SECTION 6. [
-4.
4
f
••
P.eview of Formulas and Techniques
f
2u
---du
- -3
u 2 + 1·
4
3
1
-~-
u2
+1
513
du
.
= InCu 2 + 1) - 4: tan- u + c
J
Example 1.5 was tedious, but reasonably straightforward. The issue in integration is to
recognize what pieces are present in a given integral and to see how you might rewrite the
integral in a more familiar form.
(9WRITING EXERCISES
I. In example 1.2, explain how you should know to write the de­
nominator as a 2
[1 + (~)2l Would this still be a good first
step if the numerator were x instead of I? What would you do
if the denominator were );2-_ xl?
2, In both examples 1.4 and 1.5, we completed the square and
found antiderivatives involYing sin- x, tan- x and In(x' + 1).
Briefly describe how the presence of an x in the numerator or a
square root in the denominator affects which of these functions
wi]] be in the antiderivati ve.
l
J
I
'I
f
1"
t .
eH'd x
4.
x /3(l+x'l3)
l
9.
11.
sin,IX
-dx
,IX
cOsxeslllxdx
0
13.
6.
dx
8.
10.
12.
_3_ dx
16 + x=
16.
~"x6 dx
18.
J
29.
31.
33.
o
4x
24. I
dx
26.
, dx
5 + 2x + x-
I
I
/-1
1
f
28.
.
1
.,13 - 2x - x 2
30.
dx
1 +x
-1~+ x 2 dx
32.
e 1n (x'+]) dx
34.
xJx - 3dx
36.
- -eX-2x
dx
38.
J4 -x 2-+d1 x
40.
I~-l-dx
A-xl
21. I ____
x . dx
.,Ii -x 4
f
I ~dx
I '+
I
l
1
J+
I +
22.
37.
• 0
e6x
39.
--~-dx
4
X l/
x
cos(l/x)
o
x'
o
2
sec x e
7f/'2
1
--2
sin X
tanx
dx
41.
dx
42.
- -2- d x
4 4x 1
1+ e
.
_. eX- - dx
.,II ­ e 21
I J~~'X4
dx
x+l
x" +-2x + 4 dx
x(.x:-7 + 4;-, dx
x + I
dx
2x - x 2
- -1- d r
.
/3 e
llnx dx
1
1°
xCx - 3)1 dx
eX cot(e') csc(e
X
)
dx
~l
j
,0
1ft
xe-
x2
dx
-2
J + x"
I
_5_ dx
44.
45.
and
3
and
sin2x dx
5
43. I Inx dx
-x d6x ;
1 x
­
3
r
I
I
Jv3 I fi +x
dx
In exercises 41-46, you are given a pair of integrals. Evaluate
the integral that can be worked using the techniques covered so
far (the other cannot).
dx
4
/
4x+4
5 + 2x + x 2
1
3
x sec x 2 tan Xl dx
,7/4
20.
4
35.
lT
14.
17. I I
19.
4.
smx_ dx
cos 2 X
-][/4
15.
4
5 +2x +X"
27. I(X l +4)" dx
2. I 3 cos4x dx
3. / sec 2x tan 2x dx
i.
I
I
-1
sin6x dx
..Ii
25.
l
In exercises 1-40, evaluate the integral.
l.
23.
I
I
e-
x2
dx
_5_ - dx
3 + Xl
sin' x dx
Ilnx
and
- x- d
x
1+ x8
I
J
-dx
2x
I
and
and
I
4
-x
x d
8
1+x
xe- x.2 dx
5/4
46.
CHAPTER 6
J
sec x dx
47. Finel
.,
Integration Tec.:hniques
J
2
sec x dx
and
1
2
o
I(x) dx, where I(x)
=
48. Finct12 j(x) dx, where I(x) =
2. FindJ-X-.4
1+x
{x/(X2+l)
,
example
1.5
by
J
7
r/(x-+
{x:x
2
1)
ffi
if x > 1
x
rewntmg
the
1ntegral
4
J~4
+
1
x
dx for any odd positive integer n
.
J
3. Use a CAS to find xe-x'dx, J x 3 e- x' dx and J xSe-x'dx,
Verify that each an~iderivative is correct. Generalize to give
the fonn of J x" e- x - dx for any odd positive integer n.
4. In many situations, the integral as we've defined if
must be extended to the Riemann-Stieltjes integral COn.
sidered in this exercise. For functions j and g, let p
be a regular partition of [a, b] and define the sums
as
4x + 4
dx
3
dx and complet2x 2 + 4x + 10
2x 2 + 4x + 10
ing the square in the second integral.
J
dx,J~dxandJ~dx.Generali'
1+ x
1+ x
ce
to give the form of
ifx:":l
if <0
ifx::o:O
,
rex
-2
49. Rework
6·6
n
R(f, g, P)
=L
jCCj)[g(Xj) - g(Xi-l)].
The
integral
i=l
Jab/Cx)dg(x) equals the limit of the sums R(f.g,p)
as n -+ 00, if the limit exists and equals the same
number for all evaluation points Ci (a) Show that jf
g' exists, then ~: j(x)dg(x) =~: j(x)gi(x)dx. (b) If
I a:::x:::c
g (x) - {
for some constant C with a < c < b.
-- 2 c<x:::b
$- EXPLORATORY EXERCISES
1. find
J__I_
l+x 2
dx,
J~
dx, r~ dx andJ~2 dx.
l+x
"l+x
l+x
Generalize to give the fonn of
J~
+r
1
J:
dx for any positive
f(x) dg(x). (c) Find a function g(x) such that
evaluate
J~l ~ dg(x) exists.
integer n, as completely as you can.
INTEGRATION BY PARTS
At this point, you will have recognized that there aTe many integrals that cannot be evaluated
using our basic formulas or integration by substitution. For instance,
fx
HISTORICAL NOTES
cannot be evaluated with what you presently know. We improve this situation in the current
section by introducing a powerful tool called integration by parts.
We have observed that every differentiation rule gives rise to a corresponding integration
rule. So, for the product rule:
Brook Taylor (1685-1 73 I)
An English mathematician who is
credited with devising integration
by parts. Taylor made important
contributions to probability, the
theory of magnetism and the use
of vanishing lines in linear perspec­
tive. However, he is best known
for Taylor's Theorem (see section
8.7), in which he generalized
results of Newton, Halley, the
Bernoullis and others. Personal
tragedy (both his wives died
during childbirth) and poor health
limited, the mathematical output
of this brilliant mathematician.
sinx dx
d
= j'(x)g(x) + f(x)g'(x),
~[f(x)g(x)J
dx
integrating both sides of this equation gives us
J
:x [f(x)g(x)] dx
=
f j'(x)g(x) dx + f f(x)g'(x) dx.
Ignoring the constam of integration, the integral on the left-hand side is simply !(x)g(J).
Solving for the second integral on the right-hand side then yields
f
f(x)g'(x)dx
=
f(x)g(x) -
J
j'(x)g(x) dx.
This rule is called integration by parts. You're probably wondering about the significance
of this new rule. In short, it lets us replace a given integral with an easier one. We'll let
6-7