Section 5 – 5A:
Integration by Substitution
1
• x n+1 + C , n ∈ ℜ ≠ −1 This theorem requires that
n +1
the power function must be exactly ∫ x n dx . It cannot be a function like ∫ (2x − 3) n dx or
The integral of the power function is ∫ x n dx =
∫
(
)n
3x − 1
“like” ∫ x
n
dx or ∫ (sin(x) ) n dx . This section will develop ways to integrate many functions that are
dx .
We can extend the integral ∫ x n dx where we must have the exact expression x n to state that
∫u
n
du =
1
• u n+1 + C for any function where u = f (x) and du = f ′ (x) dx
n +1
Theorem
Let f (x) be a contiouos function and g(x) be differential then
If ∫ f ( g(x)) • g′(x) dx
and u = g(x) and du = g′ (x) dx
then ∫ f (g(x) ) • g′(x)dx = ∫ f (u) du
and ∫ f (u) du = F (u) + C
This process allows us to transform an integral in terms of f (x) and dx into an integral in terms of f (u)
and du. In many cases the integral in terms of f (x) could not be integrated but the new integral in
terms of f (u) can be integrated. The process of changing an integral in terms of f (x) and dx into an
integral in terms of f (u) and du is called integration by substitution.
The integral ∫ ( 3x − 2)
4
Example 1
3 dx cannot be integrated using ∫ x n dx
If we let u = 3x − 2 and take the derivative of u
du
(3x − 2) = 3 and du = 3 dx
then
dx
Substitute u in for 3x − 2 and du in for 3 dx
4
u 4
6
47
8
du
}
3
(
)
3x
−
2
•
3dx
∫
We get an integral that we can integrate in tems of u and du
1
integrate ∫ u 4 du = u 5 + C
5
substitute 3x − 2 in place of u
1
= (3x − 2 ) 5 + C
5
The key first step was recognizing that if u = 3x − 2 the first part of the integral would become u4
Section 5 – 5A
Page 1
© 2015 Eitel
The integral ∫ ( 4x − 5)
1/2
Example 2
4 dx cannot be integrated using ∫ x n dx
If we let u = 4 x − 5 and take the derivative of u
du
(4 x − 5) = 4 and du = 4 dx
dx
Substitute u in for 4x − 5 and du in for 4 dx
1/2
64u748
du
}
1/2
• 4 dx
∫ (4 x − 5)
We get an integral that we can integrate in tems of u and du
2
integrate ∫ u1/2 du = u3/2 + C
3
substitute 4x − 5 in place of u
=
2
(4x − 5 ) 3/2 + C
3
The key first step was recognizing that if u = 4x − 5 the first part of the integral would become u1/2
(
)
Example 3
3
The integral ∫ x 2 − 1 2x dx cannot be integrated using ∫ x n dx
If we let u = x 2 − 1 and take the derivative of u
du 2
x − 1 = 2x and du = 2x dx
dx
(
)
Substitute u in for x 2 − 1 and du in for 2 x dx
3
u 4
6
47
8
∫
(
x2
)
−1
du8
67
• 2x dx
3
We get an integral that we can integrate in tems of u and du
1
integrate ∫ u 3 du = u 4 + C
4
substitute x 2 − 1 in place of u
(
)
4
1 2
x −1 + C
4
The key first step was recognizing that if u = x 2 − 1 the first part of the integral would become u3
=
Section 5 – 5A
Page 2
© 2015 Eitel
U Substitution Examples
The examples below are displayed in pairs. The examples on the left are cases where one part of
the function is declared to be u and all the rest of the function is exactly du. In these cases, direct
substitution of u and du results in an integral that can be integrated in terms of u. Integrate that function
and substitute the expression for u in terms of x back in answer to have the final answer in terms of x.
The examples on the right are cases where one part of the function is declared to be your u but the
rest of the function is NOT exactly du. The remaining expression is off by a constant factor. In
these cases you must multiply the expression by a constant and divide the integral by the reciprocal
of the constant. Now direct substitution of u and du results in an integral that can be integrated in terms
of u. Integrate that function and substitute the expression for u in terms of x back in answer to have the
final answer in terms of x.
