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Unformatted text preview: 392  CHAPTER 5 INTEGRALS TABLE OF INDEFINITE INTEGRALS 1 y cf c y f x dx x dx y k dx kx n dx xn 1
n1 ye x dx ex 2 1
2 1 y sinh x d x y 1 1
dx
x ya tan x dx n C y sec x tan x d x
yx C cos x y sec x d x fx yf t x dx x dx ytx dx C yx y sin x d x y C sin x y csc x d x
2 C
sec x cosh x ax
ln a dx C y cos x d x C tan 1x x ln x C cot x y csc x cot x d x C y C 1
s1 x2 csc x dx y cosh x d x C C sin 1x sinh x C
C C Recall from Theorem 4.9.1 that the most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative. We adopt the convention that
when a formula for a general indeﬁnite integral is given, it is valid only on an interval. Thus we write
1
1
C
y x 2 dx
x
with the understanding that it is valid on the interval 0,
or on the interval
, 0 . This
is true despite the fact that the general antiderivative of the function f x
1 x 2, x 0, is
1
x
1
x Fx
The indeﬁnite integral in Example 1 is graphed
in Figure 1 for several values of C. The value of
C is the yintercept. C1 if x 0 C2 if x 0 N EXAMPLE 1 Find the general indeﬁnite integral y 4 10 x 4 2 sec 2x d x S O L U T I O N Using our convention and Table 1, we have
_1.5 1.5 _4 FIGURE 1 y 10 x 4 2 sec2x d x 10 y x 4 d x
10 x5
5 2 y sec2x d x 2 tan x You should check this answer by differentiating it. C 2x 5 2 tan x C
M 392  CHAPTER 5 INTEGRALS TABLE OF INDEFINITE INTEGRALS 1 y cf c y f x dx x dx y k dx kx n dx xn 1
n1 ye x dx ex 2 1
2 1 y sinh x d x y 1 1
dx
x ya tan x dx n C y sec x tan x d x
yx C cos x y sec x d x fx yf t x dx x dx ytx dx C yx y sin x d x y C sin x y csc x d x
2 C
sec x cosh x ax
ln a dx C y cos x d x C tan 1x x ln x C cot x y csc x cot x d x C y C 1
s1 x2 csc x dx y cosh x d x C C sin 1x sinh x C
C C Recall from Theorem 4.9.1 that the most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative. We adopt the convention that
when a formula for a general indeﬁnite integral is given, it is valid only on an interval. Thus we write
1
1
C
y x 2 dx
x
with the understanding that it is valid on the interval 0,
or on the interval
, 0 . This
is true despite the fact that the general antiderivative of the function f x
1 x 2, x 0, is
1
x
1
x Fx
The indeﬁnite integral in Example 1 is graphed
in Figure 1 for several values of C. The value of
C is the yintercept. C1 if x 0 C2 if x 0 N EXAMPLE 1 Find the general indeﬁnite integral y 4 10 x 4 2 sec 2x d x S O L U T I O N Using our convention and Table 1, we have
_1.5 1.5 _4 FIGURE 1 y 10 x 4 2 sec2x d x 10 y x 4 d x
10 x5
5 2 y sec2x d x 2 tan x You should check this answer by differentiating it. C 2x 5 2 tan x C
M S E C T I O N 5 . 5 T H E S U B S T I T U T I O N RU L E  401 because, by the Chain Rule,
d
Ftx
dx F tx t x
t x , then from Equation 3 If we make the “change of variable” or “substitution” u
we have yF
or, writing F t x t x dx Ftx C Fu C yF u du f , we get yf yf t x t x dx u du Thus we have proved the following rule.
4 THE SUBSTITUTION RULE If u
t x is a differentiable function whose range
is an interval I and f is continuous on I , then yf yf t x t x dx u du Notice that the Substitution Rule for integration was proved using the Chain Rule for
differentiation. Notice also that if u t x , then du t x d x, so a way to remember the
Substitution Rule is to think of d x and du in (4) as differentials.
Thus the Substitution Rule says: It is permissible to operate with d x and du after
integral signs as if they were differentials. EXAMPLE 1 Find yx 3 cos x 4 2 d x. x 4 2 because its differential is du 4 x 3 d x,
which, apart from the constant factor 4, occurs in the integral. Thus, using x 3 d x du 4
and the Substitution Rule, we have S O L U T I O N We make the substitution u yx cos x 4 2 dx y cos u 1
4 1
4 3 C 1
4
N Check the answer by differentiating it. sin u
sin x 4 du 2 1
4 y cos u du C Notice that at the ﬁnal stage we had to return to the original variable x. M The idea behind the Substitution Rule is to replace a relatively complicated integral
by a simpler integral. This is accomplished by changing from the original variable x to
a new variable u that is a function of x. Thus, in Example 1, we replaced the integral
x x 3 cos x 4 2 d x by the simpler integral 1 x cos u du.
