chapter 7 techniques of integration. question : question : how to integrate, where the integrands...
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Chapter 7
Techniques of Integration
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QuestionQuestion:: How to integrate
, where the
integrands are the product of two kinds of functions? xdxexdxxxdxx x sin,sin,ln
7.1 Integration by parts
Every differentiation rule has a corresponding integration rule:
Differentiation Integration
the Chain Rule the Substitution Rule
the Product Rule the Rule for Integration by Parts
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The formula for integration by parts
(1) vduuvudv
Let u = f (x) and v = g(x) are both differentiable, then du= f’(x) dx and dv= g’(x) dx
(2) dxxgxfxgxfdxxgxf )()()()()()(
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Example 1. Find xdxxcos
Example 2. Evaluate xdxln
Example 3. Find
Example 4. Evaluate
dxex x2
xdxex sin
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Example 5. Calculate 1
01tan xdx
Example 6. Prove the reduction formula
where is an integer.
xdx
n
nxx
nxdx nnn 21 sin
1sincos
1sin
2n
The formula for definite integration by parts
(3) b
a
b
adxxgxfxgxfdxxgxf )()()()()()(
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•When the integrands are the product of two kinds of functions and neither of them is derivative of the other, we use the integration by parts. We can compare:
dttfxdgxgfdxxgxgf
dxxgxfxgxfdxxgxf
)()())(()())((
ruleonsubstitutithe
)()()()()()(
partsbynintegratiothe
SummarizeSummarize
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•First we recognize uu and vv, then confirm to be more easily integrated than . For example
xdxexdxxxdxx x sin,sin,ln
udv
vdu
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7.2 Trigonometric integralsQuestion 1: Question 1: How to evaluate ? xdxx nm cossin
• Save one cosine factor to
• Use to express the remaining factors in terms of sine:
•Then substitute u=sinx.
xx 22 sin1cos
xdxdx sincos
xdxx
xdxxxxdxx
km
kmkm
sin)sin1(sin
cos)(cossincossin
2
212
(a) If the power of cosine is odd ( n=2k+1 ).
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• Save one sine factor to
• Use to express the remaining factors in terms of sine:
•Then substitute u=cosx.
xdxdx cossin
xx 22 cos1sin
xxdx
xdxxxxdxxnk
nknk
coscos)cos1(
sincos)(sincossin2
212
(b) If the power of sine is odd ( m=2k+1 ).
)2cos1(2
1cos)2cos1(
2
1sin 22 xxxx
(c) If the powers of both sine and cosine are even, use the half-angle identities
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It is sometimes helpful to use the identity
xxx 2sincossin21
Example 1. Evaluate xdxx 25 cossin
Example 2. Evaluate
Example 3. Find xdx4sin
4
0
24 cossin
xdxx
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(a) If the power of secant is even ( n=2k).
• Save a factor of
• Use to express the remaining factors in terms of tanx:
•Then substitute u=tanx.
xdxdxx tansectosec 22
xx 22 tan1sec
xdxx
xdxxxxdxx
km
kmkm
tan)tan1(tan
sec)(sectansectan
12
2122
xdxx nm sectanQuestion 2: Question 2: How to evaluate ?
xdxx 46 sectanExample 4. Find
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Example 5. Find xdxx 75 sectan
•Save a factor of
•Use to express the remaining factors in terms of secx:
•Then substitute u=secx.
xxdx
xdxxxxxdxx
nk
nknk
secsec)1(sec
tansecsec)(tansectan
12
1212
1sectan 22 xx
xdxdxxxx sectansectotansec
(b) If the power of tangent is odd ( m=2k+1).
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xdx3tanExample 6. Find
1sectan 22 xx(c) If n = 0, only tanx occurs. Use and, if necessary, the formula
Cxxdx seclntan
(d) If n is odd and m is even, we express the integrand completely in term of secx.
Example 7. Find
Example 8. Find
xdxsec
xdx3sec
Power of secx may require integration by parts.
