Steward – Techniques of Integration
7 TECHNIQUES OF
INTEGRATION
Simpson’s Rule estimates integrals by approximating graphs with parabolas.
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 dx 苷
dx 苷 ln ⱍ x
dx 苷 e x
a y x dx 苷
ln a
sec y 2 x dx 苷
y csc 2
y tan x dx 苷 ln ⱍ sec x ⱍ
y cot x dx 苷 ln ⱍ sin x ⱍ
y 2 2 sa dx 苷 sin 冉 a 冊
x 2 2 dx 苷
tan
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 we discuss a strategy for
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 and are differentiable functions, then f t
d 关 f 共x兲 dx
In the notation for indefinite integrals this equation becomes
y 关 f 共x兲
y f 共x兲
or
We can rearrange this equation as
1 y f 共x兲
Formula 1 is called the formula for integration by parts. It is perhaps easier to remem- ber in the following notation. Let u苷f 共x兲 and v 苷 t共x兲 . Then the differentials are
共x兲 dx and d v 苷 , so, by the Substitution Rule, the formula for integration by parts becomes
2 y ud v 苷 u v y v du
EXAMPLE 1 Find . y x sin x dx
. Then 共x兲 苷 1 and
SOLUTION USING FORMULA 1 Suppose we choose f 共x兲 苷 x and
. (For we can choose any antiderivative of .) Thus, using Formula t
1, we have
y x sin x dx 苷 f 共x兲
y y cos x dx
苷x
It’s wise to check the answer by differentiating it. If we do so, we get x sin x , as expected.
CHAPTER 7 TECHNIQUES OF INTEGRATION
SOLUTION USING FORMULA 2 Let
N It is helpful to use the pattern:
y x sin x dx 苷 y x sin x dx 苷 x
y cos x dx
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 instead chosen u苷 sin x and d v 苷 x dx , then
du 苷 cos x dx and v 苷 x 2 兾2 , so integration by parts gives
x sin x dx 苷 共sin x兲
x 2 cos x dx
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 and u d v , we usually try to choose u苷f 共x兲 to
be a function that becomes simpler when differentiated (or at least not more complicated)
as long as d v 苷
can be readily integrated to give . v
y ln x dx
V EXAMPLE 2 Evaluate .
SOLUTION Here we don’t have much choice for and u d v . Let
x Integrating by parts, we get
y dx
N It’s customary to write x 1 dx as x dx .
y dx
N Check the answer by differentiating it.
Integration by parts is effective in this example because the derivative of the function
f 共x兲 苷 ln x is simpler than . f
SECTION 7.1 INTEGRATION BY PARTS
y t 2 e t dt
V EXAMPLE 3 Find .
SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e t is unchanged
when differentiated or integrated), so we choose u苷t 2 d v 苷 e t dt
Then
du 苷 2t dt
Integration by parts gives
te y t y dt
3 t 2 e t dt 苷 t 2 e t
The integral that we obtained, x te t 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苷t and
d v 苷 e t dt . Then du 苷 dt , v 苷 e t , and
te t dt 苷 te t
y t y e dt 苷
te t
Putting this in Equation 3, we get
y t 2 e t dt 苷 t 2 e t y te t dt
1 where C 1 苷
V EXAMPLE 4 Evaluate . e x y sin x dx
N An easier method, using complex numbers, is
given in Exercise 50 in Appendix H.
SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing u苷e x and d v 苷 sin x dx anyway. Then du 苷 e x dx and v 苷
, so integration by
parts gives
4 e x x y e x cos x dx y
The integral that we have obtained, e x 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苷e x and
d v 苷 cos x dx . Then du 苷 e x dx , v 苷 sin x , and
5 e x cos x dx 苷 e x e x y y sin x dx
At first glance, it appears as if we have accomplished nothing because we have arrived at
x e x sin x dx , which is where we started. However, if we put the expression for x e x cos x dx
from Equation 5 into Equation 4 we get
y e x x x y e x sin x dx
CHAPTER 7 TECHNIQUES OF INTEGRATION
N Figure 1 illustrates Example 4 by show-
This can be regarded as an equation to be solved for the unknown integral. Adding
ing the graphs of 1 f 共x兲 苷 e x sin x and
x e x sin x dx to both sides, we obtain
F 共x兲 苷 2 e . As a visual check on our work, notice that f 共x兲 苷 0 when has F a maximum or minimum.
2 e x x y x sin x
Dividing by 2 and adding the constant of integration, we get
e x sin x dx 苷 1 y x 2 e M
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 and , assuming a b and are continuous, and using the Fundamental
FIGURE 1
Theorem, we obtain
a f 共x兲
EXAMPLE 5 1 Calculate . y
0 tan x dx SOLUTION Let
So Formula 6 gives
0 tan x dx 苷 x tan x ] 0 y 0 2 dx
y 0 2 dx
苷 1 ⴢ tan
N Since tan 0 for 0 , the integral in Example 5 can be interpreted as the area of the
dx
region shown in Figure 2.
2 (since has another meaning u y=tan–!x 1 in this example). Then dt 苷 2x dx , so x dx 苷
To evaluate this integral we use the substitution t苷
2 dt . When x苷 0 , t苷 1 ; when x苷 1 ,
t苷 2 ; so
1 2 dt
1 y 0 2 dx 苷 2 y 1 2 t ln ⱍ t ⱍ] 1
2 2 ln 2
y 0 tan x dx 苷
ln 2
FIGURE 2
Therefore
dx 苷
SECTION 7.1 INTEGRATION BY PARTS
EXAMPLE 6 Prove the reduction formula
N Equation 7 is called a reduction formula because the exponent has been n reduced to
n⫺ y 1
7 sin n x dx 苷 ⫺ cos x sin n⫺ 1
y n⫺ sin 2 x dx
x⫹
n⫺ 1 and . n⫺ 2
where n艌 2 is an integer.
SOLUTION Let
so integration by parts gives
y 1 sin x
y sin 2 x cos 2 x dx
Since , we cos 2 x苷 2 x have
1 y 2 cos x sin y sin x
sin n
y sin n x 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 sin n
y sin 2 x dx
The reduction formula (7) is useful because by using it repeatedly we could eventually
express sin n x x dx in terms of x sin x dx (if is odd) or n x 0 dx 苷 x dx (if is even). n
7.1 EXERCISES
1–2 Evaluate the integral using integration by parts with the
11. y arctan 4t dt
12. y p 5 ln p dp
indicated choices of and u d v .
1. y 2 x ln x dx ; u苷 ln x , d v 苷 x 2 dx
13. 2 y s t sec 2t dt 14. y s2 ds
2. y cos d ; u苷 , d v 苷 cos d
15. y 2 dx
16. y t sinh mt dt
⫺ 3–32 Evaluate the integral. 17. e y 2 sin 3 d 18. y e cos 2 d
3. x cos 5x dx
4. y x xe dx
1 19. ⫺ t sin 3t dt y 2
0 20. ⫹ 1 y x 0 dx
5. y re r dr
6. y t sin 2t dt
22. y 4 x y 2 sin x dx
21. y 0 t cosh t dt
1 9 ln y
sy 7. dy 8.
y x cos mx dx
2 ln x
⫺ 1 23. y 1 2 dx
9. y ln
10. y sin x dx
24. x 3 cos x dx
CHAPTER 7 TECHNIQUES OF INTEGRATION
(b) Use part (a) to evaluate x 2 sin 3 x dx and x 2 sin 0 5 0 x dx y .
0 e y 1 (c) Use part (a) to show that, for odd powers of sine,
26. arctan
27. y 0 cos x dx
28. y 1 3 dx 2 sin 2 n⫹1 x dx 苷
2 ⴢ 4 ⴢ 6 ⴢ ⭈ ⭈ ⭈ ⴢ 2n
29. y cos x ln
30. y 0 2 dr
46. Prove that, for even powers of sine,
31. y 1 x
32. y 0 e sin
y 0 sin 2n x dx 苷
dx
2 ⴢ 4 ⴢ 6 ⴢ ⭈ ⭈ ⭈ ⴢ 2n 2
33–38 First make a substitution and then use integration by parts 47–50 Use integration by parts to prove the reduction formula. to evaluate the integral.
