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

d. Formula for Quadratics under a Radical Sign

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

d. Irreducible Quadratic Factors

e. Repeated Irreducible Quadratic 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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Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

### Integral Tables and Review

### Integral Tables and Review

**Example**
Calculate

**Solution**

*Set u = e*^{x}*+ 2, du = e*^{x}*dx. At x = 0, u = 3; at x = 1, u = e + 2.*

Thus

1

0 2

*x*
*x*

*e* *dx*
*e* +

### ∫

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Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

### 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

### 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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Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.

### Integration by Parts

**Example**
Calculate
**Solution**

*Setting u = x*^{2} *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* =*uv* − *v du* = −*x e*^{−} − − *xe dx*^{−} = −*x e*^{−} + *xe dx*^{−}

### ∫ ∫ ∫ ∫ ∫

2*xe dx*^{−}* ^{x}* =

*u dv*=

*uv*−

*v du*= −2

*xe*

^{−}

*− −2*

^{x}*e dx*

^{−}

*= −2*

^{x}*xe*

^{−}

*+ 2*

^{x}*e dx*

^{−}

*= −2*

^{x}*xe*

^{−}

*−2*

^{x}*e*

^{−}

*+*

^{x}*C*

### ∫ ∫ ∫ ∫ ∫

### ( )

2 2 2

2 2 2 2

*x* *x* *x* *x* *x*

*x e dx*^{−} = −*x e*^{−} − *xe*^{−} − *e*^{−} + = −*C* *x* + *x* + *e*^{−} +*C*

### ∫

### 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.

### 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*

= − = + − +

### ∫ ∫

−Main Menu

### 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:

### Powers and Products of Trigonometric Functions

**Sines and Cosines**
**Example**

Calculate

**Solution**

The relation cos^{2} *x *= 1 − sin^{2} *x enables us to express cos*^{4} *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

### sin *x* cos *x dx*

### ∫

### ( )

### ( )

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 2*x dx* = *x* − sin 2*x*+*C*

### ∫ ∫

2 1 1

2 2

sin *x* = − cos 2*x* and ^{2} ^{1} ^{1}

2 2

cos *x* = + cos 2*x*

## ( )

2 1 1 1 1

2 2 2 4

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

### ∫ ∫

and

### Powers and Products of Trigonometric Functions

**Tangents and Secants**
**Example**

Calculate

**Solution**

The relation tan^{2} *x = sec*^{2} *x *− 1 gives

tan^{4} *x = tan*^{2} *x sec*^{2} *x *− tan^{2} *x = tan*^{2} *x sec*^{2} *x *− sec^{2} *x + 1.*

Therefore

*tan x dx*4

### ∫

### ( )

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 sec^{2} *x = tan*^{2} *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 dx*3

### ∫

### ( )

3 2 2

sec *x dx* = sec*x* tan *x*+1 *dx* = sec tan*x* *x dx* + sec*x dx*

### ∫ ∫ ∫ ∫

*u = sec x, dv = sec² x dx*
*du = 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 tan^{2} *x + 1 = sec*^{2} *x and in one *
instance resorted to integration by parts. To calculate integrals that feature
cotangents and cosecants, use the identity cot^{2} *x + 1 = csc*^{2} *x and, if *

necessary, integration by parts.

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### Integrals Involving *a*

^{2}

### − *x*

^{2}

### , *a*

^{2}

### + *x*

^{2}

### , *x*

^{2}

### − *a*

^{2}

### 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

,

*a* − *x* *a* + *x* *x*^{2} −*a*^{2}

2 2

*a* − *x*

2 2

*a* + *x*

2 2

*x* −*a*

2 2 2 2 2 2

### , ,

*a* − *x* *a* + *x* *x* − *a*

Main Menu

### Integrals Involving ^{a}

^{a}

^{2}

^{−} ^{x}

^{x}

^{2}

^{,} ^{a}

^{a}

^{2}

^{+} ^{x}

^{x}

^{2}

^{,} ^{x}

^{x}

^{2}

^{−} ^{a}

^{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

### (

^{a}^{2}

^{−}

^{dx}

^{x}^{2}

### )

^{3 / 2}

^{dx}### ∫

3

2 2

1 *dx*

*a* *x*

−

### ∫

2 2

*a* − *x*

2 2

cos
*a* − *x* = *a* *u*

### Integrals Involving ^{a}

^{a}

^{2}

^{−} ^{x}

^{x}

^{2}

^{,} ^{a}

^{a}

^{2}

^{+} ^{x}

^{x}

^{2}

^{,} ^{x}

^{x}

^{2}

^{−} ^{a}

^{a}

^{2}

Main Menu

### Integrals Involving ^{a}

^{a}

^{2}

^{−} ^{x}

^{x}

^{2}

^{,} ^{a}

^{a}

^{2}

^{+} ^{x}

^{x}

^{2}

^{,} ^{x}

^{x}

^{2}

^{−} ^{a}

^{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 *^{3}∕^{2}*π. 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*

= =

−

= + +

= + − +

### ∫ ∫ ∫

### Integrals Involving ^{a}

^{a}

^{2}

^{−} ^{x}

^{x}

^{2}

^{,} ^{a}

^{a}

^{2}

^{+} ^{x}

^{x}

^{2}

^{,} ^{x}

^{x}

^{2}

^{−} ^{a}

^{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*

### 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.**

### 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.**

### 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

*u*^{2} *= x, 2u du = dx,*
*taking u *≥ 0. Then and*u* = *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:*

1−*e dx*^{x}

### ∫

1 ^{x}*u* = −*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*

*a* *f x dx* = *F b* − *F a*

### ∫

The method fails even for such simple-looking integrals as

1 1 2

0

*x* sin *x dx* or

0*e*

^{−}

^{x}*dx*

### ∫ ∫

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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:

Ω1

Ω1

### Numerical Integration

Ω1

Ω1

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:

Ω1

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### Numerical Integration

Ω1

Ω1

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:

Ω1

### 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):*

Ω1

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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):*

Ω1

### 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*

### ∫

### Numerical Integration

The theoretical error does not exceed

### ( ) ( )

^{b}

^{a}*f b* *f a*

*n*

−

−

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### Numerical Integration

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*

### Numerical Integration

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/n*^{2}*, this quantity varies as 1/n*^{4}*. 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/x*^{4}*.*
*Therefore | f *^{(4)}*(x)| ≤ 24 for all x ∈ [1, 2] and*

*This quantity is less than 0.00005 provided only that n*^{4} *> 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*

*E**n*

*n* *n*

≤ − =