |||| In the second and fourth terms we made the 1 2 1 ln x 2 ln x 1 tan x K mental substitution u x 2 1 . 2x 2 1

We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral

x 2 1 y dx xx 2 3

could be evaluated by the method of Case III, it’s much easier to observe that if u xx 2 3 x 3 3x, then du 3x 2 3 dx and so

x 2 1 y dx 1 ln x 3 3x C xx 2 3 3

Rationalizing Substitutions

Some nonrational functions can be changed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form sn tx, then the substitution u sn tx may be effective. Other instances appear in the exercises.

sx 4 EXAMPLE 9 Evaluate y dx. x

SOLUTION Let u sx 4. Then u 2 x 4,so x u 2 4 and dx 2u du. Therefore

sx 4 u u 2 y dx y 2u du 2 y du x u 2 4 u 2 4 4 2 y 1 du u 2 4

We can evaluate this integral either by factoring u 2 4 as u 2u 2 and using partial fractions or by using Formula 6 with a 2 :

sx 4 du y dx 2 y du 8 y x u 2 4 1 u 2 2u 8 ln C 2 2 u 2

sx 4 2 2sx 4 2ln C sx 4 2 504 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| 7.4 Exercises

1–6 |||| Write out the form of the partial fraction decomposition 1 1 x 31. y 3 dx 32. y 2 dx of the function (as in Example 7). Do not determine the numerical x 1 0 x 4x 13

values of the coefﬁcients. 2 3 5 x 2x x 33.y 3 2 dx 34. y 3 dx 2x 1 2 x x x 1. (a) (b) 3 4 1 3 2 x 3 3x 1 x 2x x 3 dx 1 2x 5x 35.y 4 2 36. y 4 2 dx x 1 x 1 x x 0 x x 2. (a) (b) 5 4 x 3 x 2 x 3 x x 3 x 4 1 2 37.y dx 38. y dx 2 x x 2 2x 42 xx 2 12 3. (a)2 (b) 2 x 3x 4 x 1x x 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ x 3 2x 1 4. (a) (b) 39–48 |||| Make a substitution to express the integrand as a rational 2 3 2 2 x 4x 3 x 1 x 4 function and then evaluate the integral. x 4 t 4 t 2 1 5. (a) (b) 1 1 4 2 2 2 39.y dx 40. y dx x 1 t 1 t 4 xsx 1 x sx 2 x 4 1 16 sx 1 1 6. (a)3 2 (b) 6 3 y dx y dx x xx x 3 x x 41. 42. 3 9 x 4 0 1 sx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 3 x 3 sx 43. y 3 dx 44. y 2 dx 7–38 |||| Evaluate the integral. sx 2 1 13 x x x r 2 1 7.y dx 8. y dr 45. y dx [Hint: Substitute u s6 x.] x 6 r 4 sx s3 x x 9 1 1 9.y dx 10. y dt 46. y dx [Hint: Substitute u 12sx.] x 5x 2 t 4t 1 s3 x s4 x

2x 3 1 1 x 1 e cos x 47. y dx 48. y dx 11. y 2 dx 12. y 2 dx 2x x 2 2 x 1 0 x 3x 2 e 3e 2 sin x sin x

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ax 1 13.y dx 14. y dx 2 x bx x a x b 49–50 |||| Use integration by parts, together with the techniques of 3 this section, to evaluate the integral. 1 2x 3 1 x 4x 10 15.y 2 dx 16. y 2 dx 0 x 1 0 x x 6 49.y lnx 2 x 2 dx 50. y x tan1x dx 2 2 2 4y 7y 12 x 2x 1 17. y dy 18. y 3 dx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 yy 2y 3 x x 2 1 x 2 ; 51. Use a graph of f x 1x 2x 3 to decide whether 2 19.y 2 dx 20. y 2 dx x f x dx is positive or negative. Use the graph to give a rough x 5 x 1 x 3x 2 0 estimate of the value of the integral and then use partial 5x 2 3x 2 ds fractions to ﬁnd the exact value. 21.y dx 22. y x 3 2x 2 s 2s 12 ; 52. Graph both y 1x 3 2x 2 and an antiderivative on the x 2 x 3 same screen. 23. y dx 24. y dx x 13 x 13 53–54 |||| Evaluate the integral by completing the square and using 2 10 x x 6 Formula 6. 25. y dx 26. y dx x 1x 2 9 x 3 3x dx 2x 1 3 2 2 53. y 54. y dx x x 2x 1 x 2x 1 x 2 2x 4x 2 12x 7 27.y 2 2 dx 28. y 2 2 dx x 1x 2 x 1 x 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

x 4 x 3 2x 2 x 1 55. The German mathematician Karl Weierstrass (1815Ð1897) 29. y dx 30. y dx x 2 2x 5 x 4 5x 2 4 noticed that the substitution t tanx2 will convert any SECTION 7.5 STRATEGY FOR INTEGRATION ❙❙❙❙ 505

rational function of sin x and cos x into an ordinary rational female population is related to time t by function of t . P S (a) If t tanx2 , x , sketch a right triangle or use t y dP Pr P S trigonometric identities to show that 1 Suppose an insect population with 10,000 females grows at a x 1 x t cos and sin rate of r 0.10 and 900 sterile males are added. Evaluate the 2 s1 t 2 2 s1 t 2 integral to give an equation relating the female population to (b) Show that time. (Note that the resulting equation can’t be solved explic- 1 t 2 2t itly for P .) cos x and sin x 1 t 2 1 t 2 64. Factor x 4 1 as a difference of squares by first adding and subtracting the same quantity. Use this factorization to evaluate (c) Show that x 1x 4 1 dx. 2 dx dt 1 t 2 CAS 65. (a) Use a computer algebra system to find the partial fraction decomposition of the function 56–59 |||| Use the substitution in Exercise 55 to transform the inte- 4x 3 27x 2 5x 32 grand into a rational function of t and then evaluate the integral. f x 30x 5 13x 4 50x 3 286x 2 299x 70 dx 1 56.y 57. y dx x 3 5 sin x 3 sin x 4 cos x (b) Use part (a) to find f x dx (by hand) and compare with the result of using the CAS to integrate f directly. 2 1 1 58.y dx 59. y dx Comment on any discrepancy. 3 1 sin x cos x 2 sin x sin 2x CAS 66. (a) Find the partial fraction decomposition of the function ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 12x 5 7x 3 13x 2 8 60–61 |||| Find the area of the region under the given curve from f x 6 5 4 3 2 a to b. 100x 80x 116x 80x 41x 20x 4 x 1 (b) Use part (a) to find f x dx and graph f and its indefinite 60. y ,,a 5 b 10 integral on the same screen. x 2 6x 8 (c) Use the graph of f to discover the main features of the x 1 graph of x f x dx . 61. y ,,a 2 b 3 x 1 67. Suppose that F, G , and Q are polynomials and ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Fx Gx 62. Find the volume of the resulting solid if the region under the Qx Qx curve y 1x 2 3x 2 from x 0 to x 1 is rotated about (a) the x -axis and (b) the y -axis. for all x except when Qx 0 . Prove that Fx Gx for all x. [Hint: Use continuity.] 63. One method of slowing the growth of an insect population without using pesticides is to introduce into the population a 68. If f is a quadratic function such that f 0 1 and number of sterile males that mate with fertile females but pro- f x duce no offspring. If P represents the number of female insects y dx x 2x 13 in a population,S the number of sterile males introduced each generation, and r the population’s natural growth rate, then the is a rational function, find the value of f 0 .

|||| 7.5 Strategy for Integration

As we have seen, integration is more challenging than differentiation. In ﬁnding the derivative of a function it is obvious which differentiation formula we should apply. But it may not be obvious which technique we should use to integrate a given function. Until now individual techniques have been applied in each section. For instance, we usually used substitution in Exercises 5.5, integration by parts in Exercises 7.1, and partial fractions in Exercises 7.4. But in this section we present a collection of miscellaneous integrals in random order and the main challenge is to recognize which technique or formula to use. No hard and fast rules can be given as to which method applies in a given situation, but we give some advice on strategy that you may ﬁnd useful. 506 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

A prerequisite for strategy selection is a knowledge of the basic integration formulas. In the following table we have collected the integrals from our previous list together with several additional formulas that we have learned in this chapter. Most of them should be memorized. It is useful to know them all, but the ones marked with an asterisk need not be memorized since they are easily derived. Formula 19 can be avoided by using partial fractions, and trigonometric substitutions can be used in place of Formula 20.

Table of Integration Formulas Constants of integration have been omitted.

x n1 1 1.y x n dx n 1 2. y dx ln x n 1 x a x 3.y e x dx e x 4. y a x dx ln a

5.y sin x dx cos x 6. y cos x dx sin x

7.y sec2x dx tan x 8. y csc2x dx cot x

9.y sec x tan x dx sec x 10. y csc x cot x dx csc x

11.y sec x dx ln sec x tan x 12. y csc x dx ln csc x cot x

13.y tan x dx ln sec x 14. y cot x dx ln sin x

15.y sinh x dx cosh x 16. y cosh x dx sinh x

dx 1 x dx x 17.y tan 1 18. y sin1 x 2 a 2 a a sa 2 x 2 a

dx 1 x a dx *19.y ln *20. y ln x sx 2 a 2 x 2 a 2 2a x a sx 2 a 2

Once you are armed with these basic integration formulas, if you don’t immediately see how to attack a given integral, you might try the following four-step strategy.

1. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples:

y sx (1 sx) dx y (sx x) dx

tan sin y d y cos2 d sec2 cos 1 y sin cos d 2 y sin 2 d SECTION 7.5 STRATEGY FOR INTEGRATION ❙❙❙❙ 507

y sin x cos x2 dx y sin2x 2 sin x cos x cos2x dx

y 1 2 sin x cos x dx

2. Look for an Obvious Substitution Try to ﬁnd some function u tx in the integrand whose differential du tx dx also occurs, apart from a constant factor. For instance, in the integral x y dx x 2 1 we notice that if u x 2 1 , then du 2x dx . Therefore, we use the substitution u x 2 1 instead of the method of partial fractions.

3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand f x . (a) Trigonometric functions. If f x is a product of powers of sin x and cos x , of tan x and sec x , or of cot x and csc x , then we use the substitutions recommended in Section 7.2. (b) Rational functions. If f is a rational function, we use the procedure of Sec- tion 7.4 involving partial fractions. (c) Integration by parts. If f x is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 7.1. If you look at the functions in Exer- cises 7.1, you will see that most of them are the type just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If sx 2 a 2 occurs, we use a trigonometric substitution according to the table in Section 7.3. (ii) If sn ax b occurs, we use the rationalizing substitution u sn ax b . More generally, this sometimes works for sn tx .

4. Try Again If the ﬁrst three steps have not produced the answer, remember that there are basically only two methods of integration: substitution and parts. (a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration or ingenuity (or even desperation) may suggest an appropriate substitution. (b) Try parts. Although integration by parts is used most of the time on products of the form described in Step 3(c), it is sometimes effective on single functions. Looking at Section 7.1, we see that it works on tan1x ,sin1x , and ln x , and these are all inverse functions. (c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming the integral into an easier form. These manipulations may be more substantial than in Step 1 and may involve some ingenuity. Here is an example:

dx 1 1 cos x 1 cos x y y dx y dx 1 cos x 1 cos x 1 cos x 1 cos2x

1 cos x cos x y dx y csc2x dx sin2x sin2x 508 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

(d) Relate the problem to previous problems. When you have built up some expe- rience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. For instance,x tan2x sec x dx is a challenging integral, but if we make use of the identity tan2x sec2x 1 , we can write

y tan2x sec x dx y sec3x dx y sec x dx

and if x sec3x dx has previously been evaluated (see Example 8 in Section 7.2), then that calculation can be used in the present problem. (e) Use several methods. Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions. In the following examples we indicate a method of attack but do not fully work out the integral.

tan3x EXAMPLE 1 y dx cos3x In Step 1 we rewrite the integral:

tan3x y dx y tan3x sec3x dx cos3x

The integral is now of the form x tanmx secnx dx with m odd, so we can use the advice in Section 7.2. Alternatively, if in Step 1 we had written

tan3x sin3x 1 sin3x y dx y dx y dx cos3x cos3x cos3x cos6x

then we could have continued as follows with the substitution u cos x :

sin3x 1 cos2x 1 u 2 y dx y sin x dx y du cos6x cos6x u 6 u 2 1 y du y u 4 u 6 du u 6

s EXAMPLE 2 y e x dx

According to Step 3(d)(ii) we substitute u sx . Then x u 2 , so dx 2u du and

y e sx dx 2 y ue u du

The integrand is now a product of u and the transcendental function e u so it can be integrated by parts. SECTION 7.5 STRATEGY FOR INTEGRATION ❙❙❙❙ 509

x 5 1 EXAMPLE 3 y dx x 3 3x 2 10x No algebraic simpliﬁcation or substitution is obvious, so Steps 1 and 2 don’t apply here. The integrand is a rational function so we apply the procedure of Section 7.4, remember- ing that the ﬁrst step is to divide.

dx EXAMPLE 4 y xsln x Here Step 2 is all that is needed. We substitute u ln x because its differential is du dxx, which occurs in the integral.