47. n dx 苷 x n ⫺ n n⫺ 1
y dx
y 3 sx dx 34. y t e dt
33. cos
48. x n e x dx 苷 x n e x ⫺ n x n⫺ 1 e x dx y y
35. 3 cos y 2
36. y 0 e cos t sin 2t dt
tan n⫺ 1 x
49. tan n x dx 苷
⫺ tan n⫺ 2 y x dx
37. y x ln
38. y sin
n⫺ 1 n⫺ 1 ; 39– 42 Evaluate the indefinite integral. Illustrate, and check that
your answer is reasonable, by graphing both the function and its
antiderivative (take C苷 0 ).
51. Use Exercise 47 to find
x 3 dx .
39. y x dx 40. x 3 y 2 ln x dx
52. Use Exercise 48 to find x x 4 e x dx .
53–54 y Find the area of the region bounded by the given curves.
41. x 3 2
42. y x 2 sin 2x dx
dx
53. y 苷 xe ⫺ , 0.4 x y苷 0 , x苷 5 54. y苷 5 ln x, y苷x ln x
43. (a) Use the reduction formula in Example 6 to show that
; 55–56 Use a graph to find approximate -coordinates of the y x
sin 2x
sin x dx 苷
2 4 points of intersection of the given curves. Then find (approxi- mately) the area of the region bounded by the curves.
(b) Use part (a) and the reduction formula to evaluate
sin x 4 x dx .
55. y苷x sin x , y苷
44. (a) Prove the reduction formula
56. y苷 arctan 3x , y苷 1 2 x
57–60 Use the method of cylindrical shells to find the volume
2 (b) Use part (a) to evaluate generated by rotating the region bounded by the given curves x cos x dx .
4 (c) Use parts (a) and (b) to evaluate about the specified axis. x cos x dx .
, y苷 0 , 0艋x艋1 ; about the -axis y 45. (a) Use the reduction formula in Example 6 to show that
57. y苷 cos
58. y苷e x , y苷e ⫺ x , x苷 1 ; about the -axis y 2 sin n
0 x dx 苷
sin x dx
59. y苷e , y苷 0 , x苷⫺ 1 , x苷 0 ; about x苷 1 60. y苷e x , x苷 0 , y苷 ; about the -axis x
where 2 is an integer.
SECTION 7.1 INTEGRATION BY PARTS
61. Find the average value of f 2 ln x on the interval
parts on the resulting integral to prove that
62. b A rocket accelerates by burning its onboard fuel, so its mass V苷 y
2 x f
decreases with time. Suppose the initial mass of the rocket at
liftoff (including its fuel) is , the fuel is consumed at rate , m r
and the exhaust gases are ejected with constant velocity v e y x=g(y)
y=ƒ (relative to the rocket). A model for the velocity of the rocket
at time is given by the equation t x=b
v e ln
x=a
0 a b where is the acceleration due to gravity and is not too t
t x large. If
t 苷 9.8 m s 2 , m苷 30,000 kg, r苷 160 kg s, and v e 苷 3000 m s , find the height of the rocket one minute
68. Let . I n 苷 2 x n
0 sin x dx
after liftoff.
(a) Show that I 2 n⫹2 艋 I 2 n⫹1 艋 I 2n .
(b) Use Exercise 46 to show that
63. A particle that moves along a straight line has velocity v
2n ⫹ 1 it travel during the first seconds? t
2 e meters per second after seconds. How far will t
are continuous, show that
(c) Use parts (a) and (b) to show that
0 f y 0 2n ⫹ 1 艋 I 2 n⫹1 艋 1
2n ⫹ 2
I 2n
is continuous. Find the value of x
65. Suppose that f , f ,
, and
4 and deduce that lim n ⬁ 2 n⫹1
l I I 2n
(d) Use part (c) and Exercises 45 and 46 to show that 66. (a) Use integration by parts to show that
2 2 4 4 6 6 2n 2n
n lim ⬁ 1 ⴢ
苷 3 5 f ⴢ l ⴢ 3 ⴢ 5 ⴢ 7 ⴢ⭈⭈⭈ⴢ 2n ⫺ 1 ⴢ 2n ⫹ 1
This formula is usually written as an infinite product: (b) If and are inverse functions and f t
is continuous,
prove that
b f 2 1 ⴢ 3 ⴢ 3 ⴢ 5 ⴢ 5 ⴢ 7 y ⴢ⭈⭈⭈
and is called the Wallis product. [Hint: Use part (a) and make the substitution y苷f .]
(e) We construct rectangles as follows. Start with a square of
area 1 and attach rectangles of area 1 alternately beside or b⬎a⬎ 0 , draw a diagram to give a geometric interpre-
(c) In the case where and are positive functions and f t
on top of the previous rectangle (see the figure). Find the tation of part (b).
limit of the ratios of width to height of these rectangles.
(d) Use part (b) to evaluate e x 1 ln x dx .
67. We arrived at Formula 6.3.2, V苷 b x a 2x f
, by using
cylindrical shells, but now we can use integration by parts to prove it using the slicing method of Section 6.2, at least for the case where is one-to-one and therefore has an inverse f function . Use the figure to show that t
2 2 V 苷 b d d ⫺ a c⫺
c dy
Make the substitution y苷f
and then use integration by
CHAPTER 7 TECHNIQUES OF INTEGRATION
7.2 TRIGONOMETRIC INTEGRALS In this section we use trigonometric identities to integrate certain combinations of trigo-
nometric functions. We start with powers of sine and cosine.
EXAMPLE 1 Evaluate . 3 y cos x dx
SOLUTION Simply substituting u苷 cos x isn’t helpful, since then du 苷 ⫺ 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 cos 2 x factor to an expression involving sine using the identity
sin 2 x⫹ cos 2 x苷 1 :
cos 3 x苷 cos 2 x ⴢ cos x 苷 共1 ⫺ sin 2 x 兲 cos x We can then evaluate the integral by substituting u苷 sin x , so du 苷 cos x dx and
y cos 3 x dx 苷 2 共1 ⫺ sin y 2 cos x ⴢ cos x dx 苷 y x 兲 cos x dx 苷 1 y 共1 ⫺ u
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 sin 2 x⫹ cos 2 x苷 1 enables us to convert back and forth between even powers
of sine and cosine.
5 Find . 2 y
V EXAMPLE 2
sin x cos x dx
SOLUTION
2 We could convert 2 cos x to 1 ⫺ sin x , 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 4 sin x factor in terms of cos x :
5 2 2 2 2 2 2 sin 2 x cos x苷 共sin x 兲 cos x sin x 苷 共1 ⫺ cos x 兲 cos x sin x
N Figure 1 shows the graphs of the integrand
5 2 Substituting u苷 cos x , we have du 苷 ⫺ sin x dx sin and so x cos x in Example 2 and its indefinite inte-
gral (with C苷 0 ). Which is which?
sin x cos x dx 苷 y 共sin x 兲 cos x sin x dx
2 2 2 2 4 6 苷 y 共1 ⫺ u 兲 u 共⫺du兲 苷 ⫺ y 共u ⫺ 2u ⫹ u 兲 du
⫺ 苷⫺ u 2 ⫹ u
⫹ _0.2 C
1 3 2 5 1 FIGURE 1 7 苷⫺
3 cos x⫹ 5 cos x⫺ 7 cos x⫹C
SECTION 7.2 TRIGONOMETRIC INTEGRALS
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 fol- lowing half-angle identities (see Equations 17b and 17a in Appendix D):
V EXAMPLE 3 Evaluate . 2 y
N Example 3 shows that the area of the region
0 sin x . dx
shown in Figure 2 is 兾2
SOLUTION If we write sin 2 x苷
2 x , the integral is no simpler to evaluate. Using the
1.5 half-angle formula for sin 2 x , however, we have
y=sin@ x
0 sin 2 x dx 苷 2 y 0 [ 2 ( 2 sin 2x ) ] 0
2 ( 2 ) 2 ( 2 sin 0 ) 2
Notice that we mentally made the substitution u苷 2x when integrating cos 2x . Another method for evaluating this integral was given in Exercise 43 in Section 7.1.
