Chapter 8: Techniques of Integrationmath.ncku.edu.tw/~rchen/2018-2019 Teaching/2018... · Section...

Main MenuSalas, Hille, Etgen Calculus: One and Several Variables

Section 8.1 Integral Tables and Reviewa. Important Integralsb. Examplec. Integral Tables

Section 8.2 Integration by Partsa. Formulas for Integration by Partsb. Examplec. The Definite Integrald. Log and Inverse Trigonometric Functionse. Example

Section 8.3 Powers and Products of Trigonometric Functionsa. Basic Trigonometric Identitiesb. Powers of Sines and Cosinesc. Exampled. Powers of Tangente. Powers of Secantf. Powers of Cotangent and Cosecant

Section 8.4 Integrals Involvinga. Formulas for Integration b. Trigonometric Substitutionsc. Exampled. Formula for Quadratics under a Radical Signe. Examplef. Reduction Formula

Chapter 8: Techniques of IntegrationSection 8.5 Rational Functions; Partial Fractionsa. Partial Fractionsb. Distinct Linear Factorsc. Repeated Linear Factorsd. Irreducible Quadratic Factorse. Repeated Irreducible Quadratic Factors

Section 8.6 Some Rationalizing Substitutionsa. Rationalizing Substitutionsb. Example

Section 8.7 Numerical Integrationa. Estimating Definite Integralsb. Left-endpoint Estimatec. Right-endpoint Estimated. Midpoint Estimatee. Trapezoidal Estimatef. Parabolic Estimate (Simpson’s Rule)g. Exampleh. Error Estimatesi. Theoretical Errorj. Theoretical Error for the Trapezoidal Rulek. Theoretical Error for Simpson’s Rulel. Example

2 2 2 2 2 2, ,a x a x x a− + −

Integral Tables and Review

ExampleCalculate

SolutionSet u = ex + 2, du = exdx. At x = 0, u = 3; at x = 1, u = e + 2.Thus

e dxe +∫

ExampleWe use the table to calculate

Using a Table of Integrals A table of over 100 integrals, including those listed at the beginning of this section, appears on the inside covers of this text. This is a relatively short list. Mathematical handbooks such as Burington’s Handbook of Mathematical Tables and Formulas and CRC Standard Mathematical Tables contain extensive tables; the table in the CRC reference lists 600 integrals.

SolutionClosest to what we need is Formula 90:

We can write our integral to fit the formula by settingu = 2x, du = 2 dx.

Doing this, we have

Integration by Parts

Usually we writeu = u(x), dv = v´(x) dx

du = u´(x) dx, v = v(x).

Then the formula for integration by parts reads

Integration by PartsExampleCalculate

SolutionSetting u = x2 and dv = e−x dx, we have du = 2x dx and v = −e−x .This gives

We now calculate the integral on the right, again by parts. This time we setu = 2x and dv = e−x dx which gives du = 2 dx and v = −e−x

and thus

Combining this with our earlier calculations, we have

2 xx e dx−∫

2 2 22 2x x x x xx e dx u dv uv v du x e xe dx x e xe dx− − − − −= = − = − − − = − +∫ ∫ ∫ ∫ ∫

2 2 2 2 2 2 2x x x x x x xxe dx u dv uv v du xe e dx xe e dx xe e C− − − − − − −= = − = − − − = − + = − − +∫ ∫ ∫ ∫ ∫

( )2 2 22 2 2 2x x x x xx e dx x e xe e C x x e C− − − − −= − − − + = − + + +∫

For definite integrals

Integration by PartsThrough integration by parts, we construct an antiderivative for the logarithm, for the arc sine, and for the arc tangent.

To find the integral of the arc sine, we set

arcsin ,1 ,

u x dv dx

du dx v xx

= =−

This gives

2arcsin arcsin arcsin 1

1xx dx x x dx x x x C

x= − = + − +

−∫ ∫

Powers and Products of Trigonometric FunctionsIntegrals of trigonometric powers and products can usually be reduced to elementaryintegrals by the imaginative use of the basic trigonometric identities and, here and there, some integration by parts. These are the identities that we’ll rely on:

Powers and Products of Trigonometric FunctionsSines and Cosines

ExampleCalculate

SolutionThe relation cos2 x = 1 − sin2 x enables us to express cos4 x in terms ofsin x. The integrand then becomes (a polynomial in sin x) cos x, an expression that we can integrate by the chain rule.