1 x EXAMPLE 5 y dx 1 x Although the rationalizing substitution

1 x u 1 x works here [Step 3(d)(ii)], it leads to a very complicated rational function. An easier method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiply- ing numerator and denominator by s1 x , we have

1 x 1 x y dx y dx 1 x s1 x 2 1 x y dx y dx s1 x 2 s1 x 2 sin1x s1 x 2 C

Can We Integrate All Continuous Functions?

The question arises: Will our strategy for integration enable us to ﬁnd the integral of every 2 continuous function? For example, can we use it to evaluate x e x dx ? The answer is no, at least not in terms of the functions that we are familiar with. The functions that we have been dealing with in this book are called elementary functions. These are the polynomials, rational functions, power functions x a , exponential functions a x , logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from these by the ﬁve operations of addition, subtraction, multiplication, division, and compo- sition. For instance, the function

x 2 1 f x lncosh x xe sin 2x x 3 2x 1 is an elementary function. If f is an elementary function, then f is an elementary function but x f x dx need not 2 be an elementary function. Consider f x e x . Since f is continuous, its integral exists, and if we deﬁne the function F by

x 2 Fx y e t dt 0 510 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

then we know from Part 1 of the Fundamental Theorem of Calculus that

2 Fx e x

2 Thus,f x e x has an antiderivative F , but it has been proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in evalu- 2 ating x e x dx in terms of the functions we know. (In Chapter 11, however, we will see how 2 to express x e x dx as an inﬁnite series.) The same can be said of the following integrals: e x y dx y sinx 2 dx y cose x dx x

1 sin x y sx 3 1 dx y dx y dx ln x x

In fact, the majority of elementary functions don’t have elementary antiderivatives. You may be assured, though, that the integrals in the following exercises are all elementary functions.

|||| 7.5 Exercises

2 2 1–80 |||| Evaluate the integral. 3x 2 3x 2 25.y dx 26. y dx 2 3 sin x sec x x 2x 8 x 2x 8 1.y dx 2. y tan3 d tan x 27.y cot x lnsin x dx 28. y sin sat dt 2 2t x 3.y 2 dt 4. y dx 0 s 4 t 3 3 x 5 3w 1 2 29.y dw 30. y x 2 4x dx arctan y 1 e 0 w 2 2 5. y 2 dy 6. y x csc x cot x dx 1 1 y 1 x s2x 1 y dx y dx 3 4 x 1 31. 32. 4 1 x 2x 3 7.y r ln r dr 8. y 2 dx 1 0 x 4x 5 2 1 4 cot x x 1 x 33.y s3 2x x 2 dx 34. y dx 9.y dx 10. y dx 4 4 cot x x 2 4x 5 x 4 x 2 1

1 8 3 5 35.y x sin x dx 36. y sin 4x cos 3x dx 11.y sin cos d 12. y sin x coscos x dx 1

4 4 dx s1 ln x 37. y cos2 tan2 d 38. y tan5 sec3 d 13.y 14. y dx 0 0 1 x 232 x ln x

2 x 1 12 x s22 x 39.y dx 40. y dy 15.y dx 16. y dx 1 x 2 s1 x 2 s4y 2 4y 3 0 s1 x 2 0 s1 x 2

2t e 2 2 1 17. y x sin2x dx 18. y dt 41. y tan d 42. y x tan x dx 1 e4t

xe x s3 x x x x 19.y e dx 20. y e dx 43.y e s1 e dx 44. y s1 e dx

x x 3 1 e 21.y t 3e2t dt 22. y x sin1x dx 45. y x 5e dx 46. y dx 1 e x

1 x a x s 8 2 23. y (1 x) dx 24. y ln x 1 dx 47.y 2 2 dx 48. y 4 4 dx 0 x a x a SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ❙❙❙❙ 511

1 1 3 arctanst 1 49. y dx 50. y dx s 2 s 67.y dt 68. y x x dx x 4x 1 x 4x 1 1 st 1 2e e 1 dx e 2x lnx 1 51.y dx 52. y 4 69. y dx 70. y dx xs4x 2 1 xx 1 1 e x x 2

2 2 x st 53.y x sinh mx dx 54. y x sin x dx 71.y dx 72. y dt x 4 4x 2 3 1 s3 t 1 x ln x 1 dx 55.y dx 56. y dx 73.y dx 74. y x 4 4sx 1 sx 2 1 x 2x 2 4 e x e x

3 2 57. y xsx c dx 58. y x ln1 x dx 75.y sin x sin 2x sin 3x dx 76. y x 2 bx sin 2x dx

y 1 y 1 sx sec x cos 2x 59. 3x x dx 60. 3 dx 77.y dx 78. y dx e e x sx 1 x 3 sin x sec x x 4 x 3 2 sin x cos x 61.y 10 dx 62. y 10 dx 79.y x sin x cos x dx 80. y dx x 16 x 1 sin4 x cos4 x

3 lntan x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 63. y sxesx dx 64. y dx 4 sin x cos x 2 2 81. The functions y e x and y x 2e x don’t have elementary 3 2 x 2 1 3 u 1 antiderivatives, but y 2x 1e does. Evaluate 2 65.y dx 66. y 3 2 du 2 x sx 1 sx 2 u u x 2x 1e dx.

|||| 7.6 Integration Using Tables and Computer Algebra Systems

In this section we describe how to use tables and computer algebra systems to integrate functions that have elementary antiderivatives. You should bear in mind, though, that even the most powerful computer algebra systems can’t ﬁnd explicit formulas for the antideriv- 2 atives of functions like e x or the other functions described at the end of Section 7.5.

Tables of Integrals

Tables of indeﬁnite integrals are very useful when we are confronted by an integral that is difﬁcult to evaluate by hand and we don’t have access to a computer algebra system. A relatively brief table of 120 integrals, categorized by form, is provided on the Reference Pages at the back of the book. More extensive tables are available in CRC Standard Mathe- matical Tables and Formulae, 30th ed. by Daniel Zwillinger (Boca Raton, FL: CRC Press, 1995) (581 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 6e (New York: Academic Press, 2000), which contains hundreds of pages of integrals. It should be remembered, however, that integrals do not often occur in exactly the form listed in a table. Usually we need to use substitution or algebraic manipulation to transform a given integral into one of the forms in the table.

EXAMPLE 1 The region bounded by the curves y arctan x, y 0 , and x 1 is rotated about the y -axis. Find the volume of the resulting solid.

SOLUTION Using the method of cylindrical shells, we see that the volume is

1 V y 2x arctan x dx 0 512 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| The Table of Integrals appears on the In the section of the Table of Integrals entitled Inverse Trigonometric Forms we locate Reference Pages at the back of the book. Formula 92: u 2 1 u y u tan1u du tan1u C 2 2

Thus, the volume is

2 1 1 x 1 x V 2 y x tan1x dx 2 tan1x 0 2 2 0 2 1 1 1 x 1 tan x x 0 2 tan 1 1 1 2 2 4 1 2

x 2 EXAMPLE 2 Use the Table of Integrals to ﬁnd y dx. s5 4x 2

SOLUTION If we look at the section of the table entitled Forms involving sa 2 u 2, we see that the closest entry is number 34:

u 2 u a 2 u y du sa 2 u 2 sin1 C sa 2 u 2 2 2 a

This is not exactly what we have, but we will be able to use it if we ﬁrst make the substitution u 2x:

x 2 u22 du 1 u 2 y dx y y du s5 4x 2 s5 u 2 2 8 s5 u 2

Then we use Formula 34 with a 2 5 (so a s5):

x 2 1 u 2 1 u 5 u y dx y du s5 u 2 sin1 C s5 4x 2 8 s5 u 2 8 2 2 s5 x 5 2x s5 4x 2 sin1 C 8 16 s5

EXAMPLE 3 Use the Table of Integrals to ﬁnd y x 3 sin x dx.

SOLUTION If we look in the section called Trigonometric Forms, we see that none of the entries explicitly includes a u 3 factor. However, we can use the reduction formula in entry 84 with n 3 :

y x 3 sin x dx x 3 cos x 3 y x 2 cos x dx

85. y u n cos u du We now need to evaluate x x 2 cos x dx . We can use the reduction formula in entry 85 with n 2 , followed by entry 82: u n sin u n y u n1 sin u du

y x 2 cos x dx x 2 sin x 2 y x sin x dx

x 2 sin x 2sin x x cos x K SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ❙❙❙❙ 513

Combining these calculations, we get

y x 3 sin x dx x 3 cos x 3x 2 sin x 6x cos x 6 sin x C

where C 3K.

EXAMPLE 4 Use the Table of Integrals to ﬁnd y xsx 2 2x 4 dx.

SOLUTION Since the table gives forms involving sa 2 x 2 ,sa 2 x 2 , andsx 2 a 2 , but not sax2 bx c, we ﬁrst complete the square: x 2 2x 4 x 12 3 If we make the substitution u x 1 (so x u 1 ), the integrand will involve the pattern sa 2 u 2:

y xsx 2 2x 4 dx y u 1 su 2 3 du

y usu 2 3 du y su 2 3 du

The ﬁrst integral is evaluated using the substitution t u 2 3 :

s 2 1 s 1 2 32 1 2 32 y u u 3 du 2 y t dt 2 3 t 3 u 3

u 21. y sa 2 u 2 du sa 2 u 2 For the second integral we use Formula 21 with a s3 : 2 a 2 u ln(u sa 2 u 2 ) C y su 2 3 du su 2 3 3 ln(u su 2 3) 2 2 2 Thus

y xsx 2 2x 4 dx x 1 1x 2 2x 432 sx 2 2x 4 3 ln(x 1 sx 2 2x 4 ) C 3 2 2

Computer Algebra Systems

We have seen that the use of tables involves matching the form of the given integrand with the forms of the integrands in the tables. Computers are particularly good at matching patterns. And just as we used substitutions in conjunction with tables, a CAS can perform substitutions that transform a given integral into one that occurs in its stored formulas. So it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean that integration by hand is an obsolete skill. We will see that a hand computation sometimes produces an indeﬁnite integral in a form that is more convenient than a machine answer. To begin, let’s see what happens when we ask a machine to integrate the relatively simple function y 13x 2 . Using the substitution u 3x 2 , an easy calculation by hand gives 1 y dx 1 ln 3x 2 C 3x 2 3 514 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

whereas Derive, Mathematica, and Maple all return the answer 1 3 ln 3x 2 The ﬁrst thing to notice is that computer algebra systems omit the constant of integration. In other words, they produce a particular antiderivative, not the most general one. Therefore, when making use of a machine integration, we might have to add a constant. Second, the absolute value signs are omitted in the machine answer. That is ﬁne if our 2 problem is concerned only with values of x greater than 3 . But if we are interested in other values of x , then we need to insert the absolute value symbol. In the next example we reconsider the integral of Example 4, but this time we ask a machine for the answer.

EXAMPLE 5 Use a computer algebra system to ﬁnd y xsx 2 2x 4 dx .

SOLUTION Maple responds with the answer

3 s3 1 x 2 2x 432 1 2x 2sx 2 2x 4 arcsinh 1 x 3 4 2 3

This looks different from the answer we found in Example 4, but it is equivalent because the third term can be rewritten using the identity

|||| This is Equation 3.9.3. arcsinh x ln(x sx 2 1) Thus s3 s3 1 2 arcsinh 1 x ln 1 x s| 1 x 1 3 3 3 1 ln [1 x s1 x2 3] s3 1 ln ln(x 1 sx 2 2x 4) s3 3 s The resulting extra term 2 ln(1 3) can be absorbed into the constant of integration. Mathematica gives the answer

5 x x 2 3 1 x sx 2 2x 4 arcsinh 6 6 3 2 s3

Mathematica combined the ﬁrst two terms of Example 4 (and the Maple result) into a single term by factoring. Derive gives the answer

1 s 2 2 3 s 2 6 x 2x 4 2x x 5 2 ln( x 2x 4 x 1) The ﬁrst term is like the ﬁrst term in the Mathematica answer, and the second term is identical to the last term in Example 4.