FIGURE 2
EXAMPLE 4 Find . sin 4 y x dx
SOLUTION We could evaluate this integral using the reduction formula for x sin n x dx (Equation 7.1.7) together with Example 3 (as in Exercise 43 in Section 7.1), but a better
method is to write sin 4 x苷 共sin 2 x 兲 2 and use a half-angle formula:
sin 4 x dx 苷
2 y 2 y 共sin x 兲 dx
苷 dx
4 y 2x
Since cos 2 2x occurs, we must use another half-angle formula cos 1 2 2x 苷
This gives
sin 4 x dx 苷 4 y 2
4 y ( 2 2 cos 4x ) dx
4 ( 2 8 sin 4x )
To summarize, we list guidelines to follow when evaluating integrals of the form sin m x
x n cos x dx , where and are 0 0 integers.
CHAPTER 7 TECHNIQUES OF INTEGRATION
y n sin x cos x dx
STRATEGY FOR EVALUATING
(a) If the power of cosine is odd
, save one cosine factor and use cos 2 x苷 1 ⫺ sin 2 x to express the remaining factors in terms of sine:
sin y m x cos 2k⫹1 x dx 苷 y sin m x 2 x k cos x dx
苷 y sin m x 2 x k cos x dx
Then substitute u苷 sin x . (b) If the power of sine is odd
, save one sine factor and use sin 2 x苷 1 ⫺ cos 2 x to express the remaining factors in terms of cosine:
y sin 2k⫹1 x cos n x dx 苷
y 2 x k cos n x sin x dx 苷 y 2 x k cos n x sin x dx
Then substitute u苷 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
sin 1 2 x苷 1 cos 2
2 x苷 2
It is sometimes helpful to use the identity
sin x cos x 苷 1
2 sin 2x
We can use a similar strategy to evaluate integrals of the form tan m sec n x x x dx . Since
2 x , we can separate a sec 2 x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec 2 x苷 1 ⫹ tan 2 x .
Or, since
, we can separate a sec x tan x factor and convert the
remaining (even) power of tangent to secant.
V EXAMPLE 5 Evaluate . tan 6 y x sec 4 x dx
SOLUTION If we separate one sec 2 x factor, we can express the remaining sec 2 x factor in terms of tangent using the identity sec 2 x苷 1 ⫹ tan 2 x . We can then evaluate the integral
by substituting u苷 tan x so that du 苷 sec 2 x dx :
tan x sec x dx 苷 y tan x sec x sec x dx
苷 tan 6 y x 2 x 2 x dx
苷 u 6 2 du 苷 6 ⫹u 8 y y du
⫹C
7 tan 7 x⫹ 1 9 9 tan x⫹C
SECTION 7.2 TRIGONOMETRIC INTEGRALS
EXAMPLE 6 Find . tan 5 sec y 7 d
SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with
a sec 5 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 tan 2 苷 sec 2 ⫺1 . We can then evaluate the integral by substituting u苷 sec , so du 苷 sec tan d :
tan 5 y 7 sec d 苷 y tan 4 sec 6 sec tan d
苷 2 ⫺1 2 sec y 6 sec tan d
苷 2 ⫺1 2 y 6 u du
苷 y 10 ⫺ 2u 8 ⫹u 6 du
The preceding examples demonstrate strategies for evaluating integrals of the form x tan m x sec n x dx for two cases, which we summarize here.
STRATEGY FOR EVALUATING
y m tan x sec n x dx
(a) If the power of secant is even , save a factor of sec 2 x and use sec 2 x苷 1 ⫹ tan 2 x to express the remaining factors in terms of tan x :
2k
2 k⫺ 1 y 2 tan x sec x dx 苷 y tan x x sec x dx
y m tan x 2 x k⫺ 1 sec 2 x dx
Then substitute u苷 tan x . (b) If the power of tangent is odd
, save a factor of sec x tan x and use tan 2 x苷 sec 2 x⫺ 1 to express the remaining factors in terms of sec x :
y tan 2k⫹1 x sec n x dx 苷 2 x k sec n⫺ y 1 x sec x tan x dx
苷 2 k y x⫺ 1 sec n⫺ 1 x sec x tan x dx
Then substitute u苷 sec x .
For other cases, the guidelines are not as clear-cut. We may need to use identities, inte- gration by parts, and occasionally a little ingenuity. We will sometimes need to be able to
CHAPTER 7 TECHNIQUES OF INTEGRATION
integrate tan x by using the formula established in (5.5.5):
y tan x dx 苷 ln ⱍ sec x ⱍ ⫹C
We will also need the indefinite integral of secant:
1 y sec x dx 苷 ln ⱍ sec x ⫹ tan x ⱍ ⫹C
We could verify Formula 1 by differentiating the right side, or as follows. First we multi- ply numerator and denominator by sec x ⫹ tan x :
y sec x ⫹ tan x
sec x dx 苷 y sec x
dx
sec x ⫹ tan x sec 2 x⫹ sec x tan x
苷 y dx
sec x ⫹ tan x
If we substitute u苷 sec x ⫹ tan x , then du 苷
共sec x tan x ⫹ sec 2 x 兲 dx , so the integral
becomes x 共1兾u兲 du 苷 ln ⱍ u ⱍ ⫹C . Thus we have y sec x dx 苷 ln ⱍ sec x ⫹ tan x ⱍ ⫹C
EXAMPLE 7 Find . y tan 3 x dx
SOLUTION Here only tan x occurs, so we use tan 2 x苷 sec 2 x⫺ 1 to rewrite a tan 2 x factor in
terms of sec 2 x :
y tan 3 x dx 苷 tan x tan 2 y x dx
y 2 共sec x⫺ 1 兲 dx
苷 tan x
苷 y tan x sec 2 x dx ⫺ y tan x dx
tan 2 x
⫺ ln ⱍ sec x
2 ⱍ ⫹C
In the first integral we mentally substituted u苷 tan x so that du 苷 sec 2 x 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 . sec 3 y x dx
SOLUTION Here we integrate by parts with
u 苷 sec x
d v 苷 sec 2 x dx
du 苷 sec x tan x dx
v 苷 tan x
SECTION 7.2 TRIGONOMETRIC INTEGRALS
y 2 sec 3 x dx 苷 sec x tan x ⫺ y sec x tan x dx
Then
y 2 sec x
苷 sec x tan x ⫺
共sec x⫺ 1 兲 dx
苷 sec x tan x ⫺ y sec 3 x dx ⫹ y sec x dx
Using Formula 1 and solving for the required integral, we get
sec 3 x dx 苷 2 ( sec x tan x ⫹ ln sec x ⫹ tan x ⱍ ⱍ ) ⫹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 cot m x csc n x dx can be found by similar methods because of the identity
1 ⫹ cot 2 x苷 csc 2 x .
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:
N These product identities are discussed in
(a) sin A cos B 苷 2 关sin共A ⫺ B兲 ⫹ sin共A ⫹ B兲兴
Appendix D.
(b) 1 sin A sin B 苷
2 关cos共A ⫺ B兲 ⫺ cos共A ⫹ B兲兴
(c) 1 cos A cos B 苷
2 关cos共A ⫺ B兲 ⫹ cos共A ⫹ B兲兴
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:
sin 4x cos 5x dx 苷 y 2 关sin共⫺x兲 ⫹ sin 9x兴 dx
2 ( cos x ⫺ 9 cos 9x
7.2 EXERCISES
1– 49 Evaluate the integral.
