# Chapter 8: Techniques of Integration

## Full text

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Section 8.1 Integral Tables and Review

a. Important Integrals

b. Example

c. Integral Tables

Section 8.2 Integration by Parts

a. Formulas for Integration by Parts

b. Example

c. The Definite Integral

d. Log and Inverse Trigonometric Functions

e. Example

Section 8.3 Powers and Products of Trigonometric Functions

a. Basic Trigonometric Identities

b. Powers of Sines and Cosines

c. Example

d. Powers of Tangent

e. Powers of Secant

f. Powers of Cotangent and Cosecant

Section 8.4 Integrals Involving

a. Formulas for Integration

b. Trigonometric Substitutions

c. Example

e. Example

f. Reduction Formula

### Chapter 8: Techniques of Integration

Section 8.5 Rational Functions; Partial Fractions

a. Partial Fractions

b. Distinct Linear Factors

c. Repeated Linear Factors

Section 8.6 Some Rationalizing Substitutions

a. Rationalizing Substitutions

b. Example

Section 8.7 Numerical Integration

a. Estimating Definite Integrals

b. Left-endpoint Estimate

c. Right-endpoint Estimate

d. Midpoint Estimate

e. Trapezoidal Estimate

f. Parabolic Estimate (Simpson’s Rule)

g. Example

h. Error Estimates

i. Theoretical Error

j. Theoretical Error for the Trapezoidal Rule

k. Theoretical Error for Simpson’s Rule

l. Example

2 2 2 2 2 2

, ,

a x a +x x a

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### Integral Tables and Review

Example Calculate

Solution

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

Thus

1

0 2

x x

e dx e +

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### Integral Tables and Review

Example

We 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.

Solution

Closest to what we need is Formula 90:

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

Doing this, we have

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### Integration by Parts

Usually we write

u = u(x), dv = v´(x) dx du = u´(x) dx, v = v(x).

Then the formula for integration by parts reads

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### Integration by Parts

Example Calculate Solution

Setting 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 set u = 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 x

x e dx

### ∫

2 2 2

2 2

x x x x x

x e dx = u dv =uvv du = −x e − − xe dx = −x e + xe dx

### ∫ ∫ ∫ ∫ ∫

2xe dxx = u dv=uv v du = −2xex − −2e dxx = −2xex + 2e dxx = −2xex 2ex +C

### ( )

2 2 2

2 2 2 2

x x x x x

x e dx = −x exee + = −C x + x + e +C

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### Integration by Parts

For definite integrals

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### Integration by Parts

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

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### Integration by Parts

To find the integral of the arc sine, we set

2

arcsin ,

1 ,

1

u x dv dx

du dx v x

x

= =

= =

This gives

2

arcsin arcsin 2 arcsin 1

1

x dx x x x dx x x x C

x

= − = + − +

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### Powers and Products of Trigonometric Functions

Integrals of trigonometric powers and products can usually be reduced to elementary integrals 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:

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### Powers and Products of Trigonometric Functions

Sines and Cosines Example

Calculate

Solution

The relation cos2 x = 1 − sin2 x enables us to express cos4 x in terms of

sin x. The integrand then becomes (a polynomial in sin x) cos x, an expression that we can integrate by the chain rule.

2 5

### ( )

2 5 2 4

2 2 2

2 4 6

3 5 7

sin cos sin cos cos

sin 1 sin cos

sin 2sin sin cos

1 2 1

sin sin sin

3 5 7

x x dx x x x dx

x x x dx

x x x x dx

x x x C

=

= −

= − +

= − + +

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### Powers and Products of Trig Functions

Example Calculate

Solution Since

2 2

sin x dx and cos x dx

## ( )

2 1 1 1 1

2 2 2 4

sin x dx = − cos 2x dx = x − sin 2x+C

### ∫ ∫

2 1 1

2 2

sin x = − cos 2x and 2 1 1

2 2

cos x = + cos 2x

## ( )

2 1 1 1 1

2 2 2 4

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

and

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### Powers and Products of Trigonometric Functions

Tangents and Secants Example

Calculate

Solution

The relation tan2 x = sec2 x − 1 gives

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

Therefore

tan x dx4

### ( )

4 2 2 2

1 3

3

tan tan sec sec 1

tan tan

x dx x x x dx

x x x C

= − +

= − + +

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### Powers and Products of Trig Functions

Example Calculate Solution

The 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

sec x dx3

### ( )

3 2 2

sec x dx = secx tan x+1 dx = sec tanx x dx + secx dx

### ∫ ∫ ∫ ∫

u = sec x, dv = sec² x dx du = sec x tanx dx, v = tan x, we have

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### 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.

