1
Section 7.1 Integration by Substitution
In this section and the next, we will cover the two most basic techniques for integration: Integration by Substitution and Integration by Parts. We begin with Integration by Substitution.
Consider the composition f (g(x)) of two differentiable functions f and g. By the Chain Rule, we
know that
(f (g(x)))0 = f 0 (g(x))g 0 (x),
and this gives
Z
f 0 (g(x))g 0 (x) dx = f (g(x)) + C.
2
0
0
Example 1 Find functions f and g such that 2x cos(x ) = f (g(x))g (x), and find
Z
2x cos(x2 ) dx.
In practice, it is hard to pick out f and g. Integration by Substitution is a technique with which we
can resolve this problem.
Integration by Substitution
• Step 1: Pick a function inside a function and call it u.
• Step 2: Find
du
,
dx
and express dx in terms of du.
• Step 3: Replace dx and all x’s using du and u’s. Now you have an integral of a function of u.
• Step 4: Integrate and replace all u’s with functions of x.
• Step 5: Check your answer by differentiating it.
Example 2 Following the steps described above, redo Example 1.
Z
Example 3 Integrate
(3x − 7)4 dx.
2
Z
Example 4 Integrate
3
(3x − 7)4 dx.
2
Z
Example 5 Integrate
2 +4
dx.
Z
1
dx.
x ln x
Z
1
dx.
x − 10
Z
x sin(x2 )
dx.
cos(x2 ) − 7
Z
√
x 3 − x2 dx.
Example 6 Integrate
Example 7 Integrate
Example 8 Integrate
Example 9 Integrate
xex
3
Section 7.2 Integration by Parts
By the Product Rule, we get
(uv)0 = u0 v + uv 0
or uv 0 = (uv)0 − u0 v.
Integrating both sides, we get
Z
0
uv dx = uv −
Z
Example 1 Integrate
Z
u0 v dx.
xex dx.
Z
Example 2 Integrate
x cos x dx.
Integration by Parts
• Step 1: Break up the function into the product of u and v 0 . The following LATE chart might
be helpful:
u
ln x
. . . , x−2 , x−1 , 1, x, x2 , . . .
sin x, cos x, sec2 x
• Step 2: Find u0 by differentiating u. Find v by integrating v 0 .
• Step 3: Integrate u0 v.
Z
Z
0
• Step 4: Finally, get uv dx = uv − u0 v dx.
ex
v0
4
Z
Example 3 Integrate
(2x − 5)ex dx.
Z
Example 4 Integrate
x ln x dx.
Z
Example 5 Integrate
x2 ln x dx.
Z
Example 6 Integrate
ln x dx.
Z
Example 7 Integrate
Z
Example 8 Integrate
ex sin x dx.
x sec2 x dx.
5
Some integrals require both Integration by Substitution and Integration by Parts.
Z
Example 9 Find 4x sin(2x + 1) dx.
Z
Example 10 Find
z
dz.
e3z
Z
Example 11 Find
x ln(x + 5) dx.
We end this section with some special shortcuts in Integration by Substitution.
Theorem 1 We have the following integrals:
Z 0
f (x)
dx = ln |f (x)| + C.
a.
f (x)
Z
1
0
b. If F (x) = f (x), then f (ax + b) dx = F (ax + b) + C.
a
Proof
a.
b.
6
Example 12 Find the integral of each of the following functions:
Z
Z
2x + 4
a.
cot x dx.
d.
dx.
x2 + 4x
Z
b.
tan x dx.
Z
c.
1
dx.
x ln x
Z
x+1
dx.
x2 + 2x
Z
x+1
dx.
x2 + 1
e.
f.
Example 13 Find the integral of each of the following functions:
Z
Z
1
3
a. (5x + 19) dx.
c.
dx.
x − 10
Z
b.
Z
sin 2x dx.
d.
e2x−1 dx.
7
Section 7.4 Algebraic Identities and Trigonometric Substitution
In this section, we will study some tricks to find the integral of rational functions and functions
involving a2 + x2 or a2 − x2 .
Z 3
2
Example 1 Integrate
−
dx.
x−1 x+2
Z
Example 2 Integrate
x+8
dx.
(x − 1)(x + 2)
Cx + D
can be obtained by splitting the function into
(x − α)(x − β)
A
B
Cx + D
partial fractions: find constants A and B such that
=
+
, and then in(x − α)(x − β)
x−α x−β
tegrate.
Z
1
Example 3 Integrate
dx.
(x − 2)(x − 5)
The integral of a rational function
Z
Example 4 Integrate
x+2
dx.
x2 + x
8
Z
Example 5 Integrate
2x2 + 3x + 2
dx.
x2 + x
We close this section with trigonometric substitutions.
Trigonometric Addition Formulas
a. sin(x + y) = sin x cos y + cos x sin y
e. tan(x + y) =
tan x + tan y
1 − tan x tan y
f. tan(x − y) =
tan x − tan y
1 + tan x tan y
b. sin(x − y) = sin x cos y − cos x sin y
c. cos(x + y) = cos x cos y − sin x sin y
d. cos(x − y) = cos x cos y + sin x sin y
Remark
a. It follows that
i. sin(2x) = 2 sin x cos x.
ii. cos(2x) = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x.
b. It also follows that
1 − cos(2x)
.
2
1 + cos(2x)
ii. cos2 x =
.
2
i. sin2 x =
Example 6 Integrate
Z
a.
sin(2x) dx.
Z
b.
cos2 x dx.
9
Example 7 Integrate
Z
1
√
a.
dx.
4 − x2
Z
b.