9. sin y 4
0 共3t兲 dt
10. y 6 0 cos d
1. y sin 3 x cos 2 x dx
2. y 3 sin 6 x cos x dx
11. y 共1 ⫹ cos 兲 d
y x cos x dx
3. y 兾2 sin x cos x dx
4. y
0 cos x dx
5. y 共x兲 cos 共x兲 dx
6. y sx dx
sin 3 兾2
13. sin sin 2 5 ( ) x cos 2 x dx
7. 5 y
8. sin y 2 共2兲 d
16. cos cos ssin ␣ 共sin 兲 d y
0 y d␣
d
CHAPTER 7 TECHNIQUES OF INTEGRATION
17. cos 2 x tan y 3 x dx
18. y cot 5 sin 4 d
sec y 4 54. y dx
53. sin 3x sin 6x dx
2 cos x ⫹ sin 2x
19. y dx
20. y cos 2 x sin 2x dx
sin x 55. Find the average value of the function f 共x兲 苷 sin 2 x cos 3 x on
21. y sec x tan x dx
22. y 0 sec 4 共t兾2兲 dt
the interval 关⫺, 兴 .
56. Evaluate x sin x cos x dx by four methods:
(a) the substitution u苷 y cos x
tan x dx
24. 共tan 2 y 4 x ⫹ tan x 兲 dx
(b) the substitution u苷 sin x
(c) the identity sin 2x 苷 2 sin x cos x
25. 兾4 y 4
26. y 0 sec tan 4 d
sec t dt
(d) integration by parts Explain the different appearances of the answers.
27. 兾3 y
0 tan 5 x sec 4 x dx
28. y tan 3 共2x兲 sec 5 共2x兲 dx
57–58 Find the area of the region bounded by the given curves.
5 6 57. y苷 sin x, y苷 cos x , ⫺ y 兾4 艋 x 艋 兾4 x sec x dx 30. y
29. tan 3 兾3
0 tan x sec x dx
58. y苷 sin 3 , x y苷 cos 3 x , 兾4 艋 x 艋 5兾4
31. y tan 5 x dx
32. y tan 6 共ay兲 dy
3 ; 59–60 tan Use a graph of the integrand to guess the value of the
integral. Then use the methods of this section to prove that your cos
33. y 4 d
34. y tan 2 x sec x dx
guess is correct.
35. y x sec x tan x dx
sin
36. y 59. 3 d y 0 cos x dx
60. y sin 2 x cos 5 x dx
cos
37. 兾2 2 y 3
38. y 兾4 cot x dx
cot x dx
61–64 Find the volume obtained by rotating the region bounded by the given curves about the specified axis.
39. cot 3 ␣ csc 3 40. csc y 4 ␣ d␣ y x cot 6 x dx
61. y苷 sin x , y苷 0 , 兾2 艋 x 艋 ; about the -axis x
3 62. sin x , y苷 0 , 0艋x艋 ; about the -axis 41. x y csc x dx 42. y
63. y苷 sin x , y苷 cos x , 0艋x艋 兾4 ; about y苷 1
43. y sin 8x cos 5x dx
44. y cos x cos 4x dx
64. y苷 sec x , y苷 cos x , 0艋x艋 兾3 ; about y苷⫺ 1
45. y sin 5 sin d
cos x ⫹ sin x
46. y dx
sin 2x
65. A particle moves on a straight line with velocity function 1 ⫺ tan v 2 x dx 共t兲 苷 sin t cos 2 t . Find its position function s苷f 共t兲
y if f sec 共0兲 苷 0. cos x ⫺ 1
47. y 2 dx
49. t sec 2 共t 2 兲 tan 4 y 2 共t 兲 dt
66. Household electricity is supplied in the form of alternating
current that varies from 155
V to ⫺155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation
50. If x 兾4 0 tan 6 x sec x dx 苷 I , express the value of
E 共t兲 苷 155 sin共120 t兲
0 tan x sec x dx in terms of . I where is the time in seconds. Voltmeters read the RMS t ; 51–54 Evaluate the indefinite integral. Illustrate, and check that
(root-mean-square) voltage, which is the square root of the
your answer is reasonable, by graphing both the integrand and its average value of 关E共t兲兴 over one cycle.
antiderivative (taking C苷 0 . 兲 (a) Calculate the RMS voltage of household current. (b) Many electric stoves require an RMS voltage of 220 V. A
51. y x sin 2
Find the corresponding amplitude 共x needed for the volt- 2 兲 dx
52. y sin 3 x cos 4 x dx
age . E
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
67–69 Prove the formula, where m and are positive integers. n 70. A finite Fourier series is given by the sum
67. N y
⫺ sin mx cos nx dx 苷 0
f 兺 a n sin nx
⫺ sin mx sin nx dx 苷 再 if m 苷 n
n苷 1
68. y
0 if m 苷 n
苷a 1 sin x ⫹ a 2 sin 2x ⫹ ⭈ ⭈ ⭈ ⫹ a N sin Nx
再 if m 苷 n
Show that the th coefficient m a m is given by the formula
69. y ⫺ cos mx cos nx dx 苷
0 if m 苷 n
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 x sa 2 ⫺x 2 dx , the substitution u苷a 2 ⫺x 2 would be effective
x sa ⫺x
but, as it stands,
2 dx is more difficult. If we change the variable from to by x the substitution x苷a sin , then the identity 1 ⫺ sin 2 苷 cos 2 allows us to get rid of the
root sign because
sa 2 ⫺x 2 苷 sa 2 ⫺a 2 sin 2 苷 sa 2 共1 ⫺ sin 2 兲 苷 sa 2 cos 2 苷 a ⱍ cos ⱍ
Notice the difference between the substitution u苷a 2 ⫺x 2 (in which the new variable is
a function of the old one) and the substitution x苷a sin (the old variable is a function of the new one).
In general we can make a substitution of the form x苷 t共t兲 by using the Substitution Rule in reverse. To make our calculations simpler, we assume that has an inverse func- t tion; that is, t is one-to-one. In this case, if we replace by and by in the Substitution u x x t Rule (Equation 5.5.4), we obtain
y f 共x兲 dx 苷 y f 共t共t兲兲t⬘共t兲 dt
This kind of substitution is called inverse substitution.
We can make the inverse substitution x苷a sin provided that it defines a one-to-one function. This can be accomplished by restricting to lie in the interval 关⫺兾2, 兾2兴 . 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 Section 1.6 in defining the inverse functions.)
