Calculus II (part 1): Techniques of Integration
(by Evan Dummit, 2012, v. 1.25)
Contents
5 Techniques of Integration
1
5.1
Basic Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
5.2
Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
5.3
Integration by Parts
3
5.4
Trigonometric Substitution
5.5
Partial Fractions
5.6
The Weierstrass Substitution
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
5.7
Improper Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
5
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Techniques of Integration
We discuss a number standard techniques for computing integrals:
substitution methods, integration by parts,
partial fractions, and improper integrals.
5.1
•
Basic Antiderivatives
Here is a list of common indenite integrals that should already be familiar:
ˆ
xn dx
=
xn+1
+ C, n 6= −1
n+1
x−1 dx
ˆ
ex dx
=
ln(x) + C
ˆ
= ex + C
ˆ
sin(x) dx
= − cos(x) + C
cos(x) dx
=
sin(x) + C
sec2 (x) dx
=
tan(x) + C
sec(x) tan(x) dx
ˆ
1
√
dx
1 − x2
ˆ
1
dx
1 + x2
ˆ
1
√
dx
x x2 − 1
=
sec(x) + C
=
sin−1 (x) + C
=
tan−1 (x) + C
=
sec−1 (x) + C
ˆ
ˆ
ˆ
1
•
Here are some other slightly more dicult antiderivatives that crop up occasionally:
ˆ
ln(x) dx
= x ln(x) − x + C
ˆ
tan(x) dx
= − ln(cos(x)) + C
sec(x) dx
=
csc(x) dx
= − ln(csc(x) + cot(x)) + C
cot(x) dx
=
ˆ
ln(sec(x) + tan(x)) + C
ˆ
ˆ
5.2
ln(sin(x)) + C
Substitution
ˆ
•
f 0 (g(x)) · g 0 (x) dx = f (g(x)) + C
The general substitution formula states that
. It is just the Chain Rule,
written in terms of integration via the Fundamental Theorem of Calculus. We generally don't use the formula
written this way. To do a substitution, follow this procedure:
◦
Step 1: Choose a substitution
◦
Step 2: Compute the dierential
◦
Step 3: Rewrite the original integral in terms of
∗
∗
∗
u = g(x).
du = g 0 (x) dx.
u:
3a: Rewrite the integral to peel o what will become the new dierential
3b: Write the remaining portion of the integrand in terms of
3c: Find the new limits of integration in terms of
limits are
x=a
and
x = b,
the new ones will be
du.
u.
u, if the integral is a denite integral. [If the old
u = g(a) and u = g(b). In a very concrete sense,
these are the same points.]
∗
3d: Write down the new integral.
If the integral is indenite, substitute back in for the original
variable.
•
Substitution is best learned by doing examples:
•
Example: Evaluate
•
´3
0
2
2x ex dx.
◦
Step 1: The exponential has a 'complicated' argument
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
Example: Evaluate
◦
du = 2x dx.
x=0
corresponds to
´9
0
´3
0
x2 ,
so we try setting
u = x2 .
2
ex · (2x dx).
u = 02
ex
2
, which is just
x=3
and
eu .
corresponds to
eu du = eu |9u=0 = e9 − 1
u = 32 .
.
´ e (ln(x))2
dx.
1
x
Step 1: It might not look like any function has a 'complicated' argument, but if we think carefully we
can see that the numerator is what we get if we plug in
ln(x)
to the squaring function. So we try setting
u = ln(x).
1
dx.
x
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
du =
´3
0
2
[ln(x)] ·
2
1
dx .
x
•
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
Example: Evaluate
◦
x=1
corresponds to
´1
0
2
[ln(x)]
u = ln(1) = 0
u2 du =
, which is just
and
x=e
1
1 31
u |u=0 =
3
3
u2 .
corresponds to
u = ln(e) = 1.
.
´1 √
x 3x2 + 1 dx.
0
Step 1: Here we see that the square root function has the 'complicated' argument
3x2 + 1
so we try
u = 3x2 + 1.
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
6·
0
3x2 + 1 ·
1
· (6x dx).
6
√
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
x=0
corresponds to
Step 3d: Putting it all together gives
Example: Evaluate
´3
2
u=1
and
3x2 + 1 ·
x=1
1
,
6
which is just
corresponds to
´ 4 1 1/2
1 2
7
u du = · u3/2 |4u=1 =
1 6
6 3
9
1 1/2
u .
6
u = 4.
.
2x
dx.
−1
x2
u = x2 − 1.
