Download Chapter 7: Techniques of Integration

Chapter 7: Techniques of Integration
MATH 206-01: Calculus II
Department of Mathematics
University of Louisville
last corrected September 14, 2013
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Chapter 7: Techniques of Integration
7.1. Integration by Parts
7.1. Integration by Parts
After completing this section, students should be able to:
use integration by parts to evaluate indefinite and definite
integrals
use integration by parts multiple times
solve application problems involving integrals where
integration by parts is needed.
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Chapter 7: Techniques of Integration
7.1. Integration by Parts
Recall the Product Rule:
d
[f (x)g(x)] = f (x)g 0 (x) + f 0 (x)g(x)
dx
Integration by Parts is a method of integration derived from
the product rule.
Z
f (x)g 0 (x) + f 0 (x)g(x) dx = f (x)g(x) (ignoring C for now)
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Z
f (x)g 0 (x) dx +
Z
Z
f (x)g 0 (x) dx = f (x)g(x) −
f 0 (x)g(x) dx = f (x)g(x)
Z
f 0 (x)g(x) dx
Chapter 7: Techniques of Integration
7.1. Integration by Parts
Letting u = f (x) and v = g(x), it follows that
du = f 0 (x) dx and dv = g 0 (x) dx.
So the formula for integration by parts can also be expressed
as
Z
Z
u dv = uv − v du .
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Chapter 7: Techniques of Integration
7.1. Integration by Parts
Example 7.1.1: Evaluate the following integrals.
Z
(a)
xex dx
Z
(b)
arcsin t dt
Z
(c)
r3 sin 2r dr
Z
(d)
e3y cos y dy
Z
2
(e)
w3 e−w dw
Answers to Example 7.1.1: (a) (x − 1)ex + C (b) t arcsin t +
(c) (r3 /2 + 3r/4) cos 2r + (3r2 /4 − 3/8) sin 2r + C
2
(d) (sin y + 3 cos y)e3y /10 + C (e) −(w2 + 1)e−w /2 + C
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√
1 − t2 + C
Chapter 7: Techniques of Integration
7.1. Integration by Parts
Integration by parts can also be applied to definite integrals:
Z b
Z b
b
0
f (x)g (x) dx = [f (x)g(x)]a −
f 0 (x)g(x) dx
a
a
Z b
Z b
0
f (x)g (x) dx = f (b)g(b) − f (a)g(a) −
f 0 (x)g(x) dx
a
a
Z
Example 7.1.2: Evaluate
20
ln x dx.
4
Answer to Example 7.1.2: 20 ln 20 − 4 ln 4 − 16 ≈ 38.369
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Chapter 7: Techniques of Integration
7.1. Integration by Parts
Example 7.1.3: Find the volume of the solid obtained by rotating
the region bounded by y = sin(πx) for 0 ≤ x ≤ 21 , x = 12 , and
y = 0 about the y-axis.
Answer to Example 7.1.3: 2/π
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
7.2. Trigonometric Integrals
After completing this section, students should be able to:
integrate trignometric integrals with products of powers of
sine and cosine functions
integrate trignometric integrals with products of powers of
secant and tangent functions
integrate trignometric integrals with products of sine and
cosine functions with different angles
solve application problems involving trigonometric integrals.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
In this section, methods for integrating particular products of
trigonometric functions are considered.
In particular, we consider how to evaluate integrals of the
formsZ
sinm x cosn x dx
Z
tanm x secn x dx
for positive integers m and n.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Z
Strategy for evaluating
sinm x cosn x dx when the power
of cosine is odd (n = 2k + 1, k ≥ 0):
Save one cosine factor and express the factors in terms of sine
by using the identity cos2 x = 1 − sin2 x:
Z
Z
m
2k+1
sin x cos
x dx =
sinm x (cos2 x)k cos x dx
Z
=
sinm x (1 − sin2 x)k cos x dx
Then substitute u = sin x so that du = cos x dx to obtain
Z
um (1 − u2 )k du
This also works if m is any real number.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Z
Strategy for evaluating
sinm x cosn x dx when the power
of sine is odd (m = 2h + 1, h ≥ 0):
Save one sine factor and express the factors in terms of cosine
by using the identity sin2 x = 1 − cos2 x:
Z
Z
2h+1
n
sin
x cos x dx =
(sin2 x)h cosn x sin x dx
Z
=
(1 − cos2 x)h cosn x sin x dx
Then substitute u = cos x so that du = − sin x dx to obtain
Z
− (1 − u2 )h un du
This also works if n is any real number.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Z
Strategy for evaluating
sinm x cosn x dx when the powers
of sine and cosine are both even (m = 2h and n = 2k, h > 0,
k > 0):
Use the half-angle identities sin2 x = 21 (1 − cos 2x) and
cos2 x = 12 (1 + cos 2x) to rewrite the integral as
Z
Z
2h
2k
sin x cos x dx = (sin2 x)h (cos2 x)k dx
h k
1
1
(1 − cos 2x)
(1 + cos 2x))
dx
2
2
Z
1
(1 − cos 2x)h (1 + cos 2x)k dx
h+k
Z =
=
2
This also works either if m = 0 or if n = 0.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Example 7.2.1: Evaluate the following integrals.