TABLE OF TRIGONOMETRIC SUBSTITUTIONS
Expression
Substitution
Identity
sa 2 ⫺x 2
x苷a sin , ⫺ 艋 艋
1 ⫺ sin 2 2 2 2 苷 cos
sa 2
⫹x 2
x苷a tan , ⫺ ⬍⬍
1 ⫹ tan 2 2 2 2 苷 sec
sx 3
⫺a 2 x苷a sec , 0艋 ⬍
or 艋 ⬍
sec 2 2 2 2 ⫺ 1 苷 tan
CHAPTER 7 TECHNIQUES OF INTEGRATION
Evaluate . y
s9 ⫺ x
V EXAMPLE 1
dx
SOLUTION Let , where x苷 3 sin ⫺ 兾2 艋 艋 兾2 . Then dx 苷 3 cos d and
s9 ⫺ x s9 ⫺ 9 sin 2 苷 s9 cos 2 苷 3 cos ⱍ 苷 ⱍ 3 cos
(Note that cos 艌 0 because ⫺
.) Thus the Inverse Substitution Rule
gives
s9 ⫺ x 2 y 3 cos
x 2 y 9 sin 2
dx 苷 3 cos d
cos 2
y cot
⫺1 苷 ⫺cot ⫺ ⫹ C
Since this is an indefinite integral, we must return to the original variable . This can be x
3 done either by using trigonometric identities to express cot in terms of sin 苷 x or
by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right tri-
angle. Since sin 苷 x , we label the opposite side and the hypotenuse as having lengths x and . Then the Pythagorean Theorem gives the length of the adjacent side as 3 s9 ⫺ x 2 ,
œ„„„„„ 9-≈
so we can simply read the value of cot from the figure:
(Although ⬎0 in the diagram, this expression for cot is valid even when ⬍0 .) Since sin 苷 x , we have 苷 sin ⫺1
and so
s9 ⫺ x 2 s9 ⫺ x 2 ⫺1 y x
V EXAMPLE 2 Find the area enclosed by the ellipse x 2 y 2
2 ⫹ 2 苷 a 1 b SOLUTION Solving the equation of the ellipse for , we get y
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
y苷 sa 2 ⫺x 2 0艋x艋a
FIGURE 2
y 0 sa ⫺x a dx
and so
4 A苷
a@ b@
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
To evaluate this integral we substitute x苷a sin . Then dx 苷 a cos d . To change the limits of integration we note that when x苷 0 , sin 苷 0 , so 苷0 ; when x苷a ,
sin 苷 1 , so . 苷 兾2 Also
sa 2 ⫺x 2 苷 sa 2 ⫺a 2 sin 2 苷 sa 2 cos 2 苷 a ⱍ cos 苷 ⱍ a cos
since . 0艋艋 兾2 Therefore
b a 2 2 b 兾2
A苷4 y 0 sa ⫺x dx 苷 4 a cos ⴢ a cos d
兾2 苷 4ab 1 2 y
0 cos d 苷 4ab y 0 2 共1 ⫹ cos 2兲 d
苷 2ab [ ⫹ 2 sin 2 ] 0 苷 2ab ⫹0⫺0 苷 ab
We have shown that the area of an ellipse with semiaxes and is a b ab . In particular, taking a苷b苷r , we have proved the famous formula that the area of a circle with radius is . r 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
V EXAMPLE 3 1 Find . y 2 2
SOLUTION Let . x苷 2 tan , ⫺ 兾2 ⬍ ⬍ 兾2 Then dx 苷 2 sec 2 d and
sx 2 ⫹4 苷
2 s4 sec 苷 2 sec 苷 ⱍ 2 sec ⱍ
Thus we have
1 y sec
dx
2 sec 2 d
2 2 苷 y 苷 d
tan 2 To evaluate this trigonometric integral we put everything in terms of sin and cos :
Therefore, making the substitution u苷 sin , we have
1 y du
2 We use Figure 3 to determine that csc 苷
y x 2 2 ⫹4 sx
dx
苷 ⫺ sx
⫹C
4x
CHAPTER 7 TECHNIQUES OF INTEGRATION
EXAMPLE 4 Find . y
sx 2 ⫹4 dx
SOLUTION It would be possible to use the trigonometric substitution x苷 2 tan here (as in Example 3). But the direct substitution u苷x 2 ⫹4 is simpler, because then du 苷 2x dx
and
y 苷 su ⫹C苷 sx ⫹4 ⫹C
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.
EXAMPLE 5 Evaluate , where y 2
SOLUTION 1 We let x苷a sec , where 0 ⬍⬍ 兾2 or ⬍ ⬍ 3 兾2 . Then
dx 苷 a sec tan d and
2 2 sx sa 共sec ⫺1 兲 苷
2 ⫺a 2 苷
2 tan sa 2 苷 a ⱍ 苷 tan ⱍ a tan
Therefore
a y sec tan
dx
2 2 苷 y d
sx ⫺a
a tan
苷 y sec d 苷 ln sec ⫹ tan ⫹C ⱍ ⱍ
x œ„„„„„ ≈-a@
The triangle in Figure 4 gives tan 苷
2 ⫺a sx 2 兾a , so we have
dx
y 2 2 苷 ln ⫹
2 ⫺a ⱍ 2 sx ⫺ ln a ⫹ C ⱍ
Writing , we C 1 苷 C⫺ ln a have
dx
1 2 y 2 2 2 苷 ln x⫹ sx ⫺a ⫹C 1
sx ⫺a
SOLUTION 2 For x⬎ 0 the hyperbolic substitution x苷a cosh t can also be used. Using the
2 identity , we 2 cosh y⫺ sinh y苷 1 have
2 ⫺a 2 苷
sx 2 sa t⫺ 1 sa sinh 2 t 苷 a sinh t
Since , we dx 苷 a sinh t dt obtain
y dx
a sinh t dt
⫺a 2 y a sinh t y
苷 sx dt 苷 t ⫹ C
Since , we cosh t 苷 x have t苷 cosh ⫺1
and
dx
2 y 2 2 苷 cosh ⫺1 ⫹C
sx ⫺a
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
Although Formulas 1 and 2 look quite different, they are actually equivalent by Formula 3.11.4.
NOTE As Example 5 illustrates, hyperbolic substitutions can be used in place of trigo- nometric substitutions and sometimes they lead to simpler answers. But we usually use trigonometric substitutions because trigonometric identities are more familiar than hyper- bolic identities.
3 x 3 EXAMPLE 6 Find . y 0 2
dx
SOLUTION First we note that
2 ⫹9 3 苷 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-
2 tan , which gives dx 苷
tion 3 u苷 2x . When we combine this with the tangent substitution, we have x苷
2 sec d and
s4x 2 ⫹9 苷 2 ⫹9 s9 tan 苷 3 sec
When , x苷 0 tan 苷 0 , so ; 苷0 when x苷 3 s3 , tan 苷 s3 , so 苷 .
3 3 x 27 8 tan 3 3 y 2
0 2 ⫹9 3 dx 苷 y 0 27 sec 3 2 sec d
tan 3 3
sin 3 3
苷 16 y 0 d 苷 16 y 0 d
sec
cos 2
苷 16 y 0
Now we substitute u苷 cos so that du 苷 ⫺ sin d . When 苷0 , u苷 1 ; when 苷 1
2 . Therefore
dx 苷 ⫺ du 苷 ⫺2 du y 0 2
⫹9 3 16 y 1
u 16 2 y 1
3 x 3 1 1⫺u 3 2 3 1
u⫹
] 苷 32 M
EXAMPLE 7 Evaluate . y 2 dx
s3 ⫺ 2x ⫺ x SOLUTION We can transform the integrand into a function for which trigonometric substitu-
tion is appropriate by first completing the square under the root sign:
This suggests that we make the substitution u苷x⫹ 1 . Then du 苷 dx and x苷u⫺ 1 , so
u⫺ y 1
2 dx 苷 y
s4 ⫺ u 2 s3 ⫺ 2x ⫺ x du
CHAPTER 7 TECHNIQUES OF INTEGRATION
N Figure 5 shows the graphs of the integrand
We now substitute u苷 2 sin , giving du 苷 2 cos d and s4 ⫺ u 2 苷 2 cos , so
in Example 7 and its indefinite integral (with C苷 0 ). Which is which?
y 2 sin ⫺ 1
s3 ⫺ 2x ⫺ x dx 苷 2 y 2 cos d
7.3 EXERCISES
1–3 Evaluate the integral using the indicated trigonometric sub- 0.6 x 2 1 2 stitution. Sketch and label the associated right triangle.
21. y 0
22. y 0 sx ⫹ 1 dx
s9 ⫺ 25x 2 dx
y dt
24. sx y
x 2 2 ⫺ 9 dx ; x苷 3 sec
23. s5 ⫹ 4x ⫺ x y 2 dx
st 2 ⫺ 6t ⫹ 13
2. y x 3 s9 ⫺ x 2 dx ; x苷 3 sin
25. y
x⫹ 1 y
28. y 2 2 dx y dx ; x苷 3 tan
27. y sx 2 ⫹ 2x dx
共x ⫺ 2x ⫹ 2 sx 兲 2 ⫹ 9
29. y x s1 ⫺ x dx
30. y 0
cos t
s1 ⫹ sin 2 t dt 4 –30 Evaluate the integral.