Example 1A
If u = 3x − 5 then the remaining
expression 6x dx is the
exact expression needed for
the du substitution
2
(
)
3
2
∫ 3x − 5
Example 1B
If u = 3x − 5 then the remaining
expression x dx is not the exact expression
needed for the du substitution. It is off by a
constant factor of 6
2
(
If we let u = 3x 2 − 5 then du = 6x dx
the function has the exact
expression needed for du
substitute u and du
∫
(
3x 2
)
−5
du8
67
• 6x dx
3
3
• x dx
If we let u = 3x 2 − 5 then du = 6x dx
but we only have an x dx term
3
u 4
647
8
∫
3
u 4
647
8
)
2
∫ 3x − 5
6x dx
(
3x 2
)
−5
3
•
du
to be
dx
6needs
44
47
46x
44
8
x dx
multiply x dx by 6 and
3
= ∫ u du
integrate
1
= u4 + C
4
multiply the integral by
(
3
1
2
∫ 3x − 5 • 6x dx
6
substitute 3x 2 − 5 in place of u
=
(
)
1
6
)
4
1
3x 2 − 5
+C
4
u = 3x 2 − 5 and du = 6x dx
the function now has an exact match for du
substitute u and du
1
3
∫ u du
6
integrate
1
= (u) 4 + C
6
substitute 3x 2 − 5 in place of u
4
1
= 3x 2 − 5 + C
6
(
Section 5 – 5A
Page 3
)
© 2015 Eitel
Example 2A
∫
(
12x 2
4 x3
Example 2B
∫
)
3 dx
+2
If we let u = 4 x 3 + 2 then du = 12x 2 dx
the function has the exact
expression needed for du
substitute u and du
−3
64u748
=∫
(
4x 3
)
+2
−3
−3
64u748
(
du4
6
47
8
2
• 12x dx
2
(
4x 3
+2
)
2
−3
•
du
to7
be4
124
x 24
dx
6needs
444
8
2
x dx
multiply x 2 dx by 12 and
1
multiply the integral by
12
−3
1
• 12x 2 dx
∫ 4 x3 + 2
12
(
)
u = 4x 2 + 2 and du = 6x 2 dx
the function now has an exact match for du
substitute u and du
1
−3
∫ u du
12
integrate
1 −1 −2
(u) + C
= •
12 2
)
−1
)
3
∫ 4x +2
−2
−1
4x 3 + 2
+C
2
=
)
+2
dx
3
but we only have an x 2 dx term
substitute 4x 3 + 2 in place of u
(
4 x3
If we let u = 4 x 3 + 2 then du = 12x 2 dx
= ∫ u−3 du
integrate
−1 −2
=
•u +C
2
=
(
x2
+C
substitute 4x 3 + 2 in place of u
−2
−1
=
4x 3 + 2
+C
24
(
=
Section 5 – 5A
Page 4
(
)
−1
)
24 4 x 3 + 2
2
+C
© 2015 Eitel
Example 3A
Example 3B
∫ 4 x − 2 • 4 dx
∫ 4 x − 2 • dx
If we let u = 4 x − 2 then du = 4 dx
the function has the exact
expression needed for du
substitute u and du
If we let u = 4 x − 2 then du = 4 dx
but we only have an 1dx term
1/2
64u748
du
needs
to 4
be 4
4 8
dx
64
47
1/2
•
1 dx
∫ (4 x − 2)
1/2
64u748
du
}
1/2
= ∫ (4x − 2)
• 4 dx
= ∫ u1/2 du
integrate
2
= • u 3/2 + C
3
substitute 4x − 2 in place of u
=
2
(4x − 2 ) 3/2 + C
3
multiply 1dx by 4 and
multiply the integral by
1
1/2
∫ (4 x − 2) • 4 dx
4
u = 4x − 2 and du = 4 dx
the function now has an exact match for du
substitute u and du
1 1/2
du
∫ u
4
integrate
1 2
= • (u) 3/2 + C
4 3
substitute 4x − 2 in place of u
3/2
1
= 4 x3 + 2
+C
6
(
Section 5 – 5A
1
4
Page 5
)
© 2015 Eitel
Example 4A
2x + 4
∫
∫
(
+4x
)
−1/2
∫
dx
( x 2 + 4x)1/2
x2
Example 4B
(
)
)
−1/2
• (x + 2) dx
−1/2
u744
64
4
8
∫
(
x2
+4x
)
−1/2
du4
needs
to7
be4
2x4
+44dx
6
44
8
•
x + 2 dx
multiply x + 2 dx by 2 and