Z
(a)
Z0
(b)
π/2
sin2 θ cos5 θ dθ
cos2 x sin2 x dx
Answers to Example 7.2.1: (a) 8/105 (b) x/8 − sin(4x)/32 + C
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Z
Strategy for evaluating
tanm x secn x dx when the power of
secant is even (n = 2k, k > 0):
Save a factor sec2 x and express the remaining factors in terms
of tangent by using the identity sec2 x = tan2 x + 1:
Z
Z
m
2k
tan x sec x dx =
tanm x (sec2 x)k−1 sec2 x dx
Z
=
tanm x (tan2 x + 1)k−1 sec2 x dx
Then substitute u = tan x so that du = sec2 x dx to obtain
Z
um (u2 + 1)k−1 du
This also works if m is any real number.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Z
Strategy for evaluating
tanm x secn x dx when the power of
tangent is odd (m = 2h + 1, h ≥ 0) and :
Save a factor sec x tan x and express the remaining factors in
terms of tangent by using the identity tan2 x = sec2 x − 1:
Z
Z
tan2h+1 x secn x dx =
(tan2 x)h secn−1 x sec x tan x dx
Z
=
(sec2 x − 1)h secn−1 x sec x tan x dx
Then substitute u = sec x so that du = sec x tan x dx to
obtain
Z
(u2 − 1)h un−1 du
This also works if n is any real number.
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Example 7.2.2: Evaluate the following integrals.
Z
π/4
sec4 θ
(a)
√
3
tan θ dθ
0
Z
(b)
π/3
tan3 t dt
Z0
(c)
sec x dx
Answers to Example 7.2.2: (a) 21/20 (b) 3/2 − ln 2 (c) ln |sec x + tan x| + C
16 / 43
Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Strategy for evaluating integrals with products of sine and
cosine with different angles:
R
To evaluate the integral sin(mx) cos(nx) dx, use the
identity
1
[sin(A − B) + sin(A + B)] .
2
R
To evaluate the integral sin(mx) sin(nx) dx, use the
identity
sin A cos B =
1
[cos(A − B) − cos(A + B)] .
2
R
To evaluate the integral cos(mx) cos(nx) dx, use the
identity
sin A sin B =
1
[cos(A − B) + cos(A + B)] .
2
Odd/even functions: sin(−θ) = − sin θ and cos(−θ) = cos θ
cos A cos B =
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Chapter 7: Techniques of Integration
7.2. Trigonometric Integrals
Example 7.2.3: Evaluate the following integrals.
Z
(a)
sin(4x) cos(5x) dx
Z
(b)
π/6
sin(2x) sin x dθ
0
Answers to Example 7.2.3: (a) 1/2(cos(x) − cos(9x)/9) + C (b) 1/12
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
7.3. Trigonometric Substitutions
After completing this section, students should be able to:
use trigonometric
substitutions
for √
integrating functions
√
√
2
2
2
2
involving a − x , a + x , or x2 − a2 where a is a
positive constant
solve application problems involving trigonometric
substitutions.
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
√
a2 = |a| = a.
√
For problems involving a2 − x2 :
Let x = a sin θ for − π2 ≤ θ ≤ π2 .
Then dx = a cos θ dθ and
Assume a > 0 so that
p
p
a2 − x2 =
a2 − (a sin θ)2
p
=
a2 − a2 sin2 θ
q
=
a2 (1 − sin2 θ)
√ p
=
a2 1 − sin2 θ
√ √
2
=
a2 cos
θ
π
π
= a cos θ since cos θ ≥ 0 for − ≤ θ ≤
.
2
2
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
Example 7.3.1: Evaluate the indefinite integral
Z √
9 − x2
dx
x2
√
Answer to Example 7.3.1: − 9 − x2 /x − arcsin(x/3) + C
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
√
For problems involving a2 + x2 :
Let x = a tan θ for − π2 < θ < π2 .