EXAMPLE 6 Use a CAS to evaluate y xx 2 58 dx .

SOLUTION Maple and Mathematica give the same answer:

1 18 5 16 14 1750 12 10 8 218750 6 4 390625 2 18 x 2 x 50x 3 x 4375x 21875x 3 x 156250x 2 x SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ❙❙❙❙ 515

It’s clear that both systems must have expanded x 2 58 by the Binomial Theorem and then integrated each term. If we integrate by hand instead, using the substitution u x 2 5 , we get

2 8 1 2 9 |||| Derive and the TI-89/92 also give this answer. y x x 5 dx 18 x 5 C

For most purposes, this is a more convenient form of the answer.

EXAMPLE 7 Use a CAS to ﬁnd y sin5x cos2x dx.

SOLUTION In Example 2 in Section 7.2 we found that

5 2 1 3 2 5 1 7 1 y sin x cos x dx 3 cos x 5 cos x 7 cos x C

Derive and Maple report the answer

1 4 3 4 2 3 8 3 7 sin x cos x 35 sin x cos x 105 cos x

whereas Mathematica produces 10 5 1 3 1 64 cos x 192 cos 3x 320 cos 5x 448 cos 7x

F We suspect that there are trigonometric identities which show these three answers are equivalent. Indeed, if we ask Derive, Maple, and Mathematica to simplify their expressions using trigonometric identities, they ultimately produce the same form of the answer as in Equation 1. 0 5

4 5 FIGURE 1 EXAMPLE 8 If f x x 60 sin x cos x, ﬁnd the antiderivative F of f such that F0 0. Graph F for 0 x 5 . Where does F have extreme values and inﬂection points?

10 SOLUTION The antiderivative off produced by Maple is

fª 1 2 20 3 6 20 6 4 4 16 2 32 F f F x 2 x 3 sin x cos x 7 sin x cos x 7 cos x sin x 21 cos x sin x 21 sin x

and we note that F0 0 . This expression could probably be simpliﬁed, but there’s no 05 need to do so because a computer algebra system can graph this version of F as easily as any other version. A graph of F is shown in Figure 1. To locate the extreme values of F, we graph its derivative F f in Figure 2 and observe that F has a local maximum _7 when x 2.3 and a local minimum when x 2.5 . The graph of F f in Figure 2 FIGURE 2 shows that F has inﬂection points when x 0.7 , 1.3, 1.8, 2.4, 3.3, and 3.9.

|||| 7.6 Exercises

1–4 |||| Use the indicated entry in the Table of Integrals on the 3. y sec3x dx; entry 71 Reference Pages to evaluate the integral.

2 s7 2x 2 3x 4. y e sin 3 d; entry 98 y dx; entry 33 y dx; entry 55 1. 2 2. x s3 2x 516 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

5–30 |||| Use the Table of Integrals on the Reference Pages to CAS 35–42 |||| Use a computer algebra system to evaluate the integral. evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent. 1 1 3 1 5.y 2x cos x dx 6. y 2 dx 0 2 x s4x 2 7 35.y x 2s5 x 2 dx 36. y x 21 x 3 4 dx 7.y e3x cos 4x dx 8. y csc3x2 dx 37.y sin3x cos2x dx 38. y tan2x sec4x dx dx s2y 2 3 9. y 10. y dy x 2s4x 2 9 y 2 39.y xs1 2x dx 40. y sin4x dx 0 11.y t 2et dt 12. y x 2 cos 3x dx 1 0 41.y tan5x dx 42. y x 5sx 2 1 dx tan31z 1s 13.y dz 14. y sin x dx z 2 CAS 43. Computer algebra systems sometimes need a helping hand x x 2 2 15.y e sech e dx 16. y x sin x cos 3x dx from human beings. Ask your CAS to evaluate

5 x x x 2 y s dx 17. y ys6 4y 4y dy 18. y dx 2 4 1 x 2 s2

2 dx If it doesn’t return an answer, ask it to try 19. y sin x cos x lnsin x dx 20. y e x1 2e x x s 2x x y 2 2 1 dx e 2 21.y dx 22. y x 3s4x 2 x 4 dx 2x 3 e 0 instead. Why do you think it was successful with this form of the integrand? 23.y sec5x dx 24. y sin6 2x dx CAS 44. Try to evaluate

s 2 1 4 ln x 4 x 25. y dx 26. y x e dx y 1 ln x s1 x ln x2 dx x 0

t 27. y se 2x 1 dx 28. y e sint 3 dt with a computer algebra system. If it doesn’t return an answer, make a substitution that changes the integral into one that the CAS can evaluate. x 4 dx sec2 tan2 y 30. y d 29. s 10 s 2 x 2 9 tan CAS 45–48 |||| Use a CAS to ﬁnd an antiderivative F of f such that F0 0 . Graph f and F and locate approximately the 31. Find the volume of the solid obtained when the region under x-coordinates of the extreme points and inﬂection points of F . the curve y xs4 x 2 ,0 x 2 , is rotated about the x 2 1 45. f x y-axis. x 4 x 2 1 2 32. The region under the curve y tan x from 0 to 4 is rotated 46. f x xe x sin x, 5 x 5 about the x -axis. Find the volume of the resulting solid. 47. f x sin4x cos6x, 0 x 33. Verify Formula 53 in the Table of Integrals (a) by differentiation and (b) by using the substitution t a bu . x 3 x 48. f x 6 34. Verify Formula 31 (a) by differentiation and (b) by substituting x 1

u a sin . DISCOVERY PROJECT PATTERNS IN INTEGRALS ❙❙❙❙ 517

DISCOVERY PROJECT CAS Patterns in Integrals

In this project a computer algebra system is used to investigate indeﬁnite integrals of families of functions. By observing the patterns that occur in the integrals of several members of the family, you will ﬁrst guess, and then prove, a general formula for the integral of any member of the family. 1. (a) Use a computer algebra system to evaluate the following integrals. 1 1 (i)y dx (ii) y dx x 2x 3 x 1x 5 1 1 (iii)y dx (iv) y dx x 2x 5 x 22 (b) Based on the pattern of your responses in part (a), guess the value of the integral 1 y dx x ax b if a b. What if a b? (c) Check your guess by asking your CAS to evaluate the integral in part (b). Then prove it using partial fractions. 2. (a) Use a computer algebra system to evaluate the following integrals.

(i)y sin x cos 2x dx (ii)y sin 3x cos 7x dx (iii) y sin 8x cos 3x dx (b) Based on the pattern of your responses in part (a), guess the value of the integral

y sin ax cos bx dx

(c) Check your guess with a CAS. Then prove it using the techniques of Section 7.2. For what values of a and b is it valid? 3. (a) Use a computer algebra system to evaluate the following integrals. (i)y ln x dx (ii)y x ln x dx (iii) y x 2 ln x dx

(iv)y x 3 ln x dx (v) y x7 ln x dx

(b) Based on the pattern of your responses in part (a), guess the value of

y x n ln x dx

(c) Use integration by parts to prove the conjecture that you made in part (b). For what values of n is it valid? 4. (a) Use a computer algebra system to evaluate the following integrals. (i)y xe x dx (ii)y x 2e x dx (iii) y x 3e x dx

(iv)y x 4e x dx (v) y x 5e x dx (b) Based on the pattern of your responses in part (a), guess the value of x x 6e x dx . Then use your CAS to check your guess. (c) Based on the patterns in parts (a) and (b), make a conjecture as to the value of the integral

y x ne x dx when n is a positive integer. (d) Use mathematical induction to prove the conjecture you made in part (c). 518 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| 7.7 Approximate Integration

There are two situations in which it is impossible to find the exact value of a definite integral. xb The first situation arises from the fact that in order to evaluate a f x dx using the Fundamental Theorem of Calculus we need to know an antiderivative of f . Sometimes, however, it is difficult, or even impossible, to find an antiderivative (see Section 7.5). For example, it is impossible to evaluate the following integrals exactly:

1 1 y e x 2 dx y s1 x 3 dx 0 1

The second situation arises when the function is determined from a scientific experi- ment through instrument readings or collected data. There may be no formula for the function (see Example 5). In both cases we need to find approximate values of definite integrals. We already know one such method. Recall that the definite integral is defined as a limit of Riemann sums, so any Riemann sum could be used as an approximation to the integral: If we divide a, b y into n subintervals of equal length x b an , then we have

n b y f x dx f x*i x a i1

where x*i is any point in the i th subinterval xi1, xi . If x*i is chosen to be the left endpoint of the interval, then x*i xi1 and we have

n 0 x¸ ⁄ x™ x£ x¢ x b 1 y f x dx Ln f xi1 x (a) Left endpoint approximation a i1 y If f x 0 , then the integral represents an area and (1) represents an approximation of this area by the rectangles shown in Figure 1(a). If we choose x*i to be the right endpoint, then x*i xi and we have

n b 2 y f x dx Rn f xi x a i1

[See Figure 1(b).] The approximations Ln and Rn deﬁned by Equations 1 and 2 are called 0 x¸ ⁄ x™ x£ x¢ x the left endpoint approximation and right endpoint approximation, respectively. In Section 5.2 we also considered the case where x*i is chosen to be the midpoint xi of (b) Right endpoint approximation the subinterval xi1, xi . Figure 1(c) shows the midpoint approximation Mn , which appears to be better than either Ln or Rn . y

Midpoint Rule

b y f x dx Mn x f x1 f x2 f xn a

b a where x 0 –x¡ –––x™ x£ x¢ x n (c) Midpoint approximation 1 and xi 2 xi1 xi midpoint of xi1, xi FIGURE 1 SECTION 7.7 APPROXIMATE INTEGRATION ❙❙❙❙ 519

Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2:

n n n b 1 x y f x dx f xi 1 x f xi x f xi 1 f xi a 2 i1 i1 2 i1 x f x f x f x f x f x f x 2 0 1 1 2 n 1 n x f x 2 f x 2 f x 2 f x f x 2 0 1 2 n 1 n

y Trapezoidal Rule

b x y f x dx Tn f x0 2 f x1 2 f x2 2 f xn1 f xn a 2

where x b an and xi a i x.

0 x¸ ⁄ x™x£ x¢ x The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates the case f x 0 . The area of the trapezoid that lies above the i th subinterval is FIGURE 2 f xi1 f xi x Trapezoidal approximation x f x f x 2 2 i 1 i and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal Rule. 1 y= x EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n 5 to x2 approximate the integral 1 1 x dx . SOLUTION (a) With n 5, a 1 , and b 2 , we have x 2 15 0.2 , and so the Trape- zoidal Rule gives

2 1 0.2 y dx T5 f 1 2 f 1.2 2 f 1.4 2 f 1.6 2 f 1.8 f 2 121 x 2 1 2 2 2 2 1 FIGURE 3 0.1 1 1.2 1.4 1.6 1.8 2 0.695635

1 This approximation is illustrated in Figure 3. y= x (b) The midpoints of the ﬁve subintervals are 1.1,,,,1.3 1.5 1.7 and 1.9 , so the Midpoint Rule gives

2 1 y dx x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9 1 x 1 1 1 1 1 1 5 1.1 1.3 1.5 1.7 1.9 12 0.691908

FIGURE 4 This approximation is illustrated in Figure 4. 520 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

In Example 1 we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the Fundamental Theorem of Calculus,

2 1 2 y dx ln x]1 ln 2 0.693147... 1 x

b y f x dx approximation error The error in using an approximation is deﬁned to be the amount that needs to be added to a the approximation to make it exact. From the values in Example 1 we see that the errors in the Trapezoidal and Midpoint Rule approximations for n 5 are

ET 0.002488 and EM 0.001239

In general, we have

b b ET y f x dx Tn and EM y f x dx Mn a a

Module 5.1/5.2/7.7 allows you to com- The following tables show the results of calculations similar to those in Example 1, but pare approximation methods. for n 5, 10 , and 20 and for the left and right endpoint approximations as well as the Trapezoidal and Midpoint Rules.