1
multiply the integral by
2
−1/2
1
• (2x + 4)dx
∫ x2 + 4 x
2
du 48
647
• (2x + 4) dx
= ∫ u−1/2 du
integrate
(
)
u = x 2 + 4x and du = 2x + 4 dx
the function now has an exact match for du
substitute u and du
1
−1/2 du
∫ u
2
integrate
1
= • 2 (u)1/2 + C
2
= 2• u1/2 + C
substitute u = x 2 + 3x
(
)
dx
1/2
If we let u = x 2 + 4 x then du = 2x + 4 dx
but we only have an x + 2 dx term
−1/2
2
∫ x +4x
+ 4x
(
If we let u = x 2 + 4 x then du = 2x + 4 dx
the function has the exact
expression needed for du
substitute u and du
−1/2
x2
∫ x2 + 4 x
• 2x + 3 dx
u744
64
4
8
(
x +2
)
1/2
= 2 x 2 + 4x
+C
= 2 x2 + 4 x + C
substitute u = x 2 + 4 x
(
= x 2 + 4x
=
Section 5 – 5A
Page 6
)
1/2
+C
x2 + 4 x + C
© 2015 Eitel
Example 5A
Example 5B
4
∫ sin (4 x) • cos(4 x) dx
4
∫ sin x • cos x dx
If we let u = sin(x) then du = cos(x) dx
the function has the exact
expression needed for du
substitute u and du
If we let u = sin(4x) then du = 4 cos(4 x) dx
but we only have an cos(4 x)dx term
4
u 4
du 4
needs
to4
be
cos(4
x)4
dx
647
8
6
44
744
44
8
4
cos(4x) dx
∫ (sin(4 x)) •
4
u 4
6
47
8
du 48
647
4
∫ (sin(x)) • (cos(x)) dx
= ∫ u4 du
integrate
1
= • u5 + C
5
substitute sin(x) in place of u
=
1
(sin(x) ) 5 + C
5
multiply cos(4x) dx by 4 and
1
multiply the integral by
4
1
4
∫ (sin(4 x)) • 4(cos(4x)) dx
4
u = sin(4x) and du = 4cos(4x)dx
the function now has an exact match for du
substitute u and du
1
4
∫ u du
4
integrate
1 1
= • (u) 5 + C
4 5
1
(u) 5 + C
=
20
substitute sin(4x) in place of u
1
(sin(4 x))5 + C
=
20
=
Section 5 – 5A
Page 7
1
sin5 (4 x) + C
20
© 2015 Eitel
Further Examples of pick a u substitution for integrals
Example 6
7
2
∫ 4 tan (6x) sec (6x) dx
4∫ tan 7 (6x) sec 2 (6x) dx
If we let tan(6 x) then du = 6sec 2 (6x) dx
but we only have an sec 2 (6x)dx term
2
du4
needs
to4
be7
64
sec4
(6x)
dx
u74
647
8
6
44
44
8
4∫ (tan(6x) ) 7 •
sec 2 (6x) dx
multiply sec 2 (6x) dx by 6 and
1
multiply the integral by
6
1
4 • ∫ (tan(6x) ) 7 • 6sec 2 (6x) dx
6
u = tan(6x) and du = 6sec 2 (6x)dx
the function now has an exact match for du
substitute u and du
2
∫ u7 du
3
integrate
2 1
= • (u)8 + C
3 8
substitute tan(6 x) in place of u
1
(tan(6x) )8 + C
=
12
=
Section 5 – 5A
1
tan8 (6x) + C
12
Page 8
© 2015 Eitel
What do I do if there is an extra factor in terms of x in the integral after I substitute the u and du.
1/2
6
4u74
8
64du
74
8
∫ x • 2x − 1 • 2dx
If u = 2x − 1 then du = 2 dx
substitute u and du
1/2
∫ x • u du
there is still an x in the integeral
solve u = 2x − 1 for x
u +1
x=
and substitute for x
2
1/2
∫ x • u du
u + 1 1/2
=∫
• u du
2
1
= ∫ (u + 1)• u1/2 du
2
(
)
=
1
3/2
1/2
du
∫ u +u
2
=
1 3/2
1 1/2
∫ u du + ∫ u du
2
2
=
1 2 5/2 1 2 3/2
• •u + • •u
2 5
2 3
1 5/2 1 3/2
•u + •u
5
3
substitute 2x − 1 in place of u
1
1
= • (2x − 1) 5/2 + • (2x − 1) 3/2
5
3
=
Section 5 – 5A
Page 9
© 2015 Eitel