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To evaluate the integrals (a)
(b)
use the corresponding identity:
,cossin nxdxmx
nxdxmxcnxdxmx coscos)(,sinsin
)cos()cos(2
1coscos)(
)cos()cos(2
1sinsin)(
)sin()sin(2
1cossin)(
BABABAc
BABABAb
BABABAa
Example 9. Evaluate xdxx 5cos4sin
Question 3: Question 3: How to evaluate ? nxdxmx cossin
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7.3 Trigonometric substitution
•How to integrate
?, 222222 dxaxanddxxadxxa
In general we can make a substitution of the form x=g(t) by using the Substitution Rule in reverse(called inverse substitution):
Assume that g has an inverse function, that is, g is one-to-one, we obtain
))(()()())(()( 1)()( 1
xgFtFdttgtgfdxxfxgttgx
•How to find the area of a circle or an ellipse?
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Table of trigonometric substitution
Expression Substitution Identity
2
3
tan1sec2
0,sec
sectan122
,tan
cossin122
,sin
2222
2222
2222
oraxax
axxa
axxa
One kind of inverse substitution is trigonometric
substitution.
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Example 1. Evaluate
Example 2. Find
Example 3. Evaluate
Example 4. Evaluate
dxx
x2
29
dxxx 4
122
.0,22
awhereax
dx
dxxx
x223
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Example 6. Find the area enclosed by the ellipse
12
2
2
2
b
y
a
x
Example 7. Find dxx
x
233
0 232
3
)94(
Inverse Substitution Formula For Definite Integral Let x=g( t ) and g has an inverse function, we have
where
dttgtgfdxxf
b
a)())(()(
)()( 11 bgag
Example 5. Find )0(0
22 adxxaa
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7.4 Integration of rational functions
by partial fraction
Consider a rational function
where P and Q are polynomials. If
where then the degree of P is n and we write
deg(P) = n
)(
)()(
xQ
xPxf
011
1)( axaxaxaxP nn
nn
,0na
In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fraction(called partial fraction).
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If f is improper, that is, we can
divide Q into P by long division until a remainder R(x) is
obtained such that deg(R)<deg(Q). The division
statement is
(1)
where S and R are also polynomials.
),deg()deg( QP
)(
)()(
)(
)()(
xQ
xRxS
xQ
xPxf
We can integrate rational functions according to 33 steps:
Step 1Step 1. First express f as a sum of simpler fractions.
Provide that the degree of P is less than the degree
of Q. Such a rational function is called proper.
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Step 3Step 3. Finally express the proper rational function R(x)/Q(x) as a sum of partial fractions of the form
A theorem in algebra guarantees that it is always possible to do it.
ji cbxax
BAxor
bax
A
)()( 2
).04, 22 acbwherecbxax
Step 2Step 2. Second factor the denominator Q(x) 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
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Case . The denominator Q(x) is a product of distinct linear factors.
)())(()( 2211 kk bxabxabxaxQ
The partial fraction theorem states there exist constants
such that
(2)
These constants need to be determined.
kAAA ,, 21
kk
k
bxa
A
bxa
A
bxa
A
xQ
xR
22
2
11
1
)(
)(
We explain the details for the 44 cases that occur.
.232
1223
2
dxxxx
xx
Example 1. Evaluate
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Example 2. Find .0,22 awhere
ax
dx
Suppose the first linear factor is repeated r times; that is, occurs in the factorization of Q(x). Then instead of the single term
in Equation 2, we would use
(3)
)( 11 bxa rbxa )( 11
)( 111 bxaA
rr
bxa
A
bxa
A
bxa
A
)()( 112
11
2
11
1
Case . Q(x) is a product of linear factors, some of which are repeated.
.1
14223
24
dxxxx
xxx
Example 3. Find
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If Q(x) has the factor then, in addition to the partial fraction in Equation 2 and 3, the expression for R(x)/Q(x) will have a term of the form
(4)
where A and B are constants to be determined. We can integrate (4) by completing the square and using the formula
(5)
,04),( 22 acbwherecbxax
cbxax
BAx
2
Ca
x
aax
dx
)(tan
1 122
Case . Q(x) contains irreducible quadratic factors, none of which is repeated.
dxxx
xx
4
423
2
Example 3. Find
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If Q(x) has the factor then instead of the single partial fraction(4), the sum
(6)
occurs in the partial fraction decomposition of R(x)/Q(x). Each of the term in (6) can be integrated by completed the square and making a tangent substitution.