2 1 Approximations to y dx n Ln Rn Tn Mn 1 x 5 0.745635 0.645635 0.695635 0.691908 10 0.718771 0.668771 0.693771 0.692835 20 0.705803 0.680803 0.693303 0.693069

Corresponding errors n EL ER ET EM 5 0.052488 0.047512 0.002488 0.001239 10 0.025624 0.024376 0.000624 0.000312 20 0.012656 0.012344 0.000156 0.000078

We can make several observations from these tables:

1. In all of the methods we get more accurate approximations when we increase the value of n . (But very large values of n result in so many arithmetic operations that we have to beware of accumulated round-off error.) 2. The errors in the left and right endpoint approximations are opposite in sign and appear to decrease by a factor of about 2 when we double the value of n . |||| It turns out that these observations are true 3. The Trapezoidal and Midpoint Rules are much more accurate than the endpoint in most cases. approximations. 4. The errors in the Trapezoidal and Midpoint Rules are opposite in sign and appear to decrease by a factor of about 4 when we double the value of n . 5. The size of the error in the Midpoint Rule is about half the size of the error in the Trapezoidal Rule. Figure 5 shows why we can usually expect the Midpoint Rule to be more accurate than the Trapezoidal Rule. The area of a typical rectangle in the Midpoint Rule is the same as the trapezoid ABCD whose upper side is tangent to the graph at P . The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid AQRD used in SECTION 7.7 APPROXIMATE INTEGRATION ❙❙❙❙ 521

C the Trapezoidal Rule. [The midpoint error (shaded red) is smaller than the trapezoidal error (shaded blue).] P These observations are corroborated in the following error estimates, which are proved in books on numerical analysis. Notice that Observation 4 corresponds to the n 2 in each denominator because 2n2 4n 2 . The fact that the estimates depend on the size of the B second derivative is not surprising if you look at Figure 5, because f x measures how much the graph is curved. [Recall that f x measures how fast the slope of y f x changes.]

A D x x– x i-1 i i 3 Error Bounds Suppose f x K for a x b . If ET and EM are the errors C in the Trapezoidal and Midpoint Rules, then R Kb a3 Kb a3 ET and EM P 12n 2 24n 2

B Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If Q f x 1x, then f x 1x 2 and f x 2x 3 . Since 1 x 2 , we have 1x 1 , so

2 2 A D f x 2 x 3 13 FIGURE 5 Therefore, taking K 2, a 1, b 2 , and n 5 in the error estimate (3), we see that

|||| K can be any number larger than all the 3 values of f x , but smaller values of K 2 2 1 1 ET 0.006667 give better error bounds. 1252 150

Comparing this error estimate of 0.006667 with the actual error of about 0.002488 , we see that it can happen that the actual error is substantially less than the upper bound for the error given by (3).

EXAMPLE 2 How large should we take n in order to guarantee that the Trapezoidal and x2 Midpoint Rule approximations for 1 1 x dx are accurate to within 0.0001 ? SOLUTION We saw in the preceding calculation that f x 2 for 1 x 2 , so we can take K 2 ,a 1 , and b 2 in (3). Accuracy to within 0.0001 means that the size of the error should be less than 0.0001 . Therefore, we choose n so that

213 0.0001 12n 2

Solving the inequality for n , we get

2 n 2 120.0001

1 |||| It’s quite possible that a lower value for n or n 40.8 s would sufﬁce, but 41 is the smallest value for 0.0006 which the error bound formula can guarantee us accuracy to within 0.0001 . Thus,n 41 will ensure the desired accuracy. 522 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

For the same accuracy with the Midpoint Rule we choose n so that 213 0.0001 24n 2 1 which gives n 29 s0.0012

y EXAMPLE 3 x1 x 2 (a) Use the Midpoint Rule with n 10 to approximate the integral 0 e dx . (b) Give an upper bound for the error involved in this approximation.

SOLUTION 2 y=ex (a) Since a 0, b 1 , and n 10 , the Midpoint Rule gives

1 2 y e x dx x f 0.05 f 0.15 f 0.85 f 0.95 0 0.1e 0.0025 e 0.0225 e 0.0625 e 0.1225 e 0.2025 e 0.3025 e 0.4225 e 0.5625 e 0.7225 e 0.9025 0 1 x 1.460393

FIGURE 6 Figure 6 illustrates this approximation. 2 2 2 (b) Since f x e x , we have f x 2xe x and f x 2 4x 2 e x . Also,since 0 x 1, we have x 2 1 and so

2 0 f x 2 4x 2 e x 6e

|||| Error estimates are upper bounds for Taking K 6e ,a 0 ,b 1 , and n 10 in the error estimate (3), we see that an upper the error. They give theoretical, worst-case bound for the error is scenarios. The actual error in this case turns 6e13 e out to be about 0.0023 . 0.007 24102 400

Simpson’s Rule

Another rule for approximate integration results from using parabolas instead of straight line segments to approximate a curve. As before, we divide a, b into n subintervals of equal length h x b an , but this time we assume that n is an even number. Then on each consecutive pair of intervals we approximate the curve y f x 0 by a parabola

as shown in Figure 7. If yi f xi , then Pixi, yi is the point on the curve lying above xi .

A typical parabola passes through three consecutive points Pi, Pi1 , and Pi2 . y y

P¸ P¡ P¸(_h, y¸) P∞ Pß P¡(0, ›)

P™ P¢ P™(h, ﬁ) P£

0 a=x¸ ⁄ x™x£ x¢ x∞ xß=b x _h 0 h x

FIGURE 7 FIGURE 8 SECTION 7.7 APPROXIMATE INTEGRATION ❙❙❙❙ 523

To simplify our calculations, we ﬁrst consider the case wherex0 h, x1 0 , and x2 h. (See Figure 8.) We know that the equation of the parabola through P0, P1 , and P2 is of the form y Ax 2 Bx C and so the area under the parabola from x h to x h is

h h |||| Here we have used Theorem 5.5.7. Notice 2 2 2 y Ax Bx C dx 2 y Ax C dx that Ax C is even and Bx is odd. h 0

x 3 h 2A Cx 3 0 h 3 h 2A Ch 2Ah 2 6C 3 3

But, since the parabola passes through P0h, y0 ,P10, y1 , and P2h, y2 , we have

2 2 y0 Ah Bh C Ah Bh C

y1 C

2 y2 Ah Bh C

2 and therefore y0 4y1 y2 2Ah 6C

Thus, we can rewrite the area under the parabola as

h y 4y y 3 0 1 2

Now, by shifting this parabola horizontally we do not change the area under it. This means

that the area under the parabola through P0, P1 , and P2 from x x0 to x x2 in Figure 7 is still h y 4y y 3 0 1 2

Similarly, the area under the parabola through P2, P3, and P4 from x x2 to x x4 is

h y 4y y 3 2 3 4

If we compute the areas under all the parabolas in this manner and add the results, we get

b h h h y f x dx y0 4y1 y2 y2 4y3 y4 yn2 4yn1 yn a 3 3 3 h y 4y 2y 4y 2y 2y 4y y 3 0 1 2 3 4 n 2 n 1 n

Although we have derived this approximation for the case in which f x 0 , it is a reasonable approximation for any continuous function f and is called Simpson’s Rule after the English mathematician Thomas Simpson (1710Ð1761). Note the pattern of coefﬁ- cients:1, 4, 2, 4, 2, 4, 2, ..., 4, 2, 4, 1 . 524 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| Thomas Simpson was a weaver who taught Simpson’s Rule himself mathematics and went on to become one of the best English mathematicians of the 18th b x century. What we call Simpson’s Rule was y f x dx Sn f x0 4 f x1 2 f x2 4 f x3 a 3 actually known to Cavalieri and Gregory in the 17th century, but Simpson popularized it in his 2 f xn2 4 f xn1 f xn best-selling calculus textbook, entitled A New Treatise of Fluxions. where n is even and x b an .

x2 EXAMPLE 4 Use Simpson’s Rule with n 10 to approximate 1 1 x dx . SOLUTION Putting f x 1x, n 10 , and x 0.1 in Simpson’s Rule, we obtain

2 1 y dx S10 1 x x f 1 4 f 1.1 2 f 1.2 4 f 1.3 2 f 1.8 4 f 1.9 f 2 3 0.1 1 4 2 4 2 4 2 4 2 4 1 3 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 0.693150

Notice that, in Example 4, Simpson’s Rule gives us a much better approximation

S10 0.693150 to the true value of the integral ln 2 0.693147. . . than does the Trapezoidal Rule T10 0.693771 or the Midpoint Rule M10 0.692835 . It turns out (see Exercise 48) that the approximations in Simpson’s Rule are weighted averages of those in the Trapezoidal and Midpoint Rules: 1 2 S2n 3 Tn 3 Mn

(Recall that ET and EM usually have opposite signs and EM is about half the size of ET .) In many applications of calculus we need to evaluate an integral even if no explicit formula is known for y as a function of x. A function may be given graphically or as a table of values of collected data. If there is evidence that the values are not changing rapidly, then the Trapezoidal Rule or Simpson’s Rule can still be used to ﬁnd an approximate value xb for a y dx , the integral of y with respect to x.

EXAMPLE 5 Figure 9 shows data trafﬁc on the link from the United States to SWITCH, the Swiss academic and research network, on February 10, 1998. Dt is the data throughput, measured in megabits per second Mbs . Use Simpson’s Rule to estimate the total amount of data transmitted on the link up to noon on that day. D 8

0 t (hours) FIGURE 9 3691215182124 SECTION 7.7 APPROXIMATE INTEGRATION ❙❙❙❙ 525

SOLUTION Because we want the units to be consistent and Dt is measured in megabits per second, we convert the units for t from hours to seconds. If we let At be the amount of data (in megabits) transmitted by time t , where t is measured in seconds, then At Dt. So, by the Net Change Theorem (see Section 5.4), the total amount of data transmitted by noon (when t 12 602 43,200 ) is

43,200 A43,200 y Dt dt 0

We estimate the values of Dt at hourly intervals from the graph and compile them in the table.

t hours t seconds Dt t hours t seconds Dt 0 0 3.2 7 25,200 1.3 1 3,600 2.7 8 28,800 2.8 2 7,200 1.9 9 32,400 5.7 3 10,800 1.7 10 36,000 7.1 4 14,400 1.3 11 39,600 7.7 5 18,000 1.0 12 43,200 7.9 6 21,600 1.1

Then we use Simpson’s Rule with n 12 and t 3600 to estimate the integral:

43,200 t y At dt D0 4D3600 2D7200 4D39,600 D43,200 0 3 3600 3.2 42.7 21.9 41.7 21.3 41.0 3 21.1 41.3 22.8 45.7 27.1 47.7 7.9

143,880

Thus, the total amount of data transmitted up to noon is about 144,000 megabits, or 144 gigabits.

In Exercises 27 and 28 you are asked to demonstrate, in particular cases, that the error in Simpson’s Rule decreases by a factor of about 16 when n is doubled. That is consistent with the appearance of n 4 in the denominator of the following error estimate for Simpson’s Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but it uses the fourth derivative of f .

4 Error Bound for Simpson’s Rule Suppose that f 4 x K for a x b . If ES is the error involved in using Simpson’s Rule, then

Kb a5 ES 180n 4 526 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

EXAMPLE 6 How large should we take n in order to guarantee that the Simpson’s Rule x2 approximation for 1 1 x dx is accurate to within 0.0001 ? SOLUTION If f x 1x , then f 4 x 24x 5 . Since x 1 , we have 1x 1 and so

24 f 4x 24 x 5

|||| Many calculators and computer algebra sys- Therefore, we can take K 24 in (4). Thus, for an error less than 0.0001 we should tems have a built-in algorithm that computes an choose n so that approximation of a deﬁnite integral. Some of 2415 these machines use Simpson’s Rule; others use 0.0001 more sophisticated techniques such as adaptive 180n 4 numerical integration. This means that if a function ﬂuctuates much more on a certain part of 24 4 the interval than it does elsewhere, then that This gives n part gets divided into more subintervals. This 180 0.0001 strategy reduces the number of calculations required to achieve a prescribed accuracy. 1 or n 6.04 s4 0.00075

Therefore, n 8 (n must be even) gives the desired accuracy. (Compare this with Example 2, where we obtained n 41 for the Trapezoidal Rule and n 29 for the Midpoint Rule.)

EXAMPLE 7 x1 x2 (a) Use Simpson’s Rule with n 10 to approximate the integral 0 e dx . (b) Estimate the error involved in this approximation.