2 s3
31. (a) Use trigonometric substitution to show that s16 ⫺ x
4. y 0 2 dx
2 ⫹ 2 5. ⫹ y
6. y 1 dx
2 1 2 sx 2 ⫺ 1 dx 苷
y sx 2 ⫹ 2 ln ( x⫹ sx a ) C
t st
s2 3 2 ⫺ dt
(b) Use the hyperbolic substitution x苷a sinh t y to show that
x 2 s25 ⫺ x 2 y sx 2 ⫹ 100
7. dx
8. dx
16 2 sx 2 ⫹ a 2 冉 a 冊
dx
9. y 2 ⫹ 10. y 2 ⫹ dt
t 5 dx
sx ⫹ st y sinh 1 C
y These formulas are connected by Formula 3.11.3. y
2 11. 1 s1 ⫺ 4x dx 12. x sx 2 ⫹ 4 dx
13. y 3 dx
14. y
sx 2 ⫺ 9 du
32. Evaluate
s5 ⫺ u x 2 y 2
2 ⫹ 共x dx a 2 兲 3 兾2
y dx
16. y s2兾3
0 dx
15. a x 2 sa 2 ⫺ x 2 2 兾3
x 5 ⫺ s9x 2 1 (a) by trigonometric substitution. (b) by the hyperbolic substitution x苷a sinh t .
17. y
dx
7 y 关共ax兲 2 ⫺ b 2 兴 3 兾2
33. Find the average value of f 共x兲 苷 sx 2 ⫺ 1 兾x , 1艋x艋7 .
34. Find the area of the region bounded by the hyperbola x
s1 ⫹ x 2 19. t y dx 20. y
s25 ⫺ t
2 dt
9x 2 ⫺ 4y 2 苷 36 and the line x苷 3 .
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
39. (a) Use trigonometric substitution to verify that a circle with radius and central angle . [Hint: Assume r 0⬍⬍ 兾2 and place the center of the circle at the origin
35. Prove the formula A苷 1 2 r 2 for the area of a sector of
2 2 2 y 0 sa
2 ⫺ t 2 dt 苷 1 a 2 sin 2 ⫺ 1 共x兾a兲 ⫹ 1 2 x sa 2 ⫺ x 2
so it has the equation x ⫹ y 苷 r . Then A is the sum of the
area of the triangle POQ and the area of the region PQR in (b) Use the figure to give trigonometric interpretations of the figure.] both terms on the right side of the equation in part (a).
a y=œ„„„„„ a@-t@ ¨
36. Evaluate the integral
y dx
x 4 sx 2 ⫺ 2 40. The parabola y苷 1 2 x 2 divides the disk x 2 ⫹ y 2 艋 8 into two Graph the integrand and its indefinite integral on the same
parts. Find the areas of both parts.
screen and check that your answer is reasonable. 41. Find the area of the crescent-shaped region (called a lune)
; 37. Use a graph to approximate the roots of the equation bounded by arcs of circles with radii and . (See the figure.) r R x 2 s4 ⫺ x 2 苷 2⫺x . Then approximate the area bounded by
the curve y苷x 2 s4 ⫺ x 2 and the line y苷 2⫺x .
38. A charged rod of length produces an electric field at point L r P 共a, b兲 given by
where is the charge density per unit length on the rod and 0 is the free space permittivity (see the figure). Evaluate the
42. A water storage tank has the shape of a cylinder with diam- y eter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage
integral to determine an expression for the electric field E 共P兲 .
P (a, b)
of the total capacity is being used?
43. A torus is generated by rotating the circle x 2 ⫹ 共 y ⫺ R兲 2 苷 r 2 about the -axis. Find the volume x
enclosed by the torus.
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 2 兾共x ⫺ 1兲
and 1 兾共x ⫹ 2兲 to a common denominator we obtain
2 ⫺ 1 苷 2 共x ⫹ 2兲 ⫺ 共x ⫺ 1兲 苷
x⫹ 5
x 2 ⫹ x⫺ 2 If we now reverse the procedure, we see how to integrate the function on the right side of
x⫺ 1 x⫹ 2 共x ⫺ 1兲共x ⫹ 2兲
CHAPTER 7 TECHNIQUES OF INTEGRATION
this equation:
x⫺ 2 y 冉 x⫺ 1 x⫹ 2 冊
x⫹ 5 2 y 1
dx 苷 ⫺
dx
苷 2 ln ⱍ x⫺ 1 ⱍ ⫺ ln ⱍ x⫹ 2 ⱍ ⫹ C
To see how the method of partial fractions works in general, let’s consider a rational function P 共x兲
f 共x兲 苷 Q 共x兲
where P and Q are polynomials. It’s possible to express as a sum of simpler fractions f provided that the degree of is less than the degree of . Such a rational function is called P Q proper. Recall that if
P 共x兲 苷 a n x n ⫹ a n⫺ 1 x n⫺ 1 ⫹⭈⭈⭈⫹ a 1 x⫹a 0
where a n 苷 0 , then the degree of is and we write P n deg 共P兲 苷 n . If f is improper, that is, deg 共P兲 艌 deg共Q兲 , then we must take the preliminary step of dividing Q into P (by long division) until a remainder R 共x兲 is obtained such that deg 共R兲 ⬍ deg共Q兲 . The division statement is
where and are also polynomials. S R
As the following example illustrates, sometimes this preliminary step is all that is required.
V x 3 EXAMPLE 1 x Find . y dx
x⫺ 1
≈+x +2 SOLUTION Since the degree of the numerator is greater than the degree of the denominator, x-1 ) ˛
+x
we first perform the long division. This enables us to write
˛-≈
x⫺ 1 y 冉
x⫺ 1 冊
2x x 3 x 2x-2 2
2 ⫹ 2x ⫹ 2 ln ⱍ x⫺ 1 ⱍ C M
The next step is to factor the denominator Q 共x兲 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 ax 2 ⫹ bx ⫹ c , where b 2 ⫺ 4ac ⬍ 0 ). For
instance, if Q
共x兲 苷 x 4 ⫺ 16 , we could factor it as
2 2 共x兲 苷 共x 2 ⫺ 4 兲共x ⫹ 4 兲 苷 共x ⫺ 2兲共x ⫹ 2兲共x ⫹ 4 兲 The third step is to express the proper rational function R 共x兲兾Q共x兲 (from Equation 1) as
a sum of partial fractions of the form
A Ax ⫹ B
or
共ax ⫹ b兲 2 共ax ⫹ bx ⫹ c j 兲
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
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 N The denominator Q(x) is a product of distinct linear factors.
This means that we can write
Q 共x兲 苷 共a 1 x⫹b 1 兲共a 2 x⫹b 2 兲 ⭈ ⭈ ⭈ 共a k x⫹b k 兲 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 A 1 ,A 2 , ..., A k such that
These constants can be determined as in the following example.