2 5sin cosx x dx∫

( )( )

2 5 2 4

sin cos sin cos cos

sin 1 sin cos

sin 2sin sin cos

1 2 1sin sin sin3 5 7

x x dx x x x dx

x x x dx

x x x x dx

x x x C

= − +

= − + +

∫ ∫∫∫

Powers and Products of Trig Functions

ExampleCalculate

SolutionSince

2 2sin and cosx dx x dx∫ ∫

( )2 1 1 1 12 2 2 4

sin cos 2 sin 2x dx x dx x x C= − = − +∫ ∫

2 1 12 2

sin cos 2x x= − and 2 1 12 2

cos cos 2x x= +

( )2 1 1 1 12 2 2 4

cos cos 2 sin 2x dx x dx x x C= + = + +∫ ∫and

Powers and Products of Trigonometric FunctionsTangents and Secants

ExampleCalculate

SolutionThe relation tan2 x = sec2 x − 1 gives

tan4 x = tan2 x sec2 x − tan2 x = tan2 x sec2 x − sec2 x + 1.Therefore

4tan x dx∫

( )4 2 2 2

tan tan sec sec 1

tan tan

x dx x x x dx

x x x C

= − +

= − + +

∫ ∫

Powers and Products of Trig FunctionsExampleCalculate

SolutionThe relation sec2 x = tan2 x + 1 gives

We know the second integral, but the first integral gives us problems. Not seeing any other way to proceed, we try integration by parts on the original integral. Setting

3sec x dx∫

( )3 2 2sec sec tan 1 sec tan secx dx x x dx x x dx x dx= + = +∫ ∫ ∫ ∫

u = sec x, dv = sec² x dxdu = sec x tanx dx, v = tan x,

we have

Powers and Products of Trig Functions

Cotangents and Cosecants

The integrals in Examples 7–10 featured tangents and secants. In carrying out the integrations, we relied on the identity tan2 x + 1 = sec2 x and in one instance resorted to integration by parts. To calculate integrals that feature cotangents and cosecants, use the identity cot2 x + 1 = csc2 x and, if necessary, integration by parts.

Integrals Involving 2 2 2 2 2 2, ,a x a x x a− + −

Integrals Involving

Integrals that feature or can often be calculatedby a trigonometric substitution. Taking a > 0, we proceed as follows:

for we set x = a sin u;

for we set x = a tan u;

for we set x = a sec u.

In making such substitutions, we must make clear exactly what values of u we are using.

2 2 2 2,a x a x− + 2 2x a−

2 2a x−2 2a x+2 2x a−

2 2 2 2 2 2, ,a x a x x a− + −

ExampleTo calculate

we note that the integral can be written

This integral features . For each x between −a and a, we setx = a sin u, dx = a cosu du,

taking u between −½π and ½π. For such u, cos u > 0 and

Therefore

( )3/ 22 2

dx dxa x−

1 dxa x

− ∫

2 2a x−

2 2 cosa x a u− =

ExampleWe calculate

The domain of the integrand consists of two separated sets: all x > 1 and all x < −1. Both for x > 1 and x < −1, we set

x = sec u, dx = sec u tanu du.

For x > 1 we take u between 0 and ½π; for x < −1 we take u between π and 3∕2π. For such u, tan u > 0 and

x −∫

2 1 tanx u− =

Therefore,2

sec tan sectan1

ln sec tan

dx u u du u duux

= =−

= + − +

∫ ∫ ∫

Trigonometric substitutions can be effective in cases where the quadratic is not under a radical sign. In particular, the reduction formula

(a very useful little formula) can be obtained by setting x = a tan u, taking u between −½π and ½ π.

Rational Functions; Partial Fractions

A rational function R(x) = P(x)/Q(x) is said to be proper if the degree of thenumerator is less than the degree of the denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, then the rational function is called improper. We will focus our attention on proper rational functions because any improper rational function can be written as a sum of a polynomial and a proper rational function:

As is shown in algebra, every proper rational function can be written as the sum of partial fractions, fractions of the form

( )( ) ( ) ( )

( )P x r x

p xQ x Q x

The denominator splits into distinct linear factors.