SOLUTION (a) If n 10 , then x 0.1 and Simpson’s Rule gives

1 2 x |||| Figure 10 illustrates the calculation in y e x dx f 0 4 f 0.1 2 f 0.2 2 f 0.8 4 f 0.9 f 1 Example 7. Notice that the parabolic arcs are 0 3 2 so close to the graph of y e x that they are 0.1 practically indistinguishable from it. e 0 4e 0.01 2e 0.04 4e 0.09 2e 0.16 4e 0.25 2e 0.36 3 y 4e 0.49 2e 0.64 4e 0.81 e 1 1.462681

2 y=e ≈ (b) The fourth derivative of f x e x is

2 f 4x 12 48x 2 16x 4 e x and so, since 0 x 1 , we have 0 f 4x 12 48 16e 1 76e Therefore, putting K 76e, a 0, b 1 , and n 10 in (4), we see that the error is at 0 1 x most 76e15 0.000115 FIGURE 10 180104 (Compare this with Example 3.) Thus, correct to three decimal places, we have

1 2 y e x dx 1.463 0 SECTION 7.7 APPROXIMATE INTEGRATION ❙❙❙❙ 527

|||| 7.7 Exercises

x4 (Round your answers to six decimal places.) Compare your results 1. Let I 0 f x dx , where f is the function whose graph is shown. to the actual value to determine the error in each approximation.

(a) Use the graph to find L2, R2, and M2. 1 (b) Are these underestimates or overestimates of I ? 5. y x 2 sin x dx, n 8 6. y esx dx, n 6 0 0 (c) Use the graph to find T2 . How does it compare with I ? (d) For any value of n , list the numbers Ln, Rn, Mn, Tn, and I in increasing order. 7–18 |||| Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and y (c) Simpson’s Rule to approximate the given integral with the 3 specified value of n . (Round your answers to six decimal places.) f 2 12 7. y s4 1 x 2 dx, n 8 8. y sinx 2 dx, n 4 2 0 0

2 ln x 3 dt 1 9. y dx, n 10 10. y 2 4 , n 6 1 1 x 0 1 t t

12 4 0 1 234x 11. y sine t2 dt, n 8 12. y s1 sx dx, n 8 0 0

2. The left, right, Trapezoidal, and Midpoint Rule approximations 2 1x 4 2 13. y e dx, n 4 14. y sx sin x dx, n 8 x 1 0 were used to estimate 0 f x dx , where f is the function whose graph is shown. The estimates were 0.7811, 0.8675, 0.8632, 5 cos x 6 and 0.9540, and the same number of subintervals were used in 15. y dx, n 8 16. y lnx 3 2 dx, n 10 each case. 1 x 4

(a) Which rule produced which estimate? x 3 1 4 e (b) Between which two approximations does the true value of 17. y 5 dy, n 6 18. y dx, n 10 0 1 y 2 x x2 f x dx lie? 0

y 19. (a) Find the approximations T10 and M10 for the integral 2 x2 ex dx. 1 0 (b) Estimate the errors in the approximations of part (a). y=ƒ (c) How large do we have to choose n so that the approxima-

tions Tn and Mn to the integral in part (a) are accurate to within 0.00001? 20. (a) Find the approximations T and M for x1 cosx 2 dx . 0 2 x 8 8 0 (b) Estimate the errors involved in the approximations of part (a). x1 2 ; 3. Estimate 0 cos x dx using (a) the Trapezoidal Rule and (c) How large do we have to choose n so that the approxima- (b) the Midpoint Rule, each with n 4 . From a graph of the tions Tn and Mn to the integral in part (a) are accurate to integrand, decide whether your answers are underestimates or within 0.00001? overestimates. What can you conclude about the true value of 21. (a) Find the approximations T and S for x1 e x dx and the the integral? 10 10 0 corresponding errors ET and ES . ; 4. Draw the graph of f x sinx 22 in the viewing rectangle (b) Compare the actual errors in part (a) with the error esti- x1 mates given by (3) and (4). 0, 1 by 0, 0.5 and let I 0 f x dx. (a) Use the graph to decide whether L2, R2, M2 , and T2 under- (c) How large do we have to choose n so that the approxima- estimate or overestimate I . tions Tn , Mn , and Sn to the integral in part (a) are accurate to

(b) For any value of n , list the numbers Ln, Rn, Mn, Tn, and I in within 0.00001? increasing order. 22. How large should n be to guarantee that the Simpson’s Rule 2 (c) Compute L5, R5, M5, and T5 . From the graph, which do you approximation to x1 ex dx is accurate to within 0.00001 ? think gives the best estimate of I ? 0 CAS 23. The trouble with the error estimates is that it is often very difﬁ- 5–6 |||| Use (a) the Midpoint Rule and (b) Simpson’s Rule to cult to compute four derivatives and obtain a good upper bound approximate the given integral with the speciﬁed value of n . K for f 4x by hand. But computer algebra systems have no 528 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

problem computing f 4 and graphing it, so we can easily ﬁnd Use Simpson’s Rule to estimate the area of the pool. a value for K from a machine graph. This exercise deals with approximations to the integral I x2 f x dx , where 0 5.0 f x e cos x. 5.6 4.8 6.8 4.8 (a) Use a graph to get a good upper bound for f x . 7. 2 6.2 (b) Use M10 to approximate I. (c) Use part (a) to estimate the error in part (b). (d) Use the built-in numerical integration capability of your CAS to approximate I . 31. (a) Use the Midpoint Rule and the given data to estimate the (e) How does the actual error compare with the error estimate x3.2 value of the integral 0 f x dx . in part (c)? 4 (f) Use a graph to get a good upper bound for f x . xxf x f x (g) Use S10 to approximate I. (h) Use part (f) to estimate the error in part (g). 0.0 6.8 2.0 7.6 (i) How does the actual error compare with the error estimate 0.4 6.5 2.4 8.4 in part (h)? 0.8 6.3 2.8 8.8 (j) How large should n be to guarantee that the size of the 1.2 6.4 3.2 9.0 1.6 6.9 error in using Sn is less than 0.0001 ?

1 CAS 24. Repeat Exercise 23 for the integral y s4 x 3 dx . (b) If it is known that 4 f x 1 for all x , estimate the 1 error involved in the approximation in part (a).

25–26 |||| Find the approximations Ln, Rn, Tn , and Mn for n 4, 8 , 32. A radar gun was used to record the speed of a runner during and 16 . Then compute the corresponding errors EL, ER, ET , and EM . the ﬁrst 5 seconds of a race (see the table). Use Simpson’s (Round your answers to six decimal places. You may wish to use Rule to estimate the distance the runner covered during those the sum command on a computer algebra system.) What observa- 5 seconds. tions can you make? In particular, what happens to the errors when n is doubled? t (s)v (m s) t (s)v (m s) 0 0 3.0 10.51 1 2 25.y x 3 dx 26. y e x dx 0.5 4.67 3.5 10.67 0 0 1.0 7.34 4.0 10.76 1.5 8.86 4.5 10.81

27–28 |||| Find the approximations Tn , Mn , and Sn for n 6 and 12 . 2.0 9.73 5.0 10.81 2.5 10.22 Then compute the corresponding errors ET, EM , and ES . (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can 33. The graph of the acceleration at of a car measured in fts2 is you make? In particular, what happens to the errors when n is shown. Use Simpson’s Rule to estimate the increase in the doubled? velocity of the car during the 6-second time interval. a 4 2 27.y sx dx 28. y xe x dx 12 1 1 8 29. Estimate the area under the graph in the ﬁgure by using (a) the 4 Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule, each with n 4 . 0 2 4 6 t y 34. Water leaked from a tank at a rate of rt liters per hour, where the graph of r is as shown. Use Simpson’s Rule to estimate the total amount of water that leaked out during the ﬁrst six hours. r 4 1

0 1234x 2

30. The widths (in meters) of a kidney-shaped swimming pool 0 2 4 6 t were measured at 2-meter intervals as indicated in the ﬁgure. SECTION 7.7 APPROXIMATE INTEGRATION ❙❙❙❙ 529

35. The table (supplied by San Diego Gas and Electric) gives the 39. The region bounded by the curves y s3 1 x 3 , y 0 , x 0 , power consumption in megawatts in San Diego County from and x 2 is rotated about the x -axis. Use Simpson’s Rule with midnight to 6:00 A.M. on December 8, 1999. Use Simpson’s n 10 to estimate the volume of the resulting solid. Rule to estimate the energy used during that time period. (Use the fact that power is the derivative of energy.) CAS 40. The ﬁgure shows a pendulum with length L that makes a maximum angle 0 with the vertical. Using Newton’s Second Law tP tP it can be shown that the period T (the time for one complete swing) is given by 0:00 1814 3:30 1611 0:30 1735 4:00 1621 L 2 dx 1:00 1686 4:30 1666 T 4 y 1:30 1646 5:00 1745 t 0 s1 k 2 sin2x 2:00 1637 5:30 1886 2:30 1609 6:00 2052 1 t where k sin( 2 0 ) and is the acceleration due to gravity. 3:00 1604 If L 1 m and 0 42 , use Simpson’s Rule with n 10 to ﬁnd the period.

36. Shown is the graph of trafﬁc on an Internet service provider’s T1 data line from midnight to 8:00 A.M. D is the data throughput, measured in megabits per second. Use Simpson’s Rule to ¨¸ estimate the total amount of data transmitted during that time period.

D 0.8

41. The intensity of light with wavelength traveling through a diffraction grating with N slits at an angle is given by I N 2 sin2kk 2, where k Nd sin and d is the 0.4 distance between adjacent slits. A helium-neon laser with wavelength 632.8 109 m is emitting a narrow band of light, given by 106 106 , through a grating with 10,000 slits spaced 104 m apart. Use the Midpoint Rule with x106 n 10 to estimate the total light intensity 6 I d emerg- 0 2 468 t (hours) 10 ing from the grating.

42. Use the Trapezoidal Rule with n 10 to approximate 37. If the region shown in the ﬁgure is rotated about the y -axis to x20 cosx dx. Compare your result to the actual value. Can form a solid, use Simpson’s Rule with n 8 to estimate the 0 volume of the solid. you explain the discrepancy? 43. Sketch the graph of a continuous function on 0, 2 for which y the Trapezoidal Rule with n 2 is more accurate than the 4 Midpoint Rule.

2 44. Sketch the graph of a continuous function on 0, 2 for which the right endpoint approximation with n 2 is more accurate than Simpson’s Rule. 0 2 4 6 8 10 x 45. If f is a positive function and f x 0 for a x b , show that b 38. The table shows values of a force function f x where x is Tn y f x dx Mn measured in meters and f x in newtons. Use Simpson’s Rule a to estimate the work done by the force in moving an object a 46. Show that if f is a polynomial of degree 3 or lower, then distance of 18 m. xb Simpson’s Rule gives the exact value of a f x dx . 1 x 0369121518 47. Show that2 Tn Mn T2n . f x 9.8 9.1 8.5 8.0 7.7 7.5 7.4 1 2 48. Show that3 Tn 3 Mn S2n . 530 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

|||| 7.8 Improper Integrals

xb In defining a definite integral a f x dx we dealt with a function f defined on a finite interval a, b and we assumed that f does not have an infinite discontinuity (see Section 5.2). In this section we extend the concept of a definite integral to the case where the interval is infinite and also to the case where f has an infinite discontinuity in a, b . In either case the integral is called an improper integral. One of the most important applications of this idea, probability distributions, will be studied in Section 8.5.