EXAMPLE 2 Evaluate . y
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 苷 x 2 共2x ⫹ 3x ⫺ 2 兲 苷 x共2x ⫺ 1兲共x ⫹ 2兲 Since the denominator has three distinct linear factors, the partial fraction decomposition
of the integrand (2) has the form
x 2 ⫹ 2x ⫺ 1
x 共2x ⫺ 1兲共x ⫹ 2兲
x⫹ 2 N Another method for finding , , and A B C To determine the values of , , and , we multiply both sides of this equation by the A B C
2x ⫺ 1
is given in the note after this example.
product of the denominators, x 共2x ⫺ 1兲共x ⫹ 2兲 , obtaining
4 x 2 ⫹ 2x ⫺ 1 苷 A 共2x ⫺ 1兲共x ⫹ 2兲 ⫹ Bx共x ⫹ 2兲 ⫹ Cx共2x ⫺ 1兲 Expanding the right side of Equation 4 and writing it in the standard form for polyno-
mials, we get
共2A ⫹ B ⫹ 2C兲x 2 ⫹ 共3A ⫹ 2B ⫺ C兲x ⫺ 2A The polynomials in Equation 5 are identical, so their coefficients must be equal. The
5 x 2 ⫹ 2x ⫺ 1 苷
coefficient of x 2 on the right side, 2A ⫹ B ⫹ 2C , must equal the coefficient of x 2 on the left side—namely, 1. Likewise, the coefficients of are equal and the constant terms are x equal. This gives the following system of equations for , , and : A B C
2A ⫹ B ⫹ 2C 苷 1 3A ⫹ 2B ⫺ C 苷 2
⫺2A ⫹ 2B ⫺ 2C 苷 ⫺1
CHAPTER 7 TECHNIQUES OF INTEGRATION
1 1 Solving, we get 1 A苷
2 , B苷 5 , and C苷⫺ 10 , and so
y 2x dx 3 ⫹ 3x 2 ⫺ 2x y 冉 2 x 5 2x ⫺ 1 10 x⫹ 2 冊
N We could check our work by taking the terms
x 2 ⫹ 2x ⫺ 1
to a common denominator and adding them.
dx 苷 ⫹
N Figure 1 shows the graphs of the integrand
2 ln x ⫹ 10 ln 2x ⫺ 1 ⫺ 10 ln x⫹ 2 ⫹
in Example 2 and its indefinite integral (with
K苷 0 ). Which is which?
In integrating the middle term we have made the mental substitution u苷 2x ⫺ 1 , which
2 gives and . du 苷 2 dx dx 苷 du 兾2
NOTE We can use an alternative method to find the coefficients , A B , and C in
3 Example 2. Equation 4 is an identity; it is true for every value of . Let’s choose values of x x that simplify the equation. If we put x苷 0 in Equation 4, then the second and third terms
on the right side vanish and the equation then becomes ⫺ 2A 苷 ⫺1 , or A苷 1 2 . Likewise,
x苷 1 2 gives 5B 兾4 苷 1 1 4 1 and x苷⫺ 2 gives 10C 苷 ⫺1 , so B苷 5 and C苷⫺ 10 . (You may object that Equation 3 is not valid for x苷 0 , , or 1 2 ⫺ 2 , so why should Equation 4 be valid for those
FIGURE 1
values? In fact, Equation 4 is true for all values of , even x x苷 0 , , and 1 2 ⫺ 2 . See Exercise 69
for the reason.)
dx
EXAMPLE 3 Find , where y ⫺ a苷 0 .
SOLUTION The method of partial fractions gives
x ⫺ a 2 共x ⫺ a兲共x ⫹ a兲
x⫺a
x⫹a
and therefore
A 共x ⫹ a兲 ⫹ B共x ⫺ a兲 苷 1
Using the method of the preceding note, we put x苷a in this equation and get
A 共2a兲 苷 1 , so A苷 1 兾共2a兲 . If we put x 苷 ⫺a , we get B 共⫺2a兲 苷 1 , so B苷⫺ 1 兾共2a兲 .
Thus
a 2 2a y 冉 x⫺a x⫹a 冊
dx
dx
( ln
2a ⱍ x⫺a ⱍ ln ⱍ x⫹a ⱍ ) C
Since ln x ⫺ ln y 苷 ln 共x兾y兲 , we can write the integral as
x ⫺ a 2a 冟 x⫹a 冟
dx
6 y 2 2 苷 ln
1 x⫺a
See Exercises 55–56 for ways of using Formula 6.
CASE 11 N Q(x) is a product of linear factors, some of which are repeated.
Suppose the first linear factor 共a 1 x⫹b 1 兲 is repeated times; that is, r 共a 1 x⫹b 1 兲 r occurs in the factorization of Q 共x兲 . Then instead of the single term A 1 兾共a 1 x⫹b 1 兲 in Equation 2, we
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
would use
By way of illustration, we could write
x 3 ⫺ x⫹ 1 苷 A ⫹ B ⫹ C ⫹
x 2 x⫺ 1 2 共x ⫺ 1兲 3 共x ⫺ 1兲 共x ⫺ 1兲
but we prefer to work out in detail a simpler example.
4 x 2 ⫺ 2x ⫹ 4x ⫹ 1
EXAMPLE 4 Find . y 3
dx
x⫹ 1
SOLUTION The first step is to divide. The result of long division is
3 The second step is to factor the denominator 2 Q 共x兲 苷 x ⫺ x ⫺ x⫹ 1 . Since Q 共1兲 苷 0 ,
we know that x⫺ 1 is a factor and we obtain
x⫹ 2 1苷 共x ⫺ 1兲共x ⫺ 1 兲 苷 共x ⫺ 1兲共x ⫺ 1兲共x ⫹ 1兲
苷 2 共x ⫺ 1兲 共x ⫹ 1兲 Since the linear factor x⫺ 1 occurs twice, the partial fraction decomposition is
4x
x⫹ 1 Multiplying by the least common denominator, 2 共x ⫺ 1兲 共x ⫹ 1兲 , we get
共x ⫺ 1兲 共x ⫹ 1兲
x⫺ 1 共x ⫺ 1兲
4x 苷 A 2 共x ⫺ 1兲共x ⫹ 1兲 ⫹ B共x ⫹ 1兲 ⫹ C共x ⫺ 1兲 苷 2 共A ⫹ C兲x ⫹ 共B ⫺ 2C兲x ⫹ 共⫺A ⫹ B ⫹ C兲
N Another method for finding the coefficients:
Now we equate coefficients:
Put in x苷 1 (8): B苷 2 . Put : x苷⫺ 1 C苷⫺ 1 .
A ⫹ B ⫹ C苷0
Put : x苷 0 A苷B⫹C苷 1 .
A⫺ B ⫺ 2C 苷 4 ⫺A ⫹ B ⫹ C 苷 0 Solving, we obtain A苷 1 , B苷 2 , and C苷⫺ 1 , so
x⫹ 1 y
冋 x⫺ 1 共x ⫺ 1兲 x⫹ 1 册
2 ⱍ x⫺ 1 ⱍ
x⫺ 1 ⱍ ⱍ
ln x⫹ 1 K
2 x⫺ 1 冟 x⫹ 1 冟
x⫺ 1
x⫺
ln
CHAPTER 7 TECHNIQUES OF INTEGRATION
CASE III N Q(x) contains irreducible quadratic factors, none of which is repeated.