◦
Step 1: Try the denominator:
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
◦
Step 3b:
◦
Step 3c:
◦
Step 3d:
◦
Remark: It's possible to do this one without substitution, by using partial fractions to see that
du = 2x dx.
´3
1
· (2x dx).
x2 − 1
1
−1
The remaining portion of the integrand is
, which is just u
.
x2 − 1
We see that x = 2 corresponds to u = 3 and x = 3 corresponds to u = 8.
´ 8 −1
Putting it all together gives
u du = ln(u)|8u=3 = ln 8 − ln 3 .
3
This gives
2
Integration by Parts
The integration by parts formula states
ˆ
f 0 · g dx = f · g −
f · g 0 dx
. It is just the Product Rule, re-
arranged and rewritten in terms of integrals using the Fundamental Theorem of Calculus.
integration by parts, all that is required is to pick a function
f 0 (x) · g(x)
•
f (x)
and a function
g(x)
To perform an
such that the product
is equal to the original integrand. Examples will make everything clear.
Example: Evaluate
◦
•
2x
=
−1
x2
I = [ln(x + 1) + ln(x − 1)] |3x=2 = ln(4) + ln(2) − ln(3) = ln(8) − ln(3) as before.
ˆ
•
Note that we introduced a factor of
, which is okay since it's just multiplication by 1.
1
1
+
.
x+1 x−1
5.3
´1√
◦
◦
•
1
6
du = 6x dx.
´1
0
x ex dx.
x and ex should be f 0 .
1 2
0
x
Since x gets more complicated if we take its antiderivative (since we'd get
2 x ) we try ´f = e and
´1 x
1
x
0
x 1
x
g = x, to get f = e and g = 1. Plugging into the formula gives 0 x e dx = x e |x=0 − 0 1 · e dx =
(x ex − ex )|1x=0 = 1 .
The integrand
Example: Evaluate
x · ex
´e
1
is an obvious product, and so we need to decide which of
2
[ln(x)] dx.
3
◦
2
[ln(x)] = ln(x)·ln(x), but this doesn't help unless we remember the antiderivative of ln(x).
2
2
0
Instead we write the integrand as 1 · [ln(x)] , so as to take f = 1 and g = [ln(x)] . Then we get f = x
1
0
by the Chain Rule. So plugging in will yields
and g = 2 · ln(x) ·
x
ˆ e
ˆ e
1
2
x · 2 ln(x) · dx
[ln(x)] dx = x · ln(x)2 |e1 −
x
1
1
ˆ e
= e−
2 ln(x)
1
ˆ e
1
= e − 2x · ln x|e1 −
2x · dx
(IBP again)
x
1
= e − [2e − (2e − 2)]
We can write
=
•
Example: Evaluate
◦
We want
g
´1
0
e−2 .
(x2 − 2x + 2)e3x dx.
to be the thing which gets simpler when we dierentiate. The polynomial
much simpler if we dierentiate it, while the exponential
g = x2 − 2x + 2
and
f 0 = e3x ,
so that
g 0 = 2x − 2
and
x2 − 2x + 2
gets
3x
e stays basically the same. So we should take
f = 13 e3x . Then integrating by parts yields an
expression which we can't evaluate directly we have to integrate by parts again:
ˆ
ˆ 1
1 3x 1
1
= (x − 2x + 2) · e |0 −
(2x − 2) · e3x dx (IBP once)
3
3
0
ˆ 1
2 3 2
1
=
e −
−
(2x − 2) · e3x dx
3
3
3
0
ˆ 1
2 3 2
1 3x 1
1
=
e −
(IBP again)
− (2x − 2) · e |0 −
2 · e3x dx
3
3
9
9
0
2 3 2
2
2 3
2
=
e −
−
−
e −
3
3
9
27
27
1
2
(x − 2x + 2)e
3x
2
dx
0
=
•
Example: Evaluate
◦
´
20 3 26
.
e −
27
27
x3 sin(x) dx.
Here we just need to integrate by parts repeatedly. We get
ˆ
ˆ
x3 sin(x) dx
= −x3 cos(x) +
3x2 cos(x) dx (IBP once)
ˆ
= −x3 cos(x) + 3x2 sin(x) − 6x sin(x) dx
(IBP again)
ˆ
3
2
= −x cos(x) + 3x sin(x) − −6x cos(x) + 6 cos(x) dx (IBP
=
•
Example: Find
´
again)
−x3 cos(x) + 3x2 sin(x) + 6x cos(x) + 6 sin(x) + C
x3 sin(x2 ) dx.
u = x2 . The dierential is du = 2x dx,
´
1
1
and we can rearrange the integral as
x2 sin(x2 ) · · (2x dx) = u sin(u) · du.