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2

2

2

2

2

2

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### Integrals Involving

Integrals that feature or can often be calculated by 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

,

ax a + x x2 a2

2 2

a x

2 2

a + x

2 2

x a

2 2 2 2 2 2

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2

2

2

2

2

### −a

2

Example To calculate

we note that the integral can be written

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

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

a2 dxx2

3 / 2 dx

3

2 2

1 dx

a x

2 2

a x

2 2

cos ax = a u

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2

2

2

2

2

2

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2

2

2

2

2

### −a

2

Example We 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 32π. For such u, tan u > 0 and

2 1

dx x

### ∫

2 1 tan

x − = u

Therefore,

2

2

sec tan tan sec 1

ln sec tan

ln 1

dx u u

du u du

x u

u u C

x x C

= =

= + +

= + − +

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2

2

2

2

2

### −a

2

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.

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### Rational Functions; Partial Fractions

A rational function R(x) = P(x)/Q(x) is said to be proper if the degree of the numerator 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

Q x = p x + Q x

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### Rational Functions; Partial Fractions

The denominator splits into distinct linear factors.

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### Rational Functions; Partial Fractions

The denominator has a repeated linear factor.

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### Rational Functions; Partial Fractions

The denominator has an irreducible quadratic factor.

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### Rational Functions; Partial Fractions

The denominator has a repeated irreducible quadratic factor.

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### 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.

Example Find

Solution

To rationalize the integrand, we set

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

1 dx + x

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### Some Rationalizing Substitutions

Example Find

Solution

To rationalize the integrand, we set

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

1e dxx

1 x u = −e

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### 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

a f x dx = F bF a

### ∫

The method fails even for such simple-looking integrals as

1 1 2

0

0

x

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### Numerical Integration

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

a f x dx

### ∫

The left-endpoint estimate:

1

1

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### Numerical Integration

1

1

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

The right-endpoint rectangle:

The approximation to just considered yields the following estimate for

:

b

a f x dx

### ∫

The right-endpoint estimate:

1

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### Numerical Integration

1

1

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

The midpoint rectangle:

The approximation to just considered yields the following estimate for

:

b

a f x dx

### ∫

The midpoint estimate:

1

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### Numerical Integration

By a trapezoid:

The approximation to just considered yields the following estimate for

:

b

a f x dx

### ∫

The trapezoidal estimate (trapezoidal rule):

1

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### Numerical Integration

By a parabolic region (Figure 8.7.6): take the parabola y = Ax2 + Bx + C that passes through the three points indicated.

The approximation to just considered yields the following estimate for

:

b

a f x dx

### ∫

The parabolic estimate (Simpson’s rule):

1

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### Numerical Integration

Example

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

Solution

Each subinterval has length The partition points are

3 3

0 4+x dx

### ∫

3 0 1 6 2 b a

n

= =

0 1 2 3 4 5 6

1 3 5

2 2 2

0, , 1, , 2, , 3

x = x = x = x = x = x = x =

Then

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

4 3

f x = + x

Thus

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### Numerical Integration

Error Estimates

A 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?

b

a f x dx

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### Numerical Integration

The theoretical error does not exceed

b a

f b f a

n

 − 

 −  

  

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can be written

T b

n a n

### E = ∫ 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

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can be written

S b

n a n

### E = ∫ 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

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### Numerical Integration

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.

2 1

ln 2 1 dx

=

x

5

4 4

2 1 1

2880 24 120

S

En

n n

≤ − =

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