4
The main challenge in using the Substitution Rule is to think of an appropriate substitution. You should try to choose u to be some function in the integrand whose differential
also occurs (except for a constant factor). This was the case in Example 1. If that is not 404  CHAPTER 5 INTEGRALS This rule says that when using a substitution
in a deﬁnite integral, we must put everything in
terms of the new variable u, not only x and d x
but also the limits of integration. The new limits
of integration are the values of u that correspond
to x a and x b. THE SUBSTITUTION RULE FOR DEFINITE INTEGRALS If t is continuous on 6 N t x , then a, b and f is continuous on the range of u y b a y f t x t x dx tb ta f u du P R O O F Let F be an antiderivative of f . Then, by (3), F t x is an antiderivative of f t x t x , so by Part 2 of the Fundamental Theorem, we have y b a f t x t x dx Ftx b
a F tb F ta But, applying FTC2 a second time, we also have y tb ta EXAMPLE 7 Evaluate y 4 0 f u du s2 x Fu tb
ta F tb F ta M 1 d x using (6). S O L U T I O N Using the substitution from Solution 1 of Example 2, we have u dx when x 0, u 20 y Therefore
The geometric interpretation of Example 7 is
shown in Figure 2. The substitution u 2 x 1
stretches the interval 0, 4 by a factor of 2 and
translates it to the right by 1 unit. The Substitution Rule shows that the two areas are equal. 4 0 1 s2 x 1
1 dx 2 1 su du 93 2 13 2 1
2 2
3 24 1 9 u3 2 9
1 26
3 M œ„
u
2 1 0 4 The integral given in Example 8 is an
abbreviation for N 1
3 91
2 1 4, u 3 y=œ„„„„„
2x+1 2 2 y when x y 3 1 1 and Observe that when using (6) we do not return to the variable x after integrating. We
simply evaluate the expression in u between the appropriate values of u. y y and 1
3 N FIGURE 2 2x du 2. To ﬁnd the new limits of integration we note that 5x 2 dx y= 0 x 1 EXAMPLE 8 Evaluate y S O L U T I O N Let u 5x. Then du 3 2 1 3 9 u dx
.
5x 2
5 d x, so d x du 5. When x 1, u 2 and 7.1 I NTEGRATION BY PARTS
Every differentiation rule has a corresponding integration rule. For instance, the Substitution 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
dx f xt x txf x In the notation for indeﬁnite integrals this equation becomes y t x f x dx yf or f xt x
x t x dx ytxf f xtx x dx f xtx We can rearrange this equation as yf 1 x t x dx f xtx ytxf 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 t x . Then the differentials are
du f x d x and dv t x d x, so, by the Substitution Rule, the formula for integration
by parts becomes y u dv 2 EXAMPLE 1 Find uv y v du y x sin x d x. x and t x
sin x. Then f x
1
cos x. (For t we can choose any antiderivative of t .) Thus, using Formula SOLUTION USING FORMULA 1 Suppose we choose f x and t x
1, we have y x sin x d x f xtx
x cos x ytxf x dx y cos x d x x cos x y cos x d x 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. 453 462  CHAPTER 7 TECHNIQUES OF INTEGRATION STRATEGY FOR EVALUATING y sin m x cos n x dx (a) If the power of cosine is odd n 2 k 1 , save one cosine factor and use
cos 2x 1 sin 2x to express the remaining factors in terms of sine: y sin y sin x cos 2 k 1x d x Then substitute u m x cos 2x k cos x d x y sin m m x1 sin 2x k cos x d x sin x. (b) If the power of sine is odd m 2 k 1 , save one sine factor and use
sin 2x 1 cos 2x to express the remaining factors in terms of cosine: y sin y x cos n x d x sin 2x k cos n x sin x d x y 2k 1 1 cos 2x k cos n x sin x d x 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 halfangle identities
sin 2x 1
2 1 1
2 cos 2x cos 2 x 1 cos 2 x It is sometimes helpful to use the identity
1
2 sin x cos x sin 2 x We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since
d dx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even)
power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x.
Or, since d dx 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.
V EXAMPLE 5 Evaluate y tan 6x sec 4x d x. SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in terms of tangent using the identity sec 2x 1 tan 2x. We can then evaluate the integral
by substituting u tan x so that du sec 2x d x : y tan x sec x d x y tan x sec x sec x d x
6 4 6 2 y tan x
6 yu 6 1
7 tan 7x tan 2x sec 2x d x 1 u 2 du 1 u7
7 2 u9
9 y u6 u 8 du C
1
9 tan 9x C M S ECTION 7.2 TRIGONOMETRIC INTEGRALS EXAMPLE 6 Find y tan 5  463 sec 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
sec 2
1. We can then evaluate the
involving only secant using the identity tan 2
integral by substituting u sec , so du sec tan d : y tan 5 y tan sec 7 d 4 sec 6 sec tan d y sec 2 y u2 1 2 u 6 du y u 10 2u 8 u 6 du u9
9 u7
7 u 11
11
1
11 1 2 sec 6 sec 2 2
9 sec 11 tan d C
1
7 sec 9 sec 7 C M The preceding examples demonstrate strategies for evaluating integrals of the form x tan mx sec nx d x for two cases, which we summarize here.
