Math 1B
§ 5.5 The Substitution Rule
Overview: Consider the following integral:
∫ 2x cos( x )dx
2
This does not fit any of our indefinite (antiderivative) formulas. But, if you remember the chain rule
for differentiation, if we differentiate sin( x 2 ) we get
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d
sin( x 2 ) = cos( x 2 ) • 2x
dx
This says that sin( x 2 ) is an antiderivative (integral) of 2x cos( x 2 ) . Therefore,
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∫
2x
Derivative of
inside function
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A similar problem would
be
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cos( x 2 ) dx = sin( x 2 ) + C
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Inside
function
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∫ (1+ 3x ) cos( x + x )dx = sin( x + x ) + C
2
Derivative of
inside function
3
3
Inside
function
sin( x + x 3 ) using the chain rule, we get
because, if we differentiate
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d
sin( x + x 3 ) = cos( x + x 3 )(1+ 3x 2 )
dx
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∫ (1+ 3x ) cos( x + x )dx = sin( x + x ) + C
and so,
2
3
3
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So, you see that the integrand is the product of a composite function and the derivative of the inside
function.
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The substitution method works for integrals of this form, that is, integrals that have the form
∫ f (g( x ))g"( x )dx . We use the problem-solving strategy of introducing something new into the
problem to replace a complicated integral with a simpler integral. We’re going to substitute a new
variable, u, for g( x ) and therefore, recalling differentials, du = g"( x ) dx . Then, if we do things
right, the problem will become simpler and solvable.
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The Substitution Rule
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If u = g( x ) is a differentiable function whose range is an interval I and f is continuous on I, then
∫ f (g( x ))g"( x )dx = ∫ f (u)du
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Stewart – 7e
1
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Example: Evaluate the indefinite integral by making the given substitution.
a)
∫ 2x sin( x )dx ; u = x
2
b)
2
4 5
∫ x (2 + x )
3
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dx ; u = 2 + x 4
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Note: To figure out what to use for the substitution u, look for some function in the integrand whose
differential also occurs (except for a constant factor). If that doesn’t work, try choosing u to be some
complicated part of the integrand. Remember, you’re looking for the inner function in a composite
function. If your first choice doesn’t work, try something else.
Example: Evaluate the indefinite integrals.
a)
b)
∫ sec2θ tan2θdθ
∫
x
(x
2
+ 1)
2
dx
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c)
∫
(ln x )
x
3
d)
dx
∫
tan−1 x
dx
1+ x 2
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Stewart – 7e
2
e)
∫e
ex
dx
x
+1
f)
∫ sin t sec (cos t )dt
h)
∫
2
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g)
∫
sin y
dy
y
x
dx
x +1
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i)
∫ tan xdx
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Stewart – 7e
3
The Substitution Rule for Definite Integrals
If g" is continuous on [a, b] and f is continuous on the range of u = g( x ) , then
b
g( b)
a
g a
∫ f (g( x ))g"( x )dx = ∫ ( ) f (u)du
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Note: This means that you can change the limits of integration
from x-values to u-values.
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Example: Evaluate the following
definite integrals.
a)
∫
7
0
b)
4 + 3x dx
∫
π
0
8
sin 2 2x cos2xdx
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c)
∫
1
0
e x dx
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Stewart – 7e
4
Integrals of Symmetric Functions
Suppose f is continuous on ⎡⎣ −a,a ⎤⎦ .
If f is even, that is f ( −x ) = f ( x ) ,
a
then
∫
−a
If f is odd, that is f ( −x ) = − f ( x ) ,
a
a
f ( x ) dx = 2 ∫ f ( x ) dx
then
∫ f ( x ) dx = 0
−a
0
Example: Evaluate the following definite integrals.
2
a)
∫ (x
6
)
+ 1 dx
−2
Stewart – 7e
π
2
b)
∫
− π2
x 2 sin x
dx
1+ x 6
5
Table of Indefinite Integrals
∫ c f (x ) d x
∫k
=c
∫ f (x ) d x
dx = kx + C
n
∫ x dx =
∫ tan x dx
x n +1
+C
n +1
(n
≠ − 1)
∫x
−1
= ln sec x + C
1
dx =
∫x
x
∫ a dx =
∫ sin x dx
∫ cos x dx
2
= − cos x + C
x dx = tan x + C
∫ sec x tan x dx
∫x
2
= sec x + C
1
d x = t a n− 1 x + C
+1
∫ sinh x dx
= cosh x + C
∫ csc
2
dx = ln x + C
ax
+C
lna
x
x
∫ e dx = e + C
∫ sec
Stewart – 7e
∫ !# f (x ) + g (x )"$ d x = ∫ f (x ) d x + ∫ g (x ) d x
= sin x + C
x dx = − cot x + C
∫ csc x cot x dx
∫
1
1− x
2
= − csc x + C
d x = s i n− 1 x + C
∫ cosh x dx
= sinh x + C
6