8
Techniques of Integration
Over the next few sections we examine some techniques that are frequently successful when
seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be
apparent that the function you wish to integrate is a derivative in some straightforward
way. For example, faced with
integraldisplay
x
10
dx
we realize immediately that the derivative of
x
11
will supply an
x
10
: (
x
11
)
′
= 11
x
10
. We
don’t want the “11”, but constants are easy to alter, because differentiation “ignores” them
in certain circumstances, so
d
dx
1
11
x
11
=
1
11
11
x
10
=
x
10
.
From our knowledge of derivatives, we can immediately write down a number of an
tiderivatives. Here is a list of those most often used:
integraldisplay
x
n
dx
=
x
n
+1
n
+ 1
+
C,
if
n
negationslash
=
−
1
integraldisplay
x
−
1
dx
= ln

x

+
C
integraldisplay
e
x
dx
=
e
x
+
C
integraldisplay
sin
x dx
=
−
cos
x
+
C
149
150
Chapter 8 Techniques of Integration
integraldisplay
cos
x dx
= sin
x
+
C
integraldisplay
sec
2
x dx
= tan
x
+
C
integraldisplay
sec
x
tan
x dx
= sec
x
+
C
integraldisplay
1
1 +
x
2
dx
= arctan
x
+
C
integraldisplay
1
√
1
−
x
2
dx
= arcsin
x
+
C
Needless to say, most problems we encounter will not be so simple. Here’s a slightly more
complicated example: find
integraldisplay
2
x
cos(
x
2
)
dx.
This is not a “simple” derivative, but a little thought reveals that it must have come from
an application of the chain rule. Multiplied on the “outside” is 2
x
, which is the derivative
of the “inside” function
x
2
. Checking:
d
dx
sin(
x
2
) = cos(
x
2
)
d
dx
x
2
= 2
x
cos(
x
2
)
,
so
integraldisplay
2
x
cos(
x
2
)
dx
= sin(
x
2
) +
C.
Even when the chain rule has “produced” a certain derivative, it is not always easy to
see. Consider this problem:
integraldisplay
x
3
radicalbig
1
−
x
2
dx.
There are two factors in this expression,
x
3
and
radicalbig
1
−
x
2
, but it is not apparent that the
chain rule is involved. Some clever rearrangement reveals that it is:
integraldisplay
x
3
radicalbig
1
−
x
2
dx
=
integraldisplay
(
−
2
x
)
parenleftbigg
−
1
2
parenrightbigg
(1
−
(1
−
x
2
))
radicalbig
1
−
x
2
dx.
This looks messy, but we do now have something that looks like the result of the chain
rule: the function 1
−
x
2
has been substituted into
−
(1
/
2)(1
−
x
)
√
x
, and the derivative
8.1 Substitution
151
of 1
−
x
2
,
−
2
x
, multiplied on the outside. If we can find a function
F
(
x
) whose derivative
is
−
(1
/
2)(1
−
x
)
√
x
we’ll be done, since then
d
dx
F
(1
−
x
2
) =
−
2
xF
′
(1
−
x
2
) = (
−
2
x
)
parenleftbigg
−
1
2
parenrightbigg
(1
−
(1
−
x
2
))
radicalbig
1
−
x
2
=
x
3
radicalbig
1
−
x
2
But this isn’t hard:
integraldisplay
−
1
2
(1
−
x
)
√
x dx
=
integraldisplay
−
1
2
(
x
1
/
2
−
x
3
/
2
)
dx
(8
.
1)
=
−
1
2
parenleftbigg
2
3
x
3
/
2
−
2
5
x
5
/
2
parenrightbigg
+
C
=
parenleftbigg
1
5
x
−
1
3
parenrightbigg
x
3
/
2
+
C.