STRATEGY FOR EVALUATING y tan m x sec nx dx (a) If the power of secant is even n 2 k, k 2 , save a factor of sec 2x and use
sec 2x 1 tan 2x to express the remaining factors in terms of tan x : y tan Then substitute u y tan x sec 2 kx d x m x sec 2x y tan m m x1 k1 sec 2x d x tan 2x k1 sec 2x d x tan x. (b) If the power of tangent is odd m 2 k 1 , save a factor of sec x tan x and
use tan 2x sec 2x 1 to express the remaining factors in terms of sec x : y tan Then substitute u x sec n x d x y tan 2x k sec n 1x sec x tan x d x y 2k 1 sec 2x 1 k sec n 1x sec x tan x d x sec x. For other cases, the guidelines are not as clearcut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to 464  CHAPTER 7 TECHNIQUES OF INTEGRATION integrate tan x by using the formula established in (5.5.5): y tan x d x ln sec x C We will also need the indeﬁnite integral of secant: y sec x d x 1 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 : y sec x d x y sec x sec x
sec x y
If we substitute u
becomes x 1 u du sec x
ln u sec 2x sec x tan x
dx
sec x tan x
sec 2x d x, so the integral tan x, then du
sec x tan x
C. Thus we have y sec x d x
EXAMPLE 7 Find tan x
dx
tan x ln sec x tan x C y tan x d x.
3 SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor in 2 terms of sec x : y tan x d x y tan x tan x d x y tan x
3 2 sec 2x 1 dx y tan x sec x d x y tan x d x
2 tan 2x
2 ln sec x In the ﬁrst integral we mentally substituted u C tan x so that du sec 2x d x. M 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 sec x d x.
3 SOLUTION Here we integrate by parts with u
du sec x
sec x tan x d x dv
v sec 2x d x
tan x SECTION 7.2 TRIGONOMETRIC INTEGRALS y sec x d x sec x tan x y sec x sec x tan x 465 y sec x tan x d x sec x tan x 3 Then  y sec x d x y sec x d x 2 sec 2x 1 dx 3 Using Formula 1 and solving for the required integral, we get y sec x d x 1
2 3 (sec x tan x ln sec x tan x ) C M 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 d x can be found by similar methods because of the identity
1 cot 2x csc 2x.
Finally, we can make use of another set of trigonometric identities:
2 To evaluate the integrals (a) x sin m x cos n x d x, (b) x sin m x sin n x d x, or
(c) x cos m x cos n x d x, use the corresponding identity: (a) sin A cos B 1
2 sin A B sin A (b) sin A sin B 1
2 cos A B cos A B (c) cos A cos B These product identities are discussed in
Appendix D. N 1
2 cos A B cos A B EXAMPLE 9 Evaluate B y sin 4 x cos 5x d x. SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use
the identity in Equation 2(a) as follows: y sin 4 x cos 5x d x y
1
2 y 1
2 7.2 1
2 (cos x x sin 9x d x sin x sin 9x d x 1
9 cos 9x C M EXERCISES 1– 49 Evaluate the integral. y sin x cos x d x 2. y sin x cos x d x 3. y 4. y 3 4
2 sin 5x cos 3x d x 6 2 0 5. y sin 2 7. y 2 2 x cos 5 cos2 d x dx 6. y 8. y 0 y sin 4 3 t dt y 1 13. y 15. y ssin 3 cos 5x d x sin3 (sx )
dx
sx
2 9.
11. 1. 3 0 sin sin 2 2 d 0 0 2 cos 10.
2 d sin 2x cos 2x d x cos 5 d y 12. y x cos x d x 14. y 16. y cos 0 cos6 d
2 0 sin 2 t cos 4 t dt cos5 sin d S ECTION 7.3 TRIGONOMETRIC SUBSTITUTION y sin m x cos n x d x 68. y sin m x sin n x d x 69. y cos m x cos n x d x 467 70. A ﬁnite Fourier series is given by the sum 67–69 Prove the formula, where m and n are positive integers.