Type I: Infinite Intervals

Try painting a fence that never ends. Consider the infinite region S that lies under the curve y 1x 2 , above the x -axis, and to Resources / Module 6 the right of the line x 1 . You might think that, since S is infinite in extent, its area must / How To Calculate be infinite, but let’s take a closer look. The area of the part of S that lies to the left of the / Start of Improper Integrals line x t (shaded in Figure 1) is

t t 1 1 1 A t y 2 dx 1 1 x x 1 t

Notice that At 1 no matter how large t is chosen.

y 1 y= ≈

1 area=1 - t x=1

0 t x FIGURE 1 1

We also observe that

1 lim At lim 1 1 t l t l t

The area of the shaded region approaches 1 as t l (see Figure 2), so we say that the area of the inﬁnite region S is equal to 1 and we write

1 t 1 y 2 dx lim y 2 dx 1 1 x t l 1 x

y y y y

1 4 area= area= 2 area= area=1 2 3 5

0 1 2 x 0 1 3 x 0 1 5 x 0 1 x

FIGURE 2

Using this example as a guide, we define the integral of f (not necessarily a positive function) over an infinite interval as the limit of integrals over finite intervals. SECTION 7.8 IMPROPER INTEGRALS ❙❙❙❙ 531

1 Definition of an Improper Integral of Type 1 xt (a) If a f x dx exists for every number t a , then

t y f x dx lim y f x dx a t l a

provided this limit exists (as a ﬁnite number). xb (b) If t f x dx exists for every number t b , then

b b y f x dx lim y f x dx t l t

provided this limit exists (as a ﬁnite number). x xb The improper integrals a f x dx and f x dx are called convergent if the corresponding limit exists and divergent if the limit does not exist. x xa (c) If both a f x dx and f x dx are convergent, then we deﬁne

a y f x dx y f x dx y f x dx a In part (c) any real number a can be used (see Exercise 74).

Any of the improper integrals in Deﬁnition 1 can be interpreted as an area provided that x f is a positive function. For instance, in case (a) if f x 0 and the integral a f x dx is convergent, then we deﬁne the area of the region S x, y x a, 0 y f x in Figure 3 to be AS y f x dx a

y=ƒ

0 a x FIGURE 3

x l This is appropriate because a f x dx is the limit as t of the area under the graph of f from a to t.

x EXAMPLE 1 Determine whether the integral 1 1 x dx is convergent or divergent. SOLUTION According to part (a) of Deﬁnition 1, we have

1 t 1 t y dx lim y dx lim ln x ]1 1 x t l 1 x t l lim ln t ln 1 lim ln t t l t l x The limit does not exist as a ﬁnite number and so the improper integral 1 1 x dx is divergent. 532 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

y Let’s compare the result of Example 1 with the example given at the beginning of this 1 section: y= ≈ 1 1 y 2 dx converges y dx diverges 1 x 1 x

finite area Geometrically, this says that although the curves y 1x 2 and y 1x look very similar for x 0 , the region under y 1x 2 to the right of x 1 (the shaded region in Figure 4) 0 1 x has finite area whereas the corresponding region under y 1 x (in Figure 5) has infinite area. Note that both 1x 2 and 1x approach 0 as x l but 1x 2 approaches 0 faster than FIGURE 4 1x. The values of 1x don’t decrease fast enough for its integral to have a finite value.

0 x y EXAMPLE 2 Evaluate y xe dx. 1 y= x SOLUTION Using part (b) of Deﬁnition 1, we have

0 0 y xe x dx lim y xe x dx l inﬁnite area t t We integrate by parts with u x , dv e x dx so that du dx , v e x : 0 1 x 0 0 0 x x x y xe dx xe ]t y e dx FIGURE 5 t t te t 1 e t

We know that e t l 0 as t l , and by l’Hospital’s Rule we have

t t 1 lim te lim lim t l t l e t t l e t lim e t 0 t l Therefore

0 y xe x dx lim te t 1 e t t l 0 1 0 1

1 EXAMPLE 3 Evaluate y dx. 1 x 2

SOLUTION It’s convenient to choose a 0 in Deﬁnition 1(c):

1 0 1 1 y 2 dx y 2 dx y 2 dx 1 x 1 x 0 1 x

We must now evaluate the integrals on the right side separately:

1 t dx t 1 y 2 dx lim y 2 lim tan x]0 0 1 x t l 0 1 x t l lim tan1t tan1 0 lim tan1t t l t l 2 SECTION 7.8 IMPROPER INTEGRALS ❙❙❙❙ 533

0 1 0 dx 0 1 y dx lim y lim tan x] t 1 x 2 t l t 1 x 2 t l lim tan1 0 tan1t t l 0 2 2 Since both of these integrals are convergent, the given integral is convergent and 1 y y= 1 1+≈ y dx area=π 1 x 2 2 2

2 0 x Since 11 x 0 , the given improper integral can be interpreted as the area of the inﬁnite region that lies under the curve y 11 x 2 and above the x -axis (see FIGURE 6 Figure 6).

EXAMPLE 4 For what values of p is the integral

1 y p dx 1 x convergent?

SOLUTION We know from Example 1 that if p 1 , then the integral is divergent, so let’s assume that p 1 . Then

1 t p y p dx lim y x dx 1 x t l 1 xt xp1 lim t l p 1 x1

1 1 lim 1 t l 1 p t p 1 If p 1 , then p 1 0 , so as t l , t p1 l and 1t p1 l 0. Therefore

1 1 y p dx if p 1 1 x p 1 and so the integral converges. But if p 1 , then p 1 0 and so 1 t 1p l as t l t p1 and the integral diverges.

We summarize the result of Example 4 for future reference:

1 2 y p dx is convergent if p 1 and divergent if p 1. 1 x

Type 2: Discontinuous Integrands

Suppose that f is a positive continuous function defined on a finite interval a, b but has a vertical asymptote at b . Let S be the unbounded region under the graph of f and above the x -axis between a and b . (For Type 1 integrals, the regions extended indefinitely in a 534 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

y horizontal direction. Here the region is inﬁnite in a vertical direction.) The area of the part of S between a and t (the shaded region in Figure 7) is y=ƒ x=b t At y f x dx a If it happens that At approaches a deﬁnite number A as t l b , then we say that the 0 a t b x area of the region S is A and we write

b t FIGURE 7 y f x dx lim y f x dx a t l b a We use this equation to deﬁne an improper integral of Type 2 even when f is not a positive function, no matter what type of discontinuity f has at b .

3 Definition of an Improper Integral of Type 2 (a) If f is continuous on a, b and is discontinuous at b , then |||| Parts (b) and (c) of Deﬁnition 3 are illustrated b t in Figures 8 and 9 for the case where f x 0 y f x dx lim y f x dx and f has vertical asymptotes at a and c , a t l b a respectively. if this limit exists (as a ﬁnite number). y (b) If f is continuous on a, b and is discontinuous at a , then

b b y f x dx lim y f x dx a t l a t

if this limit exists (as a ﬁnite number). xb The improper integral a f x dx is called convergent if the corresponding limit 0 at b x exists and divergent if the limit does not exist. (c) If f has a discontinuity at c , where a c b , and both xc f x dx and FIGURE 8 a xb c f x dx are convergent, then we deﬁne y b c b y f x dx y f x dx y f x dx a a c

5 1 EXAMPLE 5 Find y dx. 2 sx 2 0 acbx SOLUTION We note first that the given integral is improper because f x 1sx 2 has the vertical asymptote x 2 . Since the infinite discontinuity occurs at the left end- FIGURE 9 point of 2, 5 , we use part (b) of Definition 3:

y5 dx y5 dx lim y 2 sx 2 t l2 t sx 2 1 5 y= s lim 2 x 2]t œ„„„x-2„ t l2 lim 2(s3 st 2) t l2 area=2œ„3 2s3 0 123 45x Thus, the given improper integral is convergent and, since the integrand is positive, we FIGURE 10 can interpret the value of the integral as the area of the shaded region in Figure 10. SECTION 7.8 IMPROPER INTEGRALS ❙❙❙❙ 535

/2 EXAMPLE 6 Determine whether y sec x dx converges or diverges. 0

SOLUTION Note that the given integral is improper because lim x l/2 sec x . Using part (a) of Deﬁnition 3 and Formula 14 from the Table of Integrals, we have

/2 t y sec x dx lim y sec x dx 0 t l /2 0 t lim ln sec x tan x ]0 t l /2 lim lnsec t tan t ln 1 t l /2

because sec t l and tan t l as t l 2 . Thus, the given improper integral is divergent.

3 dx EXAMPLE 7 Evaluate y if possible. 0 x 1

SOLUTION Observe that the line x 1 is a vertical asymptote of the integrand. Since it occurs in the middle of the interval 0, 3 , we must use part (c) of Deﬁnition 3 with c 1: 3 dx 1 dx 3 dx y y y 0 x 1 0 x 1 1 x 1

1 dx t dx t where y lim y lim ln x 1 ]0 0 x 1 t l1 0 x 1 t l1 lim (ln t 1 ln 1 ) t l1 lim ln1 t t l1

l l x1 because 1 t 0 as t 1 . Thus,0 dx x 1 is divergent. This implies that x3 x3 0 dx x 1 is divergent. [We do not need to evaluate 1 dx x 1 .]

| WARNING If we had not noticed the asymptote x 1 in Example 7 and had instead confused the integral with an ordinary integral, then we might have made the following erroneous calculation:

3 dx 3 y ln x 1 ]0 ln 2 ln 1 ln 2 0 x 1 This is wrong because the integral is improper and must be calculated in terms of limits. xb From now on, whenever you meet the symbol a f x dx you must decide, by looking at the function f on a, b , whether it is an ordinary deﬁnite integral or an improper integral.

1 EXAMPLE 8 Evaluate y ln x dx. 0

SOLUTION We know that the function f x ln x has a vertical asymptote at 0 since

lim x l 0 ln x . Thus, the given integral is improper and we have

1 1 y ln x dx lim y ln x dx 0 t l 0 t 536 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

Now we integrate by parts with u ln x ,dv dx ,du dxx , and v x :

1 1 1 y ln x dx x ln x]t y dx t t 1 ln 1 t ln t 1 t t ln t 1 t To ﬁnd the limit of the ﬁrst term we use l’Hospital’s Rule: ln t lim t ln t lim t l 0 t l 0 1t y 1t lim t l 0 1t 2

0 x lim t 0 1 t l 0 area=1 1 Therefore y ln x dx lim t ln t 1 t 0 t l 0 y=ln x 0 1 0 1 Figure 11 shows the geometric interpretation of this result. The area of the shaded region FIGURE 11 above y ln x and below the x -axis is 1 .

A Comparison Test for Improper Integrals

Sometimes it is impossible to ﬁnd the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. In such cases the following theorem is useful. Although we state it for Type 1 integrals, a similar theorem is true for Type 2 integrals.

Comparison Theorem Suppose that f and t are continuous functions with f x tx 0 for x a. x x t (a) If a f x dx is convergent, then a x dx is convergent. y x t x (b) If a x dx is divergent, then a f x dx is divergent.

f We omit the proof of the Comparison Theorem, but Figure 12 makes it seem plausible. g If the area under the top curve y f x is ﬁnite, then so is the area under the bottom curve y tx. And if the area under y tx is inﬁnite, then so is the area under y f x . x t x 0 a x [Note that the reverse is not necessarily true: If a x dx is convergent,a f x dx may x x t or may not be convergent, and if a f x dx is divergent,a x dx may or may not be FIGURE 12 divergent.]

2 EXAMPLE 9 Show that y e x dx is convergent. 0

2 SOLUTION We can’t evaluate the integral directly because the antiderivative of e x is not an elementary function (as explained in Section 7.5). We write

2 1 2 2 y ex dx y ex dx y ex dx 0 0 1 SECTION 7.8 IMPROPER INTEGRALS ❙❙❙❙ 537

y and observe that the ﬁrst integral on the right-hand side is just an ordinary deﬁnite inte- 2 2 _x2 gral. In the second integral we use the fact that for x 1 we have x x , so x x y=e 2 and therefore ex ex . (See Figure 13.) The integral of ex is easy to evaluate:

_x y=e t y ex dx lim y ex dx 1 t l 1

1 t 1 0 1 x lim e e e t l

FIGURE 13 2 Thus, taking f x ex and tx ex in the Comparison Theorem, we see that x x2 x x2 1 e dx is convergent. It follows that 0 e dx is convergent.

2 TABLE 1 x x In Example 9 we showed that 0 e dx is convergent without computing its value. In 2 t xt ex dx Exercise 70 we indicate how to show that its value is approximately 0.8862. In probabil- 0 ity theory it is important to know the exact value of this improper integral, as we will see 1 0.7468241328 in Section 8.5; using the methods of multivariable calculus it can be shown that the exact 2 0.8820813908 value is s2 . Table 1 illustrates the deﬁnition of an improper integral by showing how 3 0.8862073483 xt x2 s the (computer-generated) values of 0 e dx approach 2 as t becomes large. In fact, 4 0.8862269118 2 these values converge quite quickly because e x l 0 very rapidly as x l . 5 0.8862269255 6 0.8862269255 1 e x EXAMPLE 10 The integral y dx is divergent by the Comparison Theorem 1 x because x 1 e 1 x x x and 1 1 x dx is divergent by Example 1 [or by (2) with p 1 ].

