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
Main Menu
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 = ex + 2, du = exdx. At x = 0, u = 3; at x = 1, u = e + 2.
Thus
1
0 2
x x
e dx e +
∫
Main Menu
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
Main Menu
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 = 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 =uv − v du = −x e− − − xe dx− = −x e− + xe dx−
∫ ∫ ∫ ∫ ∫
2xe dx−x = u dv=uv− v du = −2xe−x − −2e dx−x = −2xe−x + 2e dx−x = −2xe−x −2e−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
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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 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
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
=
= −
= − +
= − + +
∫ ∫
∫
∫
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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
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
= − +
= − + +
∫ ∫
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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
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.
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
Integrals Involving a
2− x
2, a
2+ x
2, x
2− a
2Integrals 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 x2 −a2
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
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
Integrals Involving a
2− x
2, a
2+ x
2, x
2− a
2Example 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 a − x = a u
Integrals Involving a
2− x
2, a
2+ x
2, x
2− a
2Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
Integrals Involving a
2− x
2, a
2+ x
2, x
2− a
2Example 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∕2π. 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
2− x
2, a
2+ x
2, x
2− a
2Trigonometric 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.
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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.
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
Rational Functions; Partial Fractions
The denominator has a repeated linear factor.
Rational Functions; Partial Fractions
The denominator has an irreducible quadratic factor.
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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
u2 = x, 2u du = dx, taking u ≥ 0. Then andu = x
1 dx + x
∫
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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 dxx
∫
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
0e
−xdx
∫ ∫
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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 3f x = + x
Thus
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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 af b f a
n
−
−
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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/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
Main Menu
Salas, Hille, Etgen Calculus: One and Several Variables Copyright 2007 © John Wiley & Sons, Inc. All rights reserved.
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( )
54 4
2 1 1
2880 24 120
S
En
n n
≤ − =