,04,)( 22 acbwherecbxax r
rrr
cbxax
BxA
cbxax
BxA
cbxax
BxA
)()( 22222
211
dxxx
xx
344
2342
2
Example 4. Evaluate
Case . Q(x) contains a repeated irreducible quadratic factors.
dxxx
xxx
22
32
)1(
231Example 5. Evaluate
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7.5 Rationalizing substitutions
By means of appropriate substitutions, some functions can be changed into rational functions. In particular, when an integrand contains an expression of the form , then the substitution u = may be effective.
n xg )(n xg )(
Example 1. Evaluate
Let
dxx
x
4
4 xu
Example 2. Find
Let
3 xx
dx
6 xu
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Therefore
2
222
222
1
1
2sin
2coscos
1
2
1
1
12
2cos
2sin2sin
t
txxx
t
t
tt
txxx
The substitution t = tan(x/2) will convert any rational function of sinx and cosx into an ordinary rational function. This is called Weierstrass substitution.
x
xt
2tanLet
2
22
12tan
2cos
2sin
1
1
2tan1
1
2sec
1
2cos
t
txxx
txxx
Then
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(1)
Since t = tan(x/2), we have , so
Thus if we make the substitution t = tan(x/2), then we have
tx 1tan2
dtt
dx 21
2
dtt
dxt
tx
t
tx 22
2
2 1
2
1
1cos
1
2sin
Example 3. Find dxxx cos4sin3
1
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In finding the derivative of a function it is obvious which differentiation formula we should apply. But when integrating a given function, it may not be obvious which techniques we should use.
7.6 Strategy For Integration
Table of integration formulas
First First it is useful to be familiar with the basic integration formulas.
Constants of integration have been omitted.
xdxx
nn
xdxx
nn ln
1.2)1(
1.1
1
Integration is more challenging than differentiation.
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22
2222
1
22
122
22
ln.20ln2
1.19
)(sin.18)(tan1
.17
sinhcosh.16coshsinh.15
sinlncot.14seclntan.13
cotcsclncsc.12tanseclnsec.11
csccotcsc.10sectansec.9
cotcsc.8tansec.7
sincos.6cossin.5
ln.4.3
axxax
dx
ax
ax
aax
dx
a
x
xa
dx
a
x
aax
dx
xxdxxxdx
xxdxxxdx
xxxdxxxxdx
xxdxxxxdxx
xxdxxxdx
xxdxxxdx
a
adxaedxe
xxxx
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4. Try again4. Try again.
((aa)) Try substitution. Try substitution. ((bb)) Try parts. Try parts.
((cc) ) Manipulate the integrand.Manipulate the integrand.
((dd)) Relate the problem to previous problems. Relate the problem to previous problems.
((ee)) Use several methods Use several methods.
SecondlySecondly if you do not immediately see how to attack a given integral, you might try the following four-step strategy.
1. Simplify the integrand if possible.1. Simplify the integrand if possible.
2. Look for an obvious substitution.2. Look for an obvious substitution.
3. Classify the integrand according to its form.3. Classify the integrand according to its form.
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Example 1.
Example 2.
Example 3.
Example 4.
Example 5.
dxx
x3
3
cos
tan
dxe x
dxxxx
x
103
123
5
xx
dx
ln
dxx
x
1
1
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7.7 Using Tables of Integrals and
Computer Algebra Systems
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7.8 Approximation Integration
In both cases we need to find approximate values of definite integrals.
?11
132
dxxordxex•How to integrate
It is difficult, or even impossible, to find an antiderivative.
•When the function is determined from a scientific experiment through instrument readings, how to integrate such discrete function?
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Using Riemann sums
The left endpoint approximation
(1)
The right endpoint approximation
(2)
(3) Midpoint rule
xxfRdxxfn
iin
b
a
)()(
1
],[)(2
1
)]()()([)(
11
21
iiiii
nnb
a
xxofmidpoitxxxand
nabxwhere
xfxfxfxMdxxf
xxfLdxxfn
iin
b
a
)()(
11
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(4) Trapezoidal rule
xiaxandn
abxwhere
xfxf
xfxfxfx
Tdxxf
i
nn
nb
a
)]()(2
)(2)(2)([2
)(
1
210
Example 1. Use (a) the Trapezoidal Rule
(b) the Midpoint Rule with n=5
to approximate the integral .)1(2
1dxx
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Notice
The error in using an approximation is defined to be the amount that needs to be added to the approximation to make it exact.