Table 2 illustrates the divergence of the integral in Example 10. It appears that the values are not approaching any ﬁxed number.

TABLE 2 xt x t 1 1 e x dx 2 0.8636306042 5 1.8276735512 10 2.5219648704 100 4.8245541204 1000 7.1271392134 10000 9.4297243064

|||| 7.8 Exercises

1. Explain why each of the following integrals is improper. 2. Which of the following integrals are improper? Why?

4 2 2 1 1 1 (a)y x 4ex dx (b) y sec x dx (a)y dx (b) y dx 1 0 1 2x 1 0 2x 1 2 x 0 1 2 y y sin x (c) 2 dx (d) 2 dx (c)y 2 dx (d) y ln x 1 dx 0 x 5x 6 x 5 1 x 1 538 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

3. Find the area under the curve y 1x 3 from x 1 to x t 2 1 ln x 39.y z 2 ln z dz 40. y dx and evaluate it for t 10 ,100 , and 1000 . Then ﬁnd the total 0 0 sx

area under this curve for x 1 . 1.1 t 0.9 ; 4. (a) Graph the functions f x 1 x and x 1 x in the 41–46 |||| Sketch the region and ﬁnd its area (if the area is ﬁnite). viewing rectangles 0, 10 by 0, 1 and 0, 100 by 0, 1 . x (b) Find the areas under the graphs of f and t from x 1 to 41. S x, y x 1, 0 y e 4 6 10 20 x t and evaluate for t 10 ,100 ,10 ,10 ,10 , and 10 . 42. S x, y x 2, 0 y e x/2 (c) Find the total area under each curve for x 1 , if it exists. ; 43. S x, y 0 y 2x 2 9 5–40 |||| Determine whether each integral is convergent or ; S x, y x 0, 0 y xx 2 9 divergent. Evaluate those that are convergent. 44. ; 45. S x, y 0 x 2, 0 y sec2x 1 0 1 5.y 2 dx 6. y dx 1 3x 1 2x 5 ; 46. S {x, y 2 x 0, 0 y 1sx 2}

1 1 x 7. y dw 8. y 2 2 dx 2 2 s2 w 0 x 2 ; 47. (a) If tx sin xx , use your calculator or computer to xt t make a table of approximate values of 1 x dx for t 2 , 1 9.y ey2 dy 10. y e2t dt 5, 10, 100, 1000, and 10,000. Does it appear that x tx dx 1 4 is convergent? 2 x (b) Use the Comparison Theorem with f x 1x to show 4 11.y 2 dx 12. y 2 v dv x t 1 x that 1 x dx is convergent. (c) Illustrate part (b) by graphing f and t on the same screen 2 3 13. y xe x dx 14. y x 2e x dx for 1 x 10 . Use your graph to explain intuitively why x t 1 x dx is convergent. t s 15.y sin d 16. y cos2 d ; 48. (a) If x 1 ( x 1) , use your calculator or computer to 2 0 xt t make a table of approximate values of 2 x dx for t 5 , 10, 100, 1000, and 10,000. Does it appear that x tx dx is x 1 dz 2 17.y 2 dx 18. y 2 convergent or divergent? 1 x 2x 0 z 3z 2 (b) Use the Comparison Theorem with f x 1sx to show 6 that x tx dx is divergent. 19.y se 5s ds 20. y re r 3 dr 2 0 (c) Illustrate part (b) by graphing f and t on the same screen for 2 x 20 . Use your graph to explain intuitively why ln x x x t 21. y dx 22. y e dx 2 x dx is divergent. 1 x 49–54 |||| Use the Comparison Theorem to determine whether the x 2 ln x 23.y dx 24. y dx integral is convergent or divergent. 9 x 6 1 x 3 cos2x 2 e x ln x x arctan x 49.y 2 dx 50. y dx 25.y dx 26. y dx 1 1 x 1 x 1 x 2 0 1 x 22 dx x 3 3 1 1 51. y 2x 52. y dx 27. y dx 28. y dx 1 x e 1 s1 x 6 0 sx 0 xsx x /2 dx 1 e 0 1 9 1 53.y 54. y dx 29.y dx 30. y dx 0 x sin x 0 sx 1 x 2 1 s3 x 9

3 1 1 dx 31. y dx 32. y 55. The integral 4 2 2 x 0 s1 x 1 y dx 0 s 33 1 1 x 1 x 33.y x 115 dx 34. y dy 0 0 4y 1 is improper for two reasons: The interval 0, is inﬁnite and the integrand has an inﬁnite discontinuity at 0. Evaluate it by 4 1 35.y sec x dx 36. y dx expressing it as a sum of improper integrals of Type 2 and 2 0 0 x x 6 Type 1 as follows: x 1 e 2 x 3 1 1 1 1 37.y x dx 38. y dx y dx y dx y dx 1 e 1 0 2x 3 0 sx 1 x 0 sx 1 x 1 sx 1 x SECTION 7.8 IMPROPER INTEGRALS ❙❙❙❙ 539

56. Evaluate a photograph. Suppose that in a spherical cluster of radius R 1 the density of stars depends only on the distance r from the y dx center of the cluster. If the perceived star density is given by 2 xsx 2 4 ys, where s is the observed planar distance from the center of by the same method as in Exercise 55. the cluster, and xr is the actual density, it can be shown that

|||| Find the values of p for which the integral converges and R 2r 57–59 ys y xr dr evaluate the integral for those values of p . s sr 2 s 2

1 1 1 1 2 If the actual density of stars in a cluster is x r 2 R r , 57. y p dx 58. y p dx 0 x e xln x ﬁnd the perceived density ys .

1 59. y x p ln x dx 67. A manufacturer of lightbulbs wants to produce bulbs that last 0 about 700 hours but, of course, some bulbs burn out faster than others. Let Ft be the fraction of the company’s bulbs that burn x n x out before t hours, so Ft always lies between 0 and 1. 60. (a) Evaluate the integral 0 x e dx for n 0 ,1 ,2 , and 3 . x n x (a) Make a rough sketch of what you think the graph of F (b) Guess the value of 0 x e dx when n is an arbitrary positive integer. might look like. (b) What is the meaning of the derivative rt Ft ? (c) Prove your guess using mathematical induction. (c) What is the value of x rt dt ? Why? x 0 61. (a) Show that x dx is divergent. (b) Show that 68. As we will see in Section 9.4, a radioactive substance decays exponentially: The mass at time t is mt m0e kt , where t lim y x dx 0 m0 is the initial mass and k is a negative constant. The mean

t l t life M of an atom in the substance is

This shows that we can’t deﬁne M k y te kt dt 0 t y f x dx lim y f x dx t l t For the radioactive carbon isotope,14C , used in radiocarbon dating, the value of k is 0.000121 . Find the mean life of a 62. The average speed of molecules in an ideal gas is 14C atom.

32 4 M 2 v y v 3eMv 2RT dv 69. Determine how large the number a has to be so that s 2RT 0 1 where M is the molecular weight of the gas,R is the gas y 2 dx 0.001 a x 1 constant,T is the gas temperature, and v is the molecular speed. Show that x x2 70. Estimate the numerical value of 0 e dx by writing it as x4 x2 x x2 8RT the sum of 0 e dx and 4 e dx . Approximate the first inte- v M gral by using Simpson’s Rule with n 8 and show that the x 4x second integral is smaller than 4 e dx , which is less than 63. We know from Example 1 that the region 0.0000001. x, y x 1, 0 y 1x has infinite area. Show that by rotating about the x -axis we obtain a solid with 71. If f t is continuous for t 0 , the Laplace transform of f is finite volume. the function F defined by

64. Use the information and data in Exercises 29 and 30 of Sec- Fs y f te st dt tion 6.4 to find the work required to propel a 1000-kg satellite 0 out of Earth’s gravitational field. and the domain of F is the set consisting of all numbers s for 65. Find the escape velocity v0 that is needed to propel a rocket which the integral converges. Find the Laplace transforms of of mass m out of the gravitational field of a planet with mass the following functions. M and radius R . Use Newton’s Law of Gravitation (see Exer- (a)f t 1 (b)f t e t (c) f t t cise 29 in Section 6.4) and the fact that the initial kinetic 1 2 at energy of 2 mv 0 supplies the needed work. 72. Show that if 0 f t Me for t 0 , where M and a are constants, then the Laplace transform Fs exists for s a . 66. Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster from the 73. Suppose that 0 f t Me at and 0 f t Ke at for t 0 , observed (two-dimensional) density that can be analyzed from where f is continuous. If the Laplace transform of f t is Fs 540 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

and the Laplace transform of f t is Gs , show that 77. Find the value of the constant C for which the integral G s sF s f 0 s a 1 C dx x y 2 74. If f x dx is convergent and a and b are real numbers, 0 sx 4 x 2 show that converges. Evaluate the integral for this value of C . a b y f x dx y f x dx y f x dx y f x dx a b 78. Find the value of the constant C for which the integral

2 2 75. Show that x x 2ex dx 1 x ex dx . x C 0 2 0 dx y 2 2 x 1 3x 1 x x x1 s 0 76. Show that 0 e dx 0 ln y dy by interpreting the integrals as areas. converges. Evaluate the integral for this value of C .

7 Review |||| CONCEPT CHECK

1. State the rule for integration by parts. In practice, how do you xb 5. State the rules for approximating the definite integral a f x dx use it? with the Midpoint Rule, the Trapezoidal Rule, and Simpson’s Rule. Which would you expect to give the best estimate? How x m n 2. How do you evaluate sin x cos x dx if m is odd? What if n is do you approximate the error for each rule? odd? What if m and n are both even? 6. Define the following improper integrals. If the expression sa 2 x 2 occurs in an integral, what sub- b 3. (a)y f x dx (b)y f x dx (c) y f x dx stitution might you try? What if sa 2 x 2 occurs? What if a 2 2 sx a occurs? xb 7. Define the improper integral a f x dx for each of the follow- 4. What is the form of the partial fraction expansion of a rational ing cases. function PxQx if the degree of P is less than the degree of (a)f has an infinite discontinuity at a . Q and Qx has only distinct linear factors? What if a linear (b)f has an infinite discontinuity at b . factor is repeated? What if Qx has an irreducible quadratic (c)f has an infinite discontinuity at c , where a c b . factor (not repeated)? What if the quadratic factor is repeated? 8. State the Comparison Theorem for improper integrals.

TRUE-FALSE QUIZ

Determine whether the statement is true or false. If it is true, explain why. 8. The Midpoint Rule is always more accurate than the If it is false, explain why or give an example that disproves the statement. Trapezoidal Rule. xx 2 4 A B 1. can be put in the form . 9. (a) Every elementary function has an elementary derivative. x 2 4 x 2 x 2 (b) Every elementary function has an elementary antiderivative. 2 x 4 A B C x 2. can be put in the form . 10. If f is continuous on 0, and 1 f x dx is convergent, then xx 2 4 x x 2 x 2 x 0 f x dx is convergent. x 2 4 A B 3. 2 can be put in the form 2 . 11. If f is a continuous, decreasing function on 1, and x x 4 x x 4 l x limx f x 0, then1 f x dx is convergent. x 2 4 A B 4. 2 can be put in the form 2 . x x t xx 4 x x 4 12. If a f x dx and a x dx are both convergent, then x f x tx dx is convergent. 4 x a 1 5. y 2 dx 2 ln 15 0 x 1 x x t 13. If a f x dx and a x dx are both divergent, then x t 1 a f x x dx is divergent. 6. y s2 dx is convergent. 1 x 14. If f x tx and x tx dx diverges, then x f x dx also t 0 0 x l x 7. If f is continuous,then f x dx limt t f x dx . diverges. CHAPTER 7 REVIEW ❙❙❙❙ 541

EXERCISES

Note: Additional practice in techniques of integration is provided in 2x s 1 2 xe 3 tan Exercises 7.5. 39.y 2 dx 40. y d 0 1 2x 4 sin 2

1–40 |||| Evaluate the integral.