2
2
´ 1
1
1
◦ Now we integrate by parts, to get − u cos(u) +
cos(u) du = [−u cos(u) + sin(u)] + C .
2
2
2
1 2
◦ Finally substitute back for x to get
−x cos(x2 ) + sin(x2 ) + C .
2
◦
First we look for a substitution. We try the argument of the sine:
´
4
5.4
•
Trigonometric Substitution
Some kinds of integrals require a more clever sort of substitution to evaluate, one that's sort of 'backwards'
from the usual way we try to do substitutions: instead of
(trigonometric) function
f.
u = f (x)
we try
x = f (u) for some appropriate
sin2 (x) + cos2 (x) = 1)
The idea is to use one of the Pythagorean relations (e.g.,
to simplify something more complicated. As always, examples make everything clear.
•
´√
Example: Evaluate
1 − x2 dx.
◦
Traditional substitution along the lines of
◦
Instead we try
x = sin(u);
´q
dx = cos(u) du.
´p
´
1 − sin2 (u) · cos(u) du =
cos2 (u) · cos(u) du = cos2 (u) du.
◦
So we get
◦
Remembering
◦
Finally substitute back for
cos2 (u) =
Example: Evaluate
◦
1 + cos(2u)
2
´
we can evaluate the integral to get
u = sin−1 (x)
◦
√
sin−1 (x) x 1 − x2
+
+C
2
2
. If desired, we can
.
1
dx.
(1 + x2 )2
x = tan(u). Then dx = sec2 (u) du.
´
· sec2 (u) du = cos2 (u) du.
This time we think of the arctangent antiderivative and try
We obtain
´
Finally substitute back for
u = tan−1 (x)
simplify this to the equivalent form
•
u sin(2u)
+
+ C.
2
4
sin−1 (x) sin(2 sin−1 (x))
+
+C
2
4
to obtain
´
1
1
· sec2 (u) du =
2
4
2
sec (u)
(1 + tan (u))
1
+
cos(2u)
◦ Remembering the identity cos2 (u) =
we
2
◦
doesn't work like we'd hope.
then
simplify this to the equivalent form
•
u = 1 − x2
Example: Evaluate
´
to obtain
can evaluate the integral to get
tan−1 (x) sin(2 tan−1 (u))
+
+C
2
4
tan−1 (x)
x
+
+C
2
2(1 + x2 )
u sin(2u)
+
+ C.
2
4
. If desired, we can
.
2−x
√
dx.
4 − x2
√
x = sin(u) substitution again but it doesn't quite work, since then 4 − x2 isn't nice.
q
√
to try x = 2 sin(u), since then
4 − x2 = 4 − 4 sin2 (u) = 2 cos(u), and we're much
◦
We'd like to do the
◦
Instead we have
happier.
◦
Then we have
dx = 2 cos(u) du, so we get
´ 2 − 2 sin(u)
´
·2 cos(u) du = [2 − 2 sin(u)] du = 2u+2 cos(u)+
2 cos(u)
C.
◦
5.5
•
Substituting back yields
, or equivalently,
2 sin−1 (x) + 2
p
1 − x2 + C
.
Partial Fractions
Generally, it is dicult to integrate rational functions without rearranging them in some way rst.
nding the antiderivative of
•
2 sin−1 (x) + 2 cos(sin−1 (x)) + C
2x
x3 + x2 + x + 1
(Try
directly.)
There is a general technique for evaluating such integrals, called partial fraction decomposition (PFD): the
idea is to break down rational functions into simpler parts which we know how to integrate. To nd partial
fraction decompositions, follow these steps:
5
◦
Step 1: Factor the denominator.
◦
Step 2: Find the form of the PFD:
For each term
∗
For each non-factorable term
··· +
◦
(x + a)n
∗
Cn x + Dn
.
(x2 + ax + b)n
Step 3: Solve for the coecients
∗
C2
Cn
C1
+
+ ··· +
.
(x + a)n
x + a (x + a)2
C1 x + D1
C2 x + D2
(x2 + ax + b)n , the PFD has terms 2
+
+ 2
(x
+ ax + b)2
x + ax + b
the PFD has terms
C1 , C2 , . . . , Cn .
