Chapter 7

Techniques of Integration

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

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 )()()()()()(

Example 1. Find xdxxcos

Example 2. Evaluate xdxln

Example 3. Find

Example 4. Evaluate

dxex x2

xdxex sin

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 )()()()()()(

•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

•First we recognize uu and vv, then confirm to be more easily integrated than . For example

xdxexdxxxdxx x sin,sin,ln

udv

vdu

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 ).

• 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

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

(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

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).

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.

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

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?

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.

Example 1. Evaluate

Example 2. Find

Example 3. Evaluate

Example 4. Evaluate

dxx

x2

29

dxxx 4

122

.0,22

awhereax

dx

dxxx

x223

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

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).

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.

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

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

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

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

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

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

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

(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

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.

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

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.

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

7.7 Using Tables of Integrals and

Computer Algebra Systems

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?

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

(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

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

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

(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

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

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

(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

)()()(

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

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

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

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