Techniques of integration --1

OVERVIEW The Fundamental Theorem connects antiderivatives and the definite integral.
Evaluating the indefinite integral
is equivalent to finding a function F such that and then adding an
arbitrary constant C:
In this chapter we study a number of important techniques for finding indefinite
integrals of more complicated functions than those seen before. The goal of this chapter
is to show how to change unfamiliar integrals into integrals we can recognize, find in a
table, or evaluate with a computer. We also extend the idea of the definite integral to
improper integrals for which the integrand may be unbounded over the interval of integration,
or the interval itself may no longer be finite.
L ƒsxd dx = Fsxd + C.
F?sxd = ƒsxd,
L ƒsxd dx
553
TECHNIQUES OF
INTEGRATION
Chapter
8
Basic Integration Formulas
To help us in the search for finding indefinite integrals, it is useful to build up a table of
integral formulas by inverting formulas for derivatives, as we have done in previous chapters.
Then we try to match any integral that confronts us against one of the standard types.
This usually involves a certain amount of algebraic manipulation as well as use of the Substitution
Rule.
Recall the Substitution Rule from Section 5.5:
where is a differentiable function whose range is an interval I and ƒ is continuous
on I. Success in integration often hinges on the ability to spot what part of the integrand
should be called u in order that one will also have du, so that a known formula can be
applied. This means that the first requirement for skill in integration is a thorough mastery of
the formulas for differentiation.
u = gsxd
L ƒsgsxddg?sxd dx = L ƒsud du
8.1
TABLE 8.1 Basic integration formulas
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
= ln ƒ sec u ƒ + C
L tan u du = -ln ƒ cos u ƒ + C
L csc u cot u du = -csc u + C
L sec u tan u du = sec u + C
L csc2 u du = -cot u + C
L sec2 u du = tan u + C
L cos u du = sin u + C
L sin u du = -cos u + C
L
du
u = ln ƒ u ƒ + C
L un du = un+1
n + 1 + C sn Z -1d
L sdu + dyd = L du + L dy
L
k du = ku + C sany number kd
L du = u + C 13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
L
du
2u2 - a2 = cosh-1 au
a b + C su 7 a 7 0d
L
du
2a2 + u2 = sinh-1 aua
b + C sa 7 0d
L
du
u2u2 - a2 = 1a
sec-1 ` ua
` + C
L
du
a2 + u2 = 1a
tan-1 aua
b + C
L
du
2a2 - u2 = sin-1 aua
b + C
L cosh u du = sinh u + C
L sinh u du = cosh u + C
L
au du = au
ln a + C sa 7 0, a Z 1d
L
eu du = eu + C
= -ln ƒ csc u ƒ + C
L cot u du = ln ƒ sin u ƒ + C
554 Chapter 8: Techniques of Integration
We often have to rewrite an integral to match it to a standard formula.
EXAMPLE 1 Making a Simplifying Substitution
Evaluate
L
2x - 9
2x2 - 9x + 1
dx.