Math 2015
Lesson 12
Integrals by Substitution
Today we will discuss a new technique for finding antiderivatives. In particular, we
discuss a technique similar to the chain rule for dealing with functions composed with
other functions.
The Chain Rule
Recall the chain rule for derivatives:
d
f ( g( x )) = f ′( g(x)) g′(x)
dx
It says that to find a derivative of one function inside another, we just take the derivative
of the outside and multiply by the derivative of the inside. For example:
( )
d
sin x 2 =
dx
Therefore, if we ever see the pattern f ′( g(x)) g′(x) in a function we are trying to find an
antiderivative for, we know the answer is just f(g(x)).
Example:
Find
∫ x cos( x ) dx .
2
Note that this looks very much like cos( x 2 ) 2x , which we found was the
derivative of sin( x 2 ) . We just need to move the x at the end, which is
perfectly legal, and get a 2 into the integral:
∫ x cos ( x ) dx =
2
In the last step, we multiplied by 2 on the inside of the integral, but
cancelled this out by multiplying by 1/2 on the outside of the integral.
(This is OK because the constant multiple rule says that constant multiples
can be moved through the integral sign.) The important thing here is that
all the equal signs are true; each step is equal to the step before it.
The final step lets us write
Therefore,
∫ x cos ( x ) dx = .
2
(We can of course check this by taking the derivative:
( )
d 1
⎡sin x 2 ⎤ = .
⎦
dx 2 ⎣
Note that the chain rule gave the 2x just like we said it would.)
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Math 2015
Lesson 12
Integrals by Substitution
We now develop the ideas above into an easier, automated method for dealing with
integrals. If we start with an integral like
∫ f (g( x )) g′(x) dx
which has a g(x) inside the function f, we will set u =
, and get:
∫ f ( g ( x )) g′(x)dx =
We know to solve this, we just need to find the antiderivative of f with the u plugged in.
We can make the rest of the integral reflect this by noting that since u = , then
du
=
dx
, so that du =
So we finally have:
∫ f ( g ( x )) g′(x)dx =
(Note: Our book seems to like to use w for substitution, but almost everyone else uses u.
It doesn’t really make any difference.)
Let’s suppose we are attempting to find
The inside function here is
∫ (3x
2
+ 1) x dx .
3
, so we set
. We then get
du
then
=
dx
du = . In general, du is just (du/dx)(dx). (Recall that the quantities du and dx are
referred to as differentials, and like ∆u or ∆x represent small increments of u or x.)
We next need to convert the x dx into something involving du. Since
So if we put an extra 6 into the integral (and cancel it with a 1/6 so that nothing is really
changed) we get the following:
∫ u x dx =
3
(Because (6x dx) = du.) But the last integral we can solve!
1 3
u du =
6∫
Now we put this into context of what we wanted to know:
∫ (3x
2
)
3
+1 x dx =
(If we hadn’t changed back to x, we would never have answered the original question!)
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Math 2015
Lesson 12
Therefore, to solve an integral by substitution, do the following:
(1) Choose u.
⎛ du ⎞
(2) Calculate du. ( du = ⎜ ⎟ dx .)
⎝ dx ⎠
(3) Substitute u in the integral, and arrange to have du in your integral. (This may
require multiplying by a constant and dividing by that constant outside the
integral.) Make sure all of the x’s (including dx) are replaced at this stage.
(4) Solve the new integral.
(5) Substitute back in for u to get the original variable back.
The technique works very well for linear substitutions:
Example:
Find ∫ e 3x +2 dx . Here we notice that this would just be eu, if we made the
substitution u = 3x + 2. If we do this, we get du = 3 dx, and so the
substitution becomes:
∫e
3x + 2
dx =
The result of the integral then becomes
(Don’t forget to substitute back in to get x again!)
A Second Approach to Substitution
If rearranging the integral to find the correct du doesn’t make sense to you, you could
alternatively try solving for dx (or dt, or whatever), and then substituting for it inside your
integral. You may need to do some simplifying afterwards, and when you are done you
still must have only u’s and du’s in your integral.