Then dx = a sec2 θ dθ and
p
p
a2 + x2 =
a2 + (a tan θ)2
p
=
a2 + a2 tan2 θ
q
a2 (1 + tan2 θ)
=
√ √
=
a2 sec2 θ
π
π
= a sec θ since sec θ > 0 for − < θ <
.
2
2
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
Example 7.3.2: Evaluate the definite integral
Z 4
dx
√
9 + x2
0
Answer to Example 7.3.2: ln 3
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
√
For problems involving x2 − a2 :
Let x = a sec θ for 0 ≤ θ < π2 or π ≤ θ <
Then dx = a sec θ tan θ dθ and
p
x2 − a2 =
=
=
=
=
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3π
2 .
p
(a sec θ)2 − a2
p
a2 sec2 θ − a2
p
a2 (sec2 θ − 1)
√ √
2
a2 tan
θ
π
3π
a tan θ since tan θ > 0 for 0 ≤ θ < or π ≤ θ <
2
2
Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
Example 7.3.3: Evaluate the following integrals.
(x2 − 3)3/2
dx
x
p
Z
y 2 + 4y
(b)
dy
y+2
Z
(a)
√
√
√
Answers to Example 7.3.3: (a) (x2 − 3)3/2 /3 − 3 x2 − 3 + 3 3 arcsec(x/ 3) + C
p
p
(b) y 2 + 4y − 2 arctan( y 2 + 4y/2) + C
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Chapter 7: Techniques of Integration
7.3. Trigonometric Substitutions
Example 7.3.4: Find the area√of the region bounded above by
y = 1, bounded below by y = 1 − x2 , and bounded on the sides
by x = 0 and x = 1.
Answer to Example 7.3.4: 1 − π/4
26 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
7.4. Integration of Rational Functions by Partial Fractions
After completing this section, students should be able to:
use long division to express a rational function as a sum of a
polynomial and a proper rational function
use the method of partial fractions to rewrite a proper rational
function
integrate rational functions using the method of partial
fractions and long division when needed
solve application problems involving integration by partial
fractions.
27 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
In this section, techniques are discussed which often help in
P (x)
the integration rational functions f (x) = Q(x)
(where P and Q are polynomials).
A rational function is said to be proper if the degree of P is
less than the degree of Q.
If a rational function is not proper, then long division can
always be used to express it as a proper function of the form
f (x) =
P (x)
R(x)
= S(x) +
Q(x)
Q(x)
where S and R are polynomials and the degree of R is less
than the degree of Q.
28 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
Example 7.4.1: Evaluate the integral
Z
x2
dx
x−1
Answer to Example 7.4.1: x2 /2 + x + ln(x − 1) + C
29 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
The method of partial fractions is a very useful technique
involved in the integration of proper rational functions.
The Fundamental Theorem of Algebra implies that any
polynomial can be expressed as a product of linear and
quadratic terms.
30 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
Case 1: Suppose Q(x) can be factored as a product of
distinct linear factors:
Q(x) = (a1 x + b1 )(a2 x + b2 ) · · · (ak x + bk ).
Then partial fractions guarantees that there are constants A1 ,
A2 , . . ., Ak such that
A1
A2
Ak
R(x)
=
+
+ ... +
.
Q(x)
a1 x + b1 a2 x + b2
ak x + bk
Example 7.4.2: Find the average value of the function
1
f (x) =
4 − x2
on the interval [−1, 1].
Answer to Example 7.4.2: ln 3/4
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Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
Case 2: Suppose Q(x) can be factored as a product of linear
factors, some of which are repeated.
Suppose a particular factor ax + b is repeated r times. Then
the partial fractions decomposition will include the terms
A1
A2
Ar
+
+ ... + ... +
.
2
ax + b (ax + b)
(ax + b)r
Example 7.4.3: Evaluate the integral
Z
x3 + 1
dx
x3 − x2
Answer to Example 7.4.3: x + 2 ln |x − 1| − ln |x| + 1/x + C
32 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
Case 3: Suppose Q(x) includes irreducible quadratic factors,
none of which are repeated.
Suppose a particular factor ax2 + bx + c is included. Then the
partial fractions decomposition will include the term
Ax + B
.
ax2 + bx + c
Example 7.4.4: Evaluate the integral
Z
1
dx
3
x +x
Answer to Example 7.4.4: ln |x| + ln(x2 + 1)/2 + C
33 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
Case 4: Suppose Q(x) includes irreducible quadratic factors,
some of which are repeated.
Suppose a particular factor ax2 + bx + c is repeated r times.