The best way is to clear all denominators and then substitute 'intelligent' values for
x (e.g., the roots
of the linear factors).
∗
Sometimes plugging in other values of
x
is necessary, in order to nd the coecients of the higher
terms.
∗
◦
One can also employ other more clever methods, such as taking derivatives.
Step 4: Evaluate the integral.
C
can be integrated directly using the Power Rule.
(x + a)n
Cx + D
E(2x + a)
F
∗ Terms of the form 2
should be separated further as
+ 2
.
n
2
n
(x + ax + b)
(x + ax + b)
(x + ax + b)n
2
The rst term can then be integrated by substituting u = x + ax + b, and the second term can be
#
"
2
x+c
2
2
2
2
2
+1 ,
integrated by completing the square as (x + a/2) + (b − a /4) = (x + c) + d = d
d
x+c
and then substituting tan(t) =
.
d
∗
Terms of the form
•
As ever, examples make the procedure clear.
•
Example: Evaluate
◦
•
so we want to write
1
A
B
= +
.
x(x + 3)
x
x+3
1 = A(x + 3) + B(x).
x = −3 to see that 1 = 3A and 1 = −3B .
´
´ 1/3
1
1
1/3
1
dx =
−
dx =
ln(x) − ln(x + 3) + C
2
x + 3x
x
x+3
3
3
Set
Then
x=0
and
´
.
1
dx.
x3 + x2
We factor and get
∗
∗
∗
∗
1
x(x + 3)
Clear denominators so that
Example: Evaluate
◦
1
dx.
x2 + 3x
We factor and get
∗
∗
◦
´
1
x2 (x + 1)
so we want to write
Clear denominators so that
1
A
B
D
= + 2+
.
x2 (x + 1)
x
x
x+1
1 = Ax(x + 1) + B(x + 1) + Dx2 .
x = 0 to get 1 = B .
x = −1 to get 1 = D.
Set x = 1 to get 1 = 2A + 2B + D so that A = −1.
´
´ −1
1
1
1
1
◦ Then
dx
=
+
+
dx = − ln(x) − + ln(x + 1) + C
x3 + x2
x
x2
x+1
x
•
Set
Set
Example: Evaluate
◦
We want
´
2x
dx.
(x + 1)(x2 + 1)
2x
A
Bx + D
=
+ 2
.
(x + 1)(x2 + 1)
x+1
x +1
6
.
2x = A(x2 + 1) + (Bx + D)(x + 1).
Set x = −1 to get −2 = 2A so that A = −1.
2
So we want x + 2x + 1 = (Bx + D)(x + 1) so we see that we need B = D = 1.
´
´
2x
−1
x
1
1
◦ Then
dx
=
+
+
dx = − ln(x + 1) + ln(x2 + 1) + tan−1 (x) + C
(x + 1)(x2 + 1)
x + 1 x2 + 1 x2 + 1
2
∗
∗
∗
5.6
•
Clear denominators so that
The Weierstrass Substitution
There is one additional substitution that deserves discussion.
◦
It is rarely covered at any length in modern textbooks, but it should be, since it allows one to integrate
sin(x)
any rational function in
•
.
The substitution is
t = tan
x
2
and
cos(x).
, which is called the Weierstrass substitution, or, occasionally, the magic
substitution, since it allows one to evaluate complicated trigonometric integrals.
1 − t2
2t
, sin(t) =
,
1 + t2
1 + t2
2
dt.
1 + t2
1 − t2
2t
ˆ
p
,
´
p(sin(x), cos(x))
2
1 + t2 1 + t2
◦ Then as we can see, the trigonometric integral
dx becomes the integral
q(sin(x), cos(x))
1 + t2
1 − t2
2t
q
,
1 + t2 1 + t2
which is a rational function of t (though complicated).
◦
With this denition, one can check that
◦
The magic substitution can also be used to reduce any problem involving trigonometric identities in
rational multiples of a single variable
x
cos(x) =
and
dx =
to a nite computation with rational functions.
(In practice,
this is not as useful as it might seem, because the polynomial algebra is usually harder than using other
techniques like the angle-addition formulas.)
•
Example: Find
◦
We set
´
1
dx.