Example:
Find ∫ cos(5x) dx . We set u = ___, so du = _____. Therefore, dx = ____.
We make this substitution:
∫ cos(5x)dx = .
Now we can complete the integral:
∫ cos(5x)dx = .
We could have used one of our rules to solve this problem, but it is good to demonstrate a
new method with a problem where we know the already know the answer.
This approach is identical to the previous approach; you can use whichever you prefer.
In either case, you must make sure to substitute for all of the variables in the problem;
you cannot have some in terms of x and some in terms of u when you are ready to
integrate.
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Math 2015
Lesson 12
Choosing u
Perhaps the hardest part of substitution is choosing what u should be. Here are two
guidelines for choosing your u:
(1) Choose u to be a function inside another function. (Remember that
substitution is the chain rule backwards.)
(2) Choose u so that du is already in the problem, except possibly for a constant
multiple (which we can always fix).
For example, in ∫ ( 3x 2 + 1) x dx , u = 3x 2 + 1 is a good choice both because it is inside of
the cube, and because du = 6x dx, and there is already an x dx in the problem.
3
In ∫ e 3x +2 dx , u = 3x + 2 is a good choice because it is inside the exponential function.
Also, du = 3 dx, which is only three times the dx we already have. (This is why linear
substitutions are easy; the derivative is only a constant, which we can always insert as
needed.)
Some Practice
Let’s try a few practice problems to get in the swing:
Example:
x
dx . (We determined last time that we could not find this
+1
integral by algebra.)
In this case, a good choice for u is
Find
∫x
2
Therefore, du =
So the integral becomes
x
dx =
+1
if we work using the first approach. If we use the second approach, we
see that dx = ______, and so we get
∫x
2
x
dx =
+1
Note that whichever way we approach the problem, we get the same end
result. Therefore,
Example:
Find
∫x
2
∫x
2
x
dx =
+1
∫ sin(7x + 2)dx .
So du =
Let u =
, and
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Math 2015
Lesson 12
Example:
Find
∫ sin(x)cos(x)dx .
Example:
Find
ex
∫ 1+ e x dx . (This one’s a little harder.)
Example:
Find
∫ x sin( x ) dx .
(Hint: There are several ways to do this.)
2
One Technique Doesn’t Fit All
Although it is a very useful technique, not every integral can be done by u-substitution.
For example, if we attempt to find ∫ sin(x 2 ) dx by u-substitution, the only reasonable
choice is u = x 2 . Then we see that u = 2x dx, or dx = u/(2x) du. Substituting, we are left
with ∫ sin(u) u /(2x) du . But then we are left with an extra x in the denominator, which
cancels with nothing!
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Math 2015
Lesson 12
Similarly, we cannot solve ∫ ( x 2 − 3) dx by making the substitution u = x 2 − 3 . If we
did, we would get du = 2x dx, and as there is no extra x in the integral, we could never get
du.
However, we can complete this integral using algebra:
3
∫ (x
2
)
− 3 dx = ∫ x 6 − 9x 4 + 27x 2 − 27 dx =
3
Morals:
(1) No one technique works for everything.
(2) Don’t forget the things we already know!
The bad news for the day is that there are lots of integrals that we will never know how to
do. (In fact, there are plenty of integrals nobody knows how to do.) When you are
dealing with a definite integral, you can then resort to numerical techniques like
Simpson’s rule. For finding antiderivatives, we can always turn to the Fundamental
Theorem version II which tells us how to build a function with a predetermined
antiderivative.
Summary
Today, we have
•
Learned how to make a u-substitution by setting u, finding du, and then
transforming to an integral we can do. When finished, we must transform back to
the original variable.
•
Discussed two approaches for completing the substitution: We can either
rearrange the integrand to find du, or solve for dx and substitute for it.
•
Found that substitution will still not allow us to solve all possible indefinite
integrals!
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