1
dx.
x2 + 9
Remark The following substitutions are useful when the expression a2 ± x2 is present:
• a2 − x2 −→ use x = a sin θ and dx = a cos θ dθ.
• a2 + x2 −→ use x = a tan θ and dx = a sec2 θ dθ.
Example 8 Integrate
Z √
a.
9 − x2 dx.
Z
Example 9 Find
x2
Z
b.
1
dx.
+ 2x + 10
x2
1
dx.
+4
10
Section 7.7 Improper Integrals
We begin with the graph of some elementary functions.
y
y
y=
1
x
y=
√
x
x
O
x
O
y
1
y
y = ex
y = ln x
1
x
O
O
In Calculus I, we studied the limit of a function as x approaches a specific number, for example,
sin x
= 1.
x→0 x
We now study the limit of a function at infinity.
lim
x
11
Definition If f (x) approaches a fixed number L as x gets sufficiently large, then we say that
f converges to L as x goes to ∞ and L is the limit of f at infinity. We write
lim f (x) = L.
x→∞
Similarly, if f (x) approaches L when x is negative and has a sufficiently large absolute value, then
we say that f converges to L as x goes to −∞ and L is the limit of f at negative infinity. We write
lim f (x) = L.
x→−∞
When the limit does not exists, we say that the function diverges.
Example 1 Find the limit of each of the following, if it exists.
1
x→∞ x
1
x→−∞ x
d. lim
a. lim
b. lim
x→∞
1
+2
x
e. lim ex
x→−∞
1
x→∞ x − 3
c. lim
f. lim ln x
x→∞
Recall the following properties of limit.
Theorem 1 Suppose that f, g, and h are functions. Suppose also that a is a real number, ∞,
or −∞.
a. If f (x) = c for all x, then lim f (x) exists and equals c.
x→a
b. If both lim f (x) and lim g(x) exist (say lim f (x) = ` and lim g(x) = k), then
x→a
x→a
x→a
x→a
(a) lim cf (x) (c is a constant) also exists and equals c`.
x→a
(b) lim (f (x) ± g(x)) also exists and equals ` ± k.
x→a
(c) lim (f (x) · g(x)) also exists and equals ` · k.
x→a
(d) If in addition k 6= 0, then lim
f (x)
x→a g(x)
also exists and equals k` .
12
Example 2 Find the limit of each of the following, if it exists.
x2 − 1
x→∞ 2x3 − 5x2 + 2
2x2 + 3x
x→∞ x2 − 4x + 1
c. lim
8x2 + x
x→−∞ 2x2 − 5x + 3
d. lim
a. lim
x4 + 2x3 − x2
x→∞ x3 − 2x + 3
b. lim
Remark The limit of lim
x→±∞
p(x)
, where p, q are polynomials with the same degree, is the ratio of
q(x)
leading coefficients.
We have a powerful tool to compute limits:
f (x)
(a can be a real number, ∞, or −∞). If the
x→a g(x)
L’Hôpital’s Theorem Consider the limit lim
limit is of the form
∞
∞
or 00 , then
f (x)
f 0 (x)
= lim 0
.
x→a g(x)
x→a g (x)
lim
Example 3 Find the limit of each of the following, if it exists.
x
x→∞ ex
c. lim
x3
x→∞ ex
d. lim
a. lim
x→∞
b. lim
x→∞
ln x
x
ln x
1
x5
Now we are ready to study improper integrals. There are various
types of improper integrals, but we
Z
∞
focus on one type, say, the improper integral of the form
f (x) dx.
Z b
Definition Suppose f is a nonnegative function. If the limit lim
f (x) dx exists, then we say
b→∞ a
Z ∞
that
f (x) dx is convergent and write
a
a
Z
∞
Z
f (x) dx = lim
a
b→∞
b
f (x) dx.
a
If the limit does not exist, then we say that the integral is divergent.
13
Z
∞
Example 4 Compute
1
Z
Example 5 Compute
1
dx, if it is convergent.
x2
∞
xe−x dx, if it is convergent.
0
Z
Example 6 Is
2
∞
1
dx convergent?
x
14
Section 7.8 Comparison of Improper Integrals
Sometimes it is difficult to find the exact value of an improper integral, but it may be possible
to determine whether an integral converges or diverges. In this section we develop a way to check
the convergence of a given improper integral.
Z ∞
Z ∞
1
1
dx is
Example 1 Recall that
dx converges. Using this result, determine whether
2
2
x
x +1
1
1
also convergent.
Z
Example 2 Recall that
also divergent.
2
∞
1
dx diverges. Using this result, determine whether
x
Z
∞
2
1
dx is
x−1
The result obtained above can be generalized as in the following:
Comparison Test for Improper Integrals Let a be a constant. Assume f and g are nonnegative
functions with f (x) ≤ g(x) for all x ≥ a.
Z ∞
Z ∞
f (x) dx also converges.
g(x) dx converges, then
a. If
a
a
Z
b. If
∞
Z
f (x) dx diverges, then
a
∞
g(x) dx also diverges.
a
Remark The following integrals are helpful when using the comparison test for integral.
Z ∞
1
•
dx converges for p > 1 and diverges for p ≤ 1.
xp
1
Z ∞
•
ekx dx converges for k < 0 and diverges for k ≥ 0.
0
Proof
15
Z
∞
x2 − 3
dx.
x4
∞
x4
dx.
x5 − 2
Example 3 Investigate the convergence of
2
Z
Example 4 Investigate the convergence of
3
Z
∞
Example 5 Investigate the convergence of
e2x
0
Z
Example 6 Investigate the convergence of
1
∞
1
dx.
+1
2 + sin x
√
dx.
x