When u
u-Substitution
Substitution Doesn’t
Doesn t Work
Consider the integral expression
Z
p
x x ¡ 1 dx
It does not appear that u-substitution is
useful, because the extra x that is
multiplied out front is not the derivative of
the x–1 inside the square root.
1
When u
u-Substitution
Substitution Doesn’t
Doesn t Work
To approach this integral
integral, assume we can
use u-substitution for the “inside” function:
u=x–1
u
x 1. Thus
Thus, du
du=dx
Z dx.
p
x u du
Everything must be replaced with
something in terms of u. Luckily, we have
defined the relationship between x and u…
2
Formal Change of Variables
Because u=x
u=x–1
1, we can rearrange this
equation to state that u+1=x. This
substitution can then be used:
Z
((u + 1))
p
u du
We have used a formal change of
variables to rewrite in terms of u.
y, this is “fancy”
y u-substitution.
Essentially,
3
Formal Change of Variables
The problem with the original integrand
was that we could not simplify it by
distributing.
distributing
After rewriting, we can now distribute the
square root: Z
((u + 1))
Z
=
¡
u
3=2
p
+u
u du
1=2
¢
du
4
Formal Change of Variables
Now we can antidifferentiate term by term
and then substitute back for u:
Z
¡
u3=2 + u
¢
1=2
du
2 5=2 2 3=2
=
u + u +C
5
3
2
2
5=2
3=2
=
(x ¡ 1) + (x ¡ 1) + C
5
3
5
Formal Change of Variables
Essentially, when solving a complicated
Essentially
integral, one should attempt to use usubstitution first
first.
If there are extra x terms, determine what
x equals in terms of u and use a formal
change of variables.
Simplifying
Si lif i will
ill often
ft b
be required
i d after
ft thi
this
point, but will usually yield an integral that
can be
b solved
l d using
i b
basic
i rules.
l
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Example
Consider the integral expression
Z
t
dt
8t + 3
A good choice for u is 8t+3, as it is in the
denominator (the more complicated part of
the integrand).
1
Thus, du = 8 dt ) du = dt .
8
7
Example
We have information to replace the 8t+3 in
the denominator and the dt. However, we
still need to replace the t in the numerator
numerator.
Use the relationship between u and t to
solve for t in terms of u:
1
u = 8t + 3 ) t = (u ¡ 3)
8
Now we can replace everything with a
function of t.
8
Example
Thus,
Thus
Z
t
dt
8t + 3
Z
u¡3 1 1
=
¢ ¢ du
8
u 8
Pull out the constants and simplify:
¶
Z μ
1
3
=
1¡
du
64
u
9
Example
Finally, antidifferentiate and backFinally
back
substitute:
¶
Z μ
1
3
1¡
du
64
u
1
(u ¡ 3 ln juj) + C
=
64
1
(8t + 3 ¡ 3 ln j8t + 3j) + C
=
64
What happened
pp
1
3
to the 3/64 term?
=
t¡
ln j8t + 3j + C
8
64
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Other Techniques
Despite the relatively short list of basic
differentiation rules, there are numerous
strategies that must be used to
antidifferentiate complicated functions.
The techniques provided in this course
only scratch the surface, but are a good
exercise in problem solving nonetheless!
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Rewriting a Perfect Square
Consider the integral expression
Z
dx
x2 + 6x
6 +9
The numerator is not the derivative of the
denominator, so this is not du/u; likewise,
formal change of variables will not work.
However, the denominator is a perfect
square:
q
x2+6x+9=(x+3)
(
)2.
12
Rewriting a Perfect Square
In rewriting the denominator
denominator, this becomes
a simple u-substitution problem. Letting
u=x+3
u
x+3, du
du=dx
dx, we can solve:
Z
dx
x2 + 6x + 9
Z
dx
=
(x + 3)2
Z
=
u¡2 du
= ¡u¡1 + C
1
= ¡
+C
x+3
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Completing the Square
Consider the integral expression
Z
dx
x2 + 6x
6 + 10
This is similar to the previous example, but
the denominator is no longer a perfect
square.
We can rewrite it by completing the
square:
q
x2+6x+10=(x
( 2+6x+9)+1=(x+3)
)
(
)2+1.
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Completing the Square
Letting u=x+3,
u=x+3 du=dx,
du=dx we can solve:
Z
=
=
=
=
dx
x2 + 6x + 10
Z
dx
(x + 3)2 + 1
Z
du
1 + u2
arctan u + C
arctan ((x + 3)) + C
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Trigonometry
Consider the integral expression
Z
sin5 μ
dμ
7
cos μ
Had there been only one sin θ, this would
be a basic u-substitution. However, in this
case, we must be creative:
Z
5
sin μ
dμ =
7
cos μ
Z
5
1
sin μ
¢
dμ =
5
2
cos μ cos μ
Z
tan5 μ sec2 μ dμ
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Trigonometry
This problem then becomes u
u-substitution
substitution,
with u=tan θ, du=sec2θ dθ:
Z
tan5 μ sec2 μ dμ
Z
=
u5 du
u6
+C
=
6
tan6 μ
+C
=
6
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A Bit of Bad News
When creative techniques are used
(properly), seemingly unsolvable integrals
suddenly become rudimentary
rudimentary.
Unfortunately, the tradeoff is that checking
by differentiation becomes more
complicated: you have to find a way to
undo the creativity that was used to
“undo”
antidifferentiate in the first place. AHHHH!
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