. errorionapproximatdxxfb
a )(
(5) Error bounds Suppose If
and are the errors in the Trapezoidal and Midpoint Rules, then
.)( bxaforKxf
TE
2
3
2
3
24
)(
12
)(
n
abKEand
n
abKE MT
ME
Mb
aMnb
aT TdxxfEandTdxxfE )()(
In general, we have
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Example 2. (a) Use the Midpoint Rule with n=10 to approximate the integral
(b) Give an upper bound for the error involved in this approximation.
.1
0
2
dxex
(6) Simpson’s Rule
where n is even and
)]()(4)(2
)(4)(2)(4)([3
)(
12
3210
nnn
nb
a
xfxfxf
xfxfxfxfx
Sdxxf
.)( nabx
Example 3. Use Simpson’s Rule with n=10 to approximation .)1(
2
1dxx
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(7) Error bound for Simpson’s Rule Suppose that
If is the error involved in using Simpson’s Rule, then
.)()4( bxaforKxf
SE
4
3
180
)(
n
abKES
Example 4. (a) Use Simpson’s Rule with n=10 to approximate the integral
(b) Estimate the error involved in this approximation.
.1
0
2
dxex
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7.9 Improper Integrals In defining a definite integral we deal with (1) the function f defined on a finite interval [a, b]; (2) f is a bounded function.
dxxfb
a )(
Consider the infinite region that lies under the curve , above the x-axis and from the line x=1 to infinite, can this area A be infinite?
21 xy
Question: How to integrate a definite integral when the interval is infinite or when f is unbounded ?
Type 1 Infinite IntervalsType 1 Infinite Intervals
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Notice
and
So the area of the infinite region is equal to 1 and we write
1)1
1(lim)(lim
11
11)(
11 2
ttA
txdx
xtA
tt
tt
11
lim1
1 21 2
dxx
dxx
t
t
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(1)Definition of An Improper Integral of Type 1(a) If exists for every number , then
provide this limit exists(as a finite number).(b) If exists for every number , then
provide this limit exists(as a finite number).
The improper integrals in (a) and (b) are called convergent if the limit exists and divergent if the limit does not exist.
(c) If both and are convergent, then we define
dxxfta )( at
dxxfdxxf tata
)(lim)(
dxxfbt )( bt
dxxfdxxf btt
b
)(lim)(
dxxfa )( dxxfa
)(
dxxfdxxfdxxf aa
)()()(
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Example 1. Determine whether the integral
is convergent or divergent.
dxx
1
1
Example 2. Evaluate .0 dxxexExample 3. Evaluate dx
x
21
1
Example 4. For what value of p is the integral
convergent?
dxx p
1
1
(2) is convergent if p >1 and divergent if dx
x p1
1.1p
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Type 2 Discontinuous IntegrandsType 2 Discontinuous Integrands
(3) Definition of An Improper Integral of Type 2
(a) If f is continuous on [a,b) and , then
if this limit exists(as a finite number).
(b) If f is continuous on (a,b] and , then
if this limit exists(as a finite number).
The improper integrals in (a) and (b) are called convergent if the limit exists and divergent if the limit does not exist.
dxxfdxxf tabt
ba
)(lim)(
dxxfdxxf btat
ba
)(lim)(
)(lim xfbt
)(lim xfat
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Example 5. Find .2
152 dx
x
Example 6. Determine whether converges or
diverges.
dxx2
1 sec
Example 7. Evaluate if possible. 30 1x
dx
Example 8. Find .ln10 dxx
(c) If , where a<c<b, and both
and are convergent, then we definedxxfca )( dxxfb
c )(
dxxfdxxf
dxxfdxxfdxxf
b
tct
t
act
b
c
c
a
b
a
)(lim)(lim
)()()(
)(lim xfct
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A Comparison Test For Improper Integrals
(4)Comparison Theorem Suppose that f and g are continuous functions with
(a) If is convergent, then is convergent.
(b) If is divergent, then is divergent.
.0)()( axforxgxf
dxxfa )(
dxxga )(
dxxga )(
dxxfa )(
Example 9. Show that (a) is convergent.
(b) is divergent.
dxe x 0
2
dxx
e x
1
1