The denominator has a repeated linear factor.

The denominator has an irreducible quadratic factor.

The denominator has a repeated irreducible quadratic factor.

Some Rationalizing Substitutions

There are integrands which are not rational functions but can be transformed into rational functions by a suitable substitution. Such substitutions are known as rationalizing substitutions.

ExampleFind

SolutionTo rationalize the integrand, we set

u2 = x, 2u du = dx,taking u ≥ 0. Then andu x=

Some Rationalizing SubstitutionsExample

SolutionTo rationalize the integrand, we set

Then 0 ≤ u < 1. To express dx in terms of u and du, we solve the equation for x:

1 xe dx−∫

1 xu e= −

Numerical Integration

To evaluate a definite integral of a continuous function by the formula

we must be able to find an antiderivative F and we must to able to evaluate this antiderivative both at a and at b. When this is not feasible, the method fails.

In the following slides we will illustrate some simple numerical methods for estimating definite integrals—methods that you can use whether or not you can find an antiderivative. All the methods described involve only simple arithmetic and are ideally suited to the computer.

( ) ( ) ( )b

af x dx F b F a= −∫

The method fails even for such simple-looking integrals as

0 0sin or xx x dx e dx−∫ ∫

The region pictured in Figure 8.7.1, can be approximated in many ways:

The left-endpoint rectangle:

The approximation to just considered yields the following estimate for

( ) :b

af x dx∫

The left-endpoint estimate:

1ΩThe region pictured in Figure 8.7.1, can be approximated in many ways:

The right-endpoint rectangle:

( ) :b

af x dx∫

The right-endpoint estimate:

1ΩThe region pictured in Figure 8.7.1, can be approximated in many ways.

The midpoint rectangle:

( ) :b

af x dx∫

The midpoint estimate:

By a trapezoid:

( ) :b

af x dx∫

The trapezoidal estimate (trapezoidal rule):

Numerical IntegrationBy a parabolic region (Figure 8.7.6): take the parabola y = Ax2 + Bx + C thatpasses through the three points indicated.

( ) :b

af x dx∫

The parabolic estimate (Simpson’s rule):

ExampleFind the approximate value of by the trapezoidal rule. Take n = 6.

SolutionEach subinterval has length

The partition points are

04 x dx+∫

3 0 16 2

b an− −

0 1 2 3 4 5 61 3 52 2 2

0, , 1, , 2, , 3x x x x x x x= = = = = = =

with . Using a calculator and rounding off to three decimal places, we have

( ) 34f x x= +

Error EstimatesA numerical estimate is useful only to the extent that we can gauge its accuracy.

Theoretical error: Error inherent in the method we use

Round-off Error: Error that accumulates from rounding off the decimals that arise during the course of computation

Consider a function f continuous and increasing on [a, b]. Subdivide [a, b] into n nonoverlapping intervals, each of length (b − a)/n. Estimate

by the left-endpoint method.

What is the theoretical error?

af x dx∫

The theoretical error does not exceed

( ) ( ) b af b f an− −

can be written

n naE f x dx T= −∫

where c is some number between a and b. Usually we cannot pinpoint c any further. However, if f´´ is bounded on [a, b], say | f´´(x)| ≤ M for a ≤ x ≤ b, then

can be written

n naE f x dx S= −∫

where, as before, c is some number between a and b. Whereas (8.7.1) varies as 1/n2, this quantity varies as 1/n4. Thus, for comparable n, we can expect greater accuracy from Simpson’s rule. In addition, if we assume that f (4)(x) is bounded on [a, b], say| f (4)(x)| ≤ M for a ≤ x ≤ b, then

To estimate

by the trapezoidal rule with theoretical error less than 0.00005, we needed to subdivide the interval [1, 2] into at least fifty-eight subintervals of equal length. To achieve the same degree of accuracy with Simpson’s rule, we need only four subintervals:

for f (x) = 1/x, f (4)(x) = 24/x4.

Therefore | f (4)(x)| ≤ 24 for all x ∈ [1, 2] and

This quantity is less than 0.00005 provided only that n4 > 167. This condition is met by n = 4.

1ln 2 dxx

2 1 1242880 120

n n−

Chapter 8: Techniques of Integrationmath.ncku.edu.tw/~rchen/2018-2019 Teaching/2018... · Section...

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