5 x 5 41–50 |||| Evaluate the integral or show that it is divergent. 1.y dx 2. y ye 0.6y dy 0 x 10 0 2 1 1 t 1 41.y 3 dx 42. y 2 dt 2 4 1 2x 1 0 t 1 cos dt 3.y d 4. y 3 0 1 t 1 sin 2 1 dx 6 y 43.y 44. y dy 1 2 x ln x 2 sy 2 5.y tan7x sec3x dx 6. y dy y 2 y 4 12 4 ln x 1 1 45.y dx 46. y dx sinln t dx 0 sx 0 2 3x 7.y dt 8. y t x 2s x 2 1 3 dx 1 x 1 47.y 2 48. y 3 dx s 0 x x 2 1 sx 4 4 32 1 arctan x 9.y x ln x dx 10. y 2 dx 1 1 0 1 x dx tan x 49.y 50. y dx 2 2 2 4x 4x 5 1 x 2 sx 1 1 sin x 11.y dx 12. y 2 dx 1 x 1 1 x

dx x 2 2 ; 51–52 |||| Evaluate the indeﬁnite integral. Illustrate and check that 13.y 14. y dx x 3 x x 2 your answer is reasonable by graphing both the function and its antiderivative (take C 0 ). 6 3 2 5 sec 15.y sin cos d 16. y d 2 x 2 51.y lnx 2x 2 dx 52. y dx tan sx 2 1 x 2 8x 3 17.y x sec x tan x dx 18. y 3 2 dx x 3x ; 53. Graph the function f x cos2x sin3x and use the graph to x2 f x dx x 1 dt guess the value of the integral 0 . Then evaluate the 19.y dx 20. y integral to conﬁrm your guess. 9x 2 6x 5 sin2t cos 2t CAS (a) How would you evaluate x x 5e 2x dx by hand? dx x 3 54. 21.y 22. y 10 dx (Don’t actually carry out the integration.) sx 2 4x x 1 (b) How would you evaluate x x 5e 2x dx using tables?

4 x (Don’t actually do it.) y y 23. csc 4x dx 24. e cos x dx (c) Use a CAS to evaluate x x 5e 2x dx . (d) Graph the integrand and the indeﬁnite integral on the same 3 2 3x x 6x 4 dx screen. 25.y 2 2 dx 26. y x x 1x 2 1 e

3 2 sx 1 27.y cos3x sin 2x dx 28. y dx 55–58 |||| Use the Table of Integrals on the Reference Pages to 3 0 sx 1 evaluate the integral.

1 5 dx x 5 29.y x sec x dx 30. y x y e s e 2x dx y t dt 1 e s1 e 2x 55. 1 56. csc

x x ln 10 e se 1 4 x sin x cot x 31.y dx 32. y dx 2 x 3 57.y sx x 1 dx 58. y dx 0 e 8 0 cos x s1 2 sin x

2 x 2 33.y dx 34. y arcsin x dx 4 x 2 3 2 59. Verify Formula 33 in the Table of Integrals (a) by differentia- 1 1 tan tion and (b) by using a trigonometric substitution. 35.y dx 36. y d sx x 32 1 tan 60. Verify Formula 62 in the Table of Integrals. 61. Is it possible to ﬁnd a number n such that x x n dx is y 2 y 1 2 0 37. cos x sin x cos 2x dx 38. x tan x dx convergent? 542 ❙❙❙❙ CHAPTER 7 TECHNIQUES OF INTEGRATION

x ax 62. For what values of a is 0 e cos x dx convergent? Evaluate the point to be 53 cm. The circumference 7 cm from each end is integral for those values ofa . 45 cm. Use Simpson’s Rule to make your estimate.

63–64 |||| Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule with n 10 to approximate the given integral. Round your answers to six decimal places.

1 2 63.y s1 x 4 dx 64. y ssin x dx 0 0 28 cm 71. Use the Comparison Theorem to determine whether the integral 65. Estimate the errors involved in Exercise 63, parts (a) and (b). 3 How large should n be in each case to guarantee an error of x y 5 dx less than 0.00001? 1 x 2 is convergent or divergent. 66. Use Simpson’s Rule with n 6 to estimate the area under the curve y e xx from x 1 to x 4. 72. Find the area of the region bounded by the hyperbola y 2 x 2 1 and the line y 3 . 67. The speedometer reading (v ) on a car was observed at 73. Find the area bounded by the curves y cos x and y cos2x 1-minute intervals and recorded in the chart. Use Simpson’s between x 0 and x . Rule to estimate the distance traveled by the car. 74. Find the area of the region bounded by the curves s s t (min)v (mih) t (min)v (mih) y 1 (2 x),y 1 (2 x) , and x 1. 75. The region under the curve y cos2x, 0 x 2 , is rotated 040656 about the x -axis. Find the volume of the resulting solid. 1 42 7 57 2 45 8 57 76. The region in Exercise 75 is rotated about the y -axis. Find the 3 49 9 55 volume of the resulting solid. 4 52 10 56 77. If f is continuous on 0, and lim x l f x 0 , show that 554 y f x dx f 0 0 68. A population of honeybees increased at a rate of rt bees per 78. We can extend our definition of average value of a continuous week, where the graph of r is as shown. Use Simpson’s Rule function to an infinite interval by defining the average value of with six subintervals to estimate the increase in the bee popula- f on the interval a, to be tion during the first 24 weeks. 1 t lim y f x dx l r t t a a 1 12000 (a) Find the average value of y tan x on the interval 0, . x (b) If f x 0 and a f x dx is divergent, show that the average value of f on the interval a, is lim l f x , if this 8000 x limit exists. x (c) If a f x dx is convergent, what is the average value of f 4000 on the interval a, ? (d) Find the average value of y sin x on the interval 0, . 79. Use the substitution u 1x to show that 0 4 8 12 16 20 24 t (weeks) ln x y 2 dx 0 0 1 x CAS 69. (a) If f x sinsin x , use a graph to find an upper bound 80. The magnitude of the repulsive force between two point charges for f 4 x. with the same sign, one of size 1 and the other of size q , is (b) Use Simpson’s Rule with n 10 to approximate q F x f x dx and use part (a) to estimate the error. 2 0 40r (c) How large should n be to guarantee that the size of the where r is the distance between the charges and 0 is a error in using Sn is less than 0.00001 ? constant. The potential V at a point P due to the charge q is 70. Suppose you are asked to estimate the volume of a football. defined to be the work expended in bringing a unit charge to P You measure and find that a football is 28 cm long. You use a from infinity along the straight line that joins q and P . Find a piece of string and measure the circumference at its widest formula for V . PROBLEMS Cover up the solution to the example and try it yourself first. PLUS EXAMPLE (a) Prove that if f is a continuous function, then

a a y f x dx y f a x dx 0 0

(b) Use part (a) to show that

n 2 sin x y n n dx 0 sin x cos x 4

for all positive numbers n .

SOLUTION (a) At first sight, the given equation may appear somewhat baffling. How is it possible |||| The principles of problem solving are to connect the left side to the right side? Connections can often be made through one of discussed on page 80. the principles of problem solving: introduce something extra. Here the extra ingredient is a new variable. We often think of introducing a new variable when we use the Substitu- tion Rule to integrate a specific function. But that technique is still useful in the present circumstance in which we have a general function f . Once we think of making a substitution, the form of the right side suggests that it should be u a x. Then du dx. When x 0,u a ; when x a,u 0 . So

a 0 a y f a x dx y f u du y f u du 0 a 0

xa But this integral on the right side is just another way of writing 0 f x dx . So the given equation is proved. (b) If we let the given integral be I and apply part (a) with a 2 , we get

n n 2 sin x 2 sin 2 x I y n n dx y n n dx 0 sin x cos x 0 sin 2 x cos 2 x

|||| The computer graphs in Figure 1 make it A well-known trigonometric identity tells us that sin2 x cos x and seem plausible that all of the integrals in the example have the same value. The graph of each cos 2 x sin x, so we get integrand is labeled with the corresponding n value of n . 2 cos x I y n n dx 0 cos x sin x 1 3 Notice that the two expressions for I are very similar. In fact, the integrands have the 4 2 1 same denominator. This suggests that we should add the two expressions. If we do so, we get

n n 2 sin x cos x 2 2I y n n dx y 1 dx 0 sin x cos x 0 2 0 π 2 FIGURE 1 Therefore,I 4 . PROBLEMS ; 1. Three mathematics students have ordered a 14-inch pizza. Instead of slicing it in the tradi- tional way, they decide to slice it by parallel cuts, as shown in the ﬁgure. Being mathematics majors, they are able to determine where to slice so that each gets the same amount of pizza. Where are the cuts made? 1 2. Evaluate y dx. x 7 x The straightforward approach would be to start with partial fractions, but that would be brutal. Try a substitution.

1 3. Evaluate y (s3 1 x7 s7 1 x 3) dx. 0

14 in 4. A man initially standing at the point O walks along a pier pulling a rowboat by a rope of length L . The man keeps the rope straight and taut. The path followed by the boat is a curve FIGURE FOR PROBLEM 1 called a tractrix and it has the property that the rope is always tangent to the curve (see the ﬁgure). (a) Show that if the path followed by the boat is the graph of the function y f x , then y dy sL 2 x 2 f x dx x

(b) Determine the function y f x .

pier L (x, y) 5. A function f is deﬁned by (L, 0) O x f x y cos t cosx t dt 0 x 2 0

FIGURE FOR PROBLEM 4 Find the minimum value of f . 6. If n is a positive integer, prove that

1 y ln xn dx 1n n! 0

7. Show that

2n 2 1 2 n! y 1 x 2 n dx 0 2n 1!

Hint: Start by showing that if In denotes the integral, then 2k 2 I I k 1 2k 3 k ; 8. Suppose that f is a positive function such that f is continuous. (a) How is the graph of y f x sin nx related to the graph of y f x ? What happens as n l ? (b) Make a guess as to the value of the limit

1 lim y f x sin nx dx n l 0 based on graphs of the integrand. (c) Using integration by parts, conﬁrm the guess that you made in part (b). [Use the fact that, since f is continuous, there is a constant M such that f x M for 0 x 1 .]

1t 1 9. If 0 a b, ﬁnd lim y bx a1 xt dx . t l 0 0 x xt1 ; 10. Graph f x sin e and use the graph to estimate the value of t such that t f x dx is a maximum. Then ﬁnd the exact value of t that maximizes this integral.

y 11. The circle with radius 1 shown in the ﬁgure touches the curve y 2x twice. Find the area of the region that lies between the two curves.

12. A rocket is ﬁred straight up, burning fuel at the constant rate of b kilograms per second. Let v vt be the velocity of the rocket at time t and suppose that the velocity u of the exhaust gas is constant. Let M Mt be the mass of the rocket at time t and note that M decreases y=| 2x | as the fuel burns. If we neglect air resistance, it follows from Newton’s Second Law that dv 0 x F M ub dt

FIGURE FOR PROBLEM 11 where the force F Mt . Thus dv 1 M ub Mt dt

Let M1 be the mass of the rocket without fuel,M2 the initial mass of the fuel, and

M0 M1 M2. Then, until the fuel runs out at time t M2b , the mass is M M0 bt .

(a) Substitute M M0 bt into Equation 1 and solve the resulting equation for v . Use the initial condition v0 0 to evaluate the constant.

(b) Determine the velocity of the rocket at time t M2b . This is called the burnout velocity. (c) Determine the height of the rocket y yt at the burnout time. (d) Find the height of the rocket at any time t . 13. Use integration by parts to show that, for all x 0 ,

sin t 2 0 y dt 0 ln1 x t ln1 x

; 14. The Chebyshev polynomials Tn are deﬁned by

Tnx cosn arccos x n 0,1 ,2 ,3 ,... (a) What are the domain and range of these functions?

(b) We know that T0x 1 and T1x x . Express T2 explicitly as a quadratic polynomial

and T3 as a cubic polynomial. (c) Show that, for n 1 ,

Tn1x 2xTnx Tn1x

(d) Use part (c) to show that Tn is a polynomial of degree n .

(e) Use parts (b) and (c) to express T4, T5, T6, and T7 explicitly as polynomials.

(f) What are the zeros of Tn ? At what numbers does Tn have local maximum and minimum values?

(g) Graph T2 ,T3 ,T4 , and T5 on a common screen.

(h) Graph T5 ,T6 , and T7 on a common screen.

(i) Based on your observations from parts (g) and (h), how are the zeros of Tn related to the

zeros of Tn1 ? What about the x -coordinates of the maximum and minimum values? x1 (j) Based on your graphs in parts (g) and (h), what can you say about 1 Tn x dx when n is odd and when n is even? (k) Use the substitution u arccos x to evaluate the integral in part (j). (l) The family of functions f x cosc arccos x are deﬁned even when c is not an integer (but then f is not a polynomial). Describe how the graph of f changes as c increases.