2 + cos(x)
t = tan
x
2
, with
ˆ
5.7
•
cos(x) =
1
dx
2 + cos(x)
◦
In this new integral we set
◦
Substituting back for
t
1 − t2
1 + t2
√
u = t/ 3
and then
x
and
dx =
2
dt,
1 + t2
to obtain
ˆ
2
1
2 · 1 + t2 dt
1−t
2+
1 + t2
ˆ
2
dt
=
2
2(1 + t ) + (1 − t2 )
ˆ
ˆ
2
2
1
√
=
dt =
·
dt
3 + t2
3 1 + (t/ 3)2
=
with
√
du = dt/ 3
√
´ 2
2
3
·
du = √ tan−1 (u) + C .
3 1 + u2
3
x
2
1
√ tan−1 √ tan
+C .
2
3
3
to obtain
yields the answer as
Improper Integration
Sometimes we like to integrate to innity, by which we mean, take the limit as a limit of integration becomes
arbitrarily large.
7
◦
•
´∞
In other words, we write
a
f (x) dx
as shorthand for
´q
q→∞ a
f (x) dx.
Other times, we like to integrate through a 'singularity' of a function that is, through a point where a
function is undened because it blows up: for example,
◦
•
lim
´1 1
dx
0 x
Again, we will write
as shorthand for
x=0
lim+
q→0
for the function
f (x) = 1/x.
´1 1
dx.
q x
Integrals with either of these two kinds of 'bad behaviors' are called improper integrals. (Perhaps the terminology stems from the fact that trying to evaluate such integrals is not something done in polite company.)
•
Typically we will be interested in asking (i) if the integral actually converges to a nite value, and (ii) if it
does, what the value is.
•
There are roughly two ways to solve problems like this:
◦
Method 1: Find the indenite integral, and then evaluate the limit. If the limit is hard, try rewriting
the function to simplify the limit.
◦
Method 2: Compare the integrand to another function whose integral is easier to evaluate. We are aided
by the following theorem:
∗
Theorem (Comparison Test for Integrals): If
a
one or both of
and
b
converges, then so does
∗
´b
a
f (x) dx.
The idea behind the theorem is to say that if
f (x)-integral
so if the
Example: Evaluate
◦
•
´∞
0
´∞
1
x2
´∞
0
q→∞
◦
Now to take the limit as
as
ln
in the interval
´b
a
´b
q→∞
.
´
x2
1
dx = ln(x) − ln(x + 1) + C .
+x
x → ∞ is not so easy, unless we notice that we can rewrite the indenite integral
.
´∞
1
x
x
1
q
ln
dx
=
lim
|
=
lim
ln
−
ln(
)
= ln(2)
x=1
q→∞
q→∞
x2 + x
x+1
x+1
2
Then we have
◦
If the problem had only asked about convergence, we could have observed that
so
[a, b] where
´b
g(x) dx
a
If
´b
f (x) dx < a g(x) dx, and
a
and conversely if the g(x)-integral
then
◦
x,
g(x) dx.
1
dx.
+x
We do partial fractions to see
x
x+1
x
e−x dx = lim (−e−x )|qx=0 = lim −e−q + 1 = 1
◦
0 < f (x) < g(x)
the g(x)-integral,
all
so does
e−x dx.
We just evaluate to see
Example: Evaluate
∞ then so must
f (x)-integral.
goes to
stays nite then so must the
•
0 < f (x) < g(x) for
´b
f (x) dx diverges,
a
can be innite, then if
1
1
1
< 2.
2
x +x
x
Then the Comparison Test would have said that, because
converges, then so does
´∞
1
1
dx.
x2 + x
.
x2 + x > x2 for positive
´∞ 1
1
=1
dx = − |∞
1 x2
x x=1
However, the Comparison Test doesn't help us in computing the
actual value.
•
Example: Determine whether the integral
◦
´∞
0
1
dx
(x + 1)(x + 2)(x + 3)(x + 4)
Doing partial fractions is too much work (although it does give the correct answer, if the trick in the
previous example is used repeatedly).
◦
converges.
But it's easier to notice that
1
1
1
1
<
<
<
.
x+4
x+3
x+2
x+1
8
◦
Then
nite
◦
•
´q
´q
1
1
1
dx = − (x + 1)−3 |qx=0 .
dx < 0
4
(x + 1)(x + 2)(x + 3)(x + 4)
(x + 1)
4
1
(in fact, it is
).
4
0
Therefore this integral
q→∞
this remains
converges .
Example: Determine whether the integral
´1 1
dx
0 x
converges.
´1 1
dx = − ln(q).
q x
◦
We integrate to get
◦
So as
◦
Therefore this integral
q → 0+
As
−∞.
we see that this diverges to
diverges to
−∞
.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2012-2015. You may not reproduce or distribute this
material without my express permission.
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