Then the partial fractions decomposition will include the terms
A1 x + B1
A2 x + B 2
Ar x + B r
+
+...+...+
.
ax2 + bx + c (ax2 + bx + c)2
(ax2 + bx + c)r
Example 7.4.5: Evaluate the integral
Z
2x2 − 3x + 2
dx
x4 + 2x2 + 1
Answer to Example 7.4.5: 2 arctan x + 3/{2(x2 + 1)} + C
34 / 43
Chapter 7: Techniques of Integration
7.4. Integration of Rational Functions by Partial Fractions
Sometimes transformations can be made to general integrals
that turn it into an integral of a rational function.
Example 7.4.6: Evaluate the integral
Z √
x+4
dx
x
√
using the transformation u = x + 4.
√
√
√
Answer to Example 7.4.6: 2 x + 4 + 2 ln | x + 4 − 2| − 2 ln( x + 4 + 2) + C
35 / 43
Chapter 7: Techniques of Integration
7.8. Improper Integrals
7.8. Improper Integrals
After completing this section, students should be able to:
rewrite an improper integral as a limit or limits, determine if it
is convergent, and evaluate it if it is convergent
rewrite an integral with a discontinuous integrand as a limit or
limits, determine if it is convergent, and evaluate it if it is
convergent
36 / 43
Chapter 7: Techniques of Integration
7.8. Improper Integrals
Z
So far, when considering definite integrals
b
f (x) dx, we
a
have assumed that the interval [a, b] is finite and the function
f is continuous on [a, b].
In this section, we extend the definition of the definite integral
to cover cases when these assumptions are violated.
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Chapter 7: Techniques of Integration
7.8. Improper Integrals
Infinite Integrals:
See illustrations on page 519.
Suppose f is a continuous function
on the interval [1, ∞) and
Z ∞
we want to find the define
f (x) dx.
1
Z t
Let A(t) =
f (x) dx be the area under the curve from
1
x = 1 to x = t.
Then
Z ∞
Z
f (x) dx = “A(∞)” = lim A(t) = lim
1
38 / 43
t→∞
t→∞ 1
t
f (x) dx.
Chapter 7: Techniques of Integration
7.8. Improper Integrals
There are three types of definitions for infinite integrals:
Z
∞
t
Z
f (x) dx = lim
t→∞
a
Z
f (x) dx
a
b
Z
f (x) dx = lim
t→−∞
−∞
Z
∞
Z
f (x) dx
t
c
f (x) dx =
−∞
b
Z
f (x) dx +
−∞
∞
f (x) dx
c
for any real number c
An improper integral is called convergent if the corresponding
limit exists. It is called divergent if the limit does not exist.
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Chapter 7: Techniques of Integration
7.8. Improper Integrals
Example 7.8.1: Determine whether each integral is convergent or
divergent. Evaluate the integrals that are convergent.
Z
∞
(a)
1
Z
1
√ dx
x
0
(b)
Z−∞
∞
(c)
−∞
xe2x dx
1
dx
4 + x2
Answer to Example 7.8.1: (a) Divergent (∞) (b) −1/4 (c) π/2
40 / 43
Chapter 7: Techniques of Integration
7.8. Improper Integrals
Discontinuous Integrands:
Suppose f is a continuous function on the interval (0, 1] but
discontinuous
at 0, and we want to find the define
Z
1
f (x) dx.
Z
Let A(t) =
0
1
f (x) dx be the area under the curve from
t
x = t to x = 1.
Then
Z 1
Z
+
f (x) dx = “A(0 )” = lim A(t) = lim
0
41 / 43
t→0+
t→0+
1
f (x) dx.
t
Chapter 7: Techniques of Integration
7.8. Improper Integrals
There are three types of definitions for integrals with discontinuous
integrands:
If f is continuous on (a, b] but discontinuous at a, then
b
Z
Z
b
f (x) dx = lim
f (x) dx.
t→a+
a
t
If f is continuous on [a, b) but discontinuous at b, then
Z
b
Z
f (x) dx = lim
t→b−
a
t
f (x) dx.
a
If f is discontinuous at c and c is in the interval (a, b), then
Z
b
Z
f (x) dx =
a
42 / 43
c
Z
f (x) dx +
a
b
f (x) dx.
c
Chapter 7: Techniques of Integration
7.8. Improper Integrals
Example 7.8.2: Determine whether each integral is convergent or
divergent. Evaluate the integrals that are convergent.
Z
1
ln x dx
(a)
0
Z
3
0
Z
x
dx
9 − x2
1
dx
x3
√
(b)
1
(c)
−1
Answer to Example 7.8.2: (a) −1 (b) 3 (c) Divergent (does not exist)
43 / 43
Chapter 7: Techniques of Integration