67.  N 0 a n sin n x fx
n1 0
0 if m
if m n
n a 1 sin x a 2 sin 2 x a N sin Nx Show that the mth coefﬁcient a m is given by the formula if m
if m n
n 7.3 TRIGONOMETRIC SUBSTITUTION am 1 y f x sin m x d x In ﬁnding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 d x arises,
where a 0. If it were x x sa 2 x 2 d x, the substitution u a 2 x 2 would be effective
but, as it stands, x sa 2 x 2 d x is more difﬁcult. If we change the variable from x to by
the substitution x a sin , then the identity 1 sin 2
cos 2 allows us to get rid of the
root sign because
sa 2 x2 sa 2 sa 2 1 a 2 sin 2 sa 2 cos 2 sin 2 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 t t by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is onetoone. In this case, if we replace u by x and x by t in the Substitution
Rule (Equation 5.5.4), we obtain yf x dx yf t t t t dt This kind of substitution is called inverse substitution.
We can make the inverse substitution x a sin provided that it deﬁnes a onetoone
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 restriction on is imposed to ensure that the function that deﬁnes the substitution is onetoone.
(These are the same intervals used in Section 1.6 in deﬁning the inverse functions.)
TABLE OF TRIGONOMETRIC SUBSTITUTIONS
Expression
Substitution sa 2 x2 x a sin , sa 2 x2 x a tan , sx 2 a2 x a sec , 2 0 1 2
2 or 3
2 sin 2 cos 2 1 2 2 Identity tan 2 sec 2 sec 2 1 tan 2 484  CHAPTER 7 TECHNIQUES OF INTEGRATION 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.
1. y x n dx xn 1
n1 3. y e x dx ex 5. y sin x d x 7. y sec x d x 9. y sec x tan x d x 2 2. tan x ln x y a x dx ax
ln a y cos x d x 8. cos x 1
dx
x 6. 1 y 4. n y csc x d x sin x 2 cot x 10. sec x y csc x cot x d x 12. y csc x d x ln csc x
ln sin x 11. y sec x d x ln sec x 13. y tan x d x ln sec x 14. y cot x d x 15. y sinh x d x cosh x 16. y cosh x d x 17. yx 1
tan
a 18. y sa *19. yx * 20. y sx dx
2 a 2 dx
2 a2 1 tan x x
a 1
x
ln
2a
x a
a x 2 dx
2 cot x sinh x dx
2 csc x a2 sin 1 ln x x
a
sx 2 a2 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 fourstep 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 s x (1
y y (s x sx ) dx tan
d
sec2 y cos
sin cos2 d y sin
y sin x cos x 2 d x x) d x cos d y sin 2x y 1 1
2 y sin 2 2 sin x cos x 2 sin x cos x d x d cos 2x d x S ECTION 7.5 STRATEGY FOR INTEGRATION  485 t x in the
t x d x also occurs, apart from a constant fac 2. Look for an Obvious Substitution Try to ﬁnd some function u integrand whose differential du
tor. For instance, in the integral y x
x 2 1 dx we notice that if u x 2 1, then du 2 x d x. 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 Section 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 Exercises 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 s x 2 a 2 occurs, we use a trigonometric substitution according to
the table in Section 7.3.
n
n
(ii) If sax b occurs, we use the rationalizing substitution u sax b .
n
More generally, this sometimes works for st x .
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 tan 1x, sin 1x, 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: y1 dx
cos x y
y 1
1 1
cos x 1
1 cos x
dx
sin 2x cos x
dx
cos x y csc 2x y 1
1 cos x
sin 2x cos x
dx
cos 2x
dx (d) Relate the problem to previous problems. When you have built up some experience 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 486  CHAPTER 7 TECHNIQUES OF INTEGRATION instance, x tan 2x sec x d x is a challenging integral, but if we make use of the identity tan 2x sec 2x 1, we can write y tan x sec x d x y sec x d x y sec x d x
2 3 and if x sec 3x d x 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.
EXAMPLE 1 y tan 3x
dx
cos 3x In Step 1 we rewrite the integral: y tan 3x
dx
cos 3x y tan x sec x d x
3 3 The integral is now of the form x tan m x sec n x d x with m odd, so we can use the advice in
Section 7.2.
Alternatively, if in Step 1 we had written
tan 3x
dx
cos 3x y sin 3x y cos x
3 1
dx
cos 3x sin 3x y cos x d x
6 then we could have continued as follows with the substitution u
sin 3x y cos x d x y 1 6 y
V EXAMPLE 2 ye sx cos 2x
sin x d x
cos 6x u2 1
u 6 y du u y
4 cos x : u2 1
u6
u 6 du du M dx According to (ii) in Step 3(d), we substitute u ye sx dx s x . Then x u 2, so dx 2u du and 2 y u e u du The integrand is now a product of u and the transcendental function e u so it can be integrated by parts.
M ...
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This note was uploaded on 01/26/2012 for the course ECON 401 taught by Professor Burbidge,john during the Fall '08 term at Waterloo.
 Fall '08
 Burbidge,John

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