If Q 共x兲 has the factor ax 2 ⫹ bx⫹c , where b 2 ⫺ 4ac ⬍ 0 , then, in addition to the partial fractions in Equations 2 and 7, the expression for R 共x兲兾Q共x兲 will have a term of the form
Ax ⫹ B
ax 2 ⫹ bx ⫹ c
where A and B are constants to be determined. For instance, the function given by
f 共x兲 苷 x兾关共x ⫺ 2兲共x 2 ⫹ 1 兲共x 2 ⫹ 4 兲兴 has a partial fraction decomposition of the form
A ⫹ ⫹ 4 x⫺ 2 x 2 ⫹ 1 x 共x ⫺ 2兲共x 2 兲共x 兲 ⫹ 4
⫹ Bx ⫹ C
Dx ⫹ E
The term given in (9) can be integrated by completing the square and using the formula
dx
10 ⫺ 苷 tan 1
EXAMPLE 5 Evaluate . y
SOLUTION Since x 3 ⫹ 4x 苷 x 2 共x ⫹ 4 兲 can’t be factored further, we write
2x 2 ⫺ x⫹ 4 A Bx ⫹ C
x 共x 2 兲 ⫹ 4 Multiplying by x 2 共x ⫹ 4 兲 , we have
2x 2 ⫺ x⫹ 4苷A 2 共x ⫹ 4 兲 ⫹ 共Bx ⫹ C兲x
共A ⫹ B兲x 2 ⫹ Cx ⫹ 4A
Equating coefficients, we obtain
A⫹B苷 2 C苷⫺ 1 4A 苷 4
Thus , A苷 1 B苷 1 , and C苷⫺ 1 and so
2x 2 ⫺ x⫹ 4 1 x⫺ y 1
In order to integrate the second term we split it into two parts:
x⫺ 1 x
dx 苷 y dx ⫺
dx
We make the substitution u苷x 2 ⫹ 4 in the first of these integrals so that du 苷 2x dx . We evaluate the second integral by means of Formula 10 with a苷 2 :
y x 2 ⫹ dx ⫺ 4 y x 2 ⫹ 共x dx 兲 4
1 1 苷 ln ⫺ ⱍ x ⫹ 1 ⱍ
2 ln 2 共x ⫹ 4 兲⫺ 2 tan 共x兾2兲 ⫹ K
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
4x 2 ⫺ 3x ⫹ 2 EXAMPLE 6 Evaluate . y 2 ⫺ dx
4x
4x ⫹ 3
SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain
Notice that the quadratic 2 4x ⫺ 4x ⫹ 3 is irreducible because its discriminant is
b 2 ⫺ 4ac 苷 ⫺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 苷 2 共2x ⫺ 1兲 ⫹ 2
This suggests that we make the substitution 1 u苷 2x ⫺ 1 . Then, du 苷 2 dx and x苷 2 共u ⫹ 1兲 , so
y 冉 4x ⫺ 4x ⫹ 3 冊
x⫺ y 1
苷x⫹ 2 y 2 ⫹
1 u⫺ 1
du 苷 x ⫹ 4 y 2
du
苷x⫹ 4 y 2 ⫹ du ⫺ 4 y 2
du
冉 s2 冊
冉 s2 冊
NOTE Example 6 illustrates the general procedure for integrating a partial fraction of the form
Ax ⫹ B
2 where b ⫺
We complete the square in the denominator and then make a substitution that brings the integral into the form
Cu ⫹ D
y 2 2 du ⫹ D 2 2 a du u ⫹ a y u ⫹ a
2 ⫹ 2 du 苷 C
⫺ Then the first integral is a logarithm and the second is expressed in terms of 1 tan .
CASE IV N Q(x) contains a repeated irreducible quadratic factor.
If 2 Q 共x兲 has the factor 共ax ⫹ bx ⫹ c 兲 , where b ⫺ 4ac ⬍ 0 , then instead of the single partial fraction (9), the sum
A 1 x⫹B 1 A ⫹ 2 x⫹B 2 ⫹⭈⭈⭈⫹
A r x⫹B r
2 ax 2 共ax bx ⫹ c 兲 共ax ⫹ bx ⫹ c r bx ⫹ c 兲
CHAPTER 7 TECHNIQUES OF INTEGRATION
occurs in the partial fraction decomposition of R 共x兲兾Q共x兲 . Each of the terms in (11) can be integrated by first completing the square.
N 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,
however, can find the numerical values very
quickly. For instance, the Maple command 2 x 共x ⫺ 1兲共x ⫹
x⫹ 2 1 ⫹ 兲共x 3 1 兲
convert 共f, parfrac, x兲
SOLUTION
or the Mathematica command Apart[f]
gives the following values:
x 共x ⫺ 1兲共x
E苷 15 1 8 3 , F苷⫺ 8 , G苷H苷 4 ,
EXAMPLE 8 Evaluate . y 2 2 dx
x 共x ⫹ 1 兲 SOLUTION The form of the partial fraction decomposition is
2 ⫹ Multiplying by 2 x 共x 1 兲 , we have
3 ⫹ 2 ⫹ 2 ⫺x 2 2x ⫺ 2 x⫹ 1苷A 共x 1 兲 ⫹ 共Bx ⫹ C兲x共x ⫹ 1 兲 ⫹ 共Dx ⫹ E兲x
4 ⫹ 2 3 ⫹ 苷A 2 共x 2x 1 兲 ⫹ B共x x 兲 ⫹ C共x x 兲 ⫹ Dx ⫹ Ex
4 3 苷 2 共A ⫹ B兲x ⫹ Cx ⫹ 共2A ⫹ B ⫹ D兲x ⫹ 共C ⫹ E兲x ⫹ A
If we equate coefficients, we get the system
A⫹B苷 0 C苷⫺ 1 2A ⫹ B ⫹ D 苷 2
C⫹E苷⫺ 1 A苷 1 which has the solution A苷 1 , B苷⫺ 1 , C苷⫺ 1 , D苷 1 , and E苷 0 . Thus
y 冉 x x ⫹ 1 共x ⫹ 1 兲 冊
1 x⫹ 1 y x
y x ⫹ dx ⫺ 1 y x 2 ⫹ 1 y 2 ⫹ 共x 2 1 兲
N In the second and fourth terms we made the
苷 ln ⱍ x ⱍ 2 ln 共x 1 兲 ⫺ tan x⫺ 2 ⫹ ⫹ K M
mental substitution u苷x ⫹ 1 .
1 兲 We note that sometimes partial fractions can be avoided when integrating a rational func-
2 共x
tion. For instance, although the integral
2 dx x 共x ⫹ 3 兲
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
could be evaluated by the method of Case III, it’s much easier to observe that if u苷x 2 ⫹ 3 3 ⫹ 3x , then and du 苷 共x 2 兲苷x 共3x ⫹ 3 兲 dx so
dx 苷 3 ln x 3 ⫹ 3x ⫹ C
x 共x
RATIONALIZING SUBSTITUTIONS Some nonrational functions can be changed into rational functions by means of appropri-
ate substitutions. In particular, when an integrand contains an expression of the form
s n t共x兲 , then the substitution u苷 s t共x兲 may be effective. Other instances appear in the exercises.
sx ⫹ 4
EXAMPLE 9 Evaluate . y dx
SOLUTION Let . u苷 sx ⫹ 4 Then u 2 苷 x⫹ 4 , so and x苷u 2 ⫺ 4 dx 苷 2u du .
Therefore
sx ⫹ 4 2 u y u
y u du 2 ⫺ 4 y u 2 ⫺ 4
We can evaluate this integral either by factoring u 2 ⫺ 4 as 共u ⫺ 2兲共u ⫹ 2兲 and using
partial fractions or by using Formula 6 with a苷 2 :
y du
sx ⫹ 4
dx 苷 2 y du ⫹ 8
2ⴢ2 冟 u⫹ 2 冟
7.4 EXERCISES
1–6 Write out the form of the partial fraction decomposition of the x 4 t 4 ⫹ t 2 ⫹ 1 function (as in Example 7). Do not determine the numerical values
1 共t 2 ⫹ 1 兲共t 2 ⫹ 4 兲 2 of the coefficients.
x 6 ⫺ x 3 共x ⫹ 3兲共3x ⫹ 1兲
x⫺ 2 x 2 ⫹ x⫹ 2 7–38 Evaluate the integral.
2 2 7. y dx
8. y dr
3 2 ⫹ 2 9. y dx
y 共t ⫹ 4兲共t ⫺ 1兲
共x ⫹ 5兲共x ⫺ 2兲
CHAPTER 7 TECHNIQUES OF INTEGRATION
11. y 2 2 dx
12. y 0 dx
x 2 47. y e 2x
13. y 2 dx
1 cos x
14. dx
y sin 2 sin x
48. dx
15. y 3 3 2 dx
16. y 0 2 dx
49. y
4 x 3 2 1 x 2 3 sec t
tan 2 x dt x
17. y dy
2 4y 2 2 e x x
18. y 3 dx
50. y x
2x
dx
19. y 2